*Ok, we are going to work out some examples on integration by parts.*0000

*The first example is the integral of arctan(x) dx.*0005

*Again, the difficult part of integration by parts is deciding what to make u and what to make dv. *0011

*In this case, arctan is an inverse trigonometric function.*0018

*So, we are going to make that the u arctan(x) and our dv is going to be the dx.*0024

*Then we have to fill in du and v.*0038

*du is something that hopefully you remember from your Calculus 1 class.*0043

*The derivative of arctan(x) is 1/x*^{2}+1 dx.0049

*V is just the integral of dx which is x.*0056

*Remember your main integration by parts formula, uv - the integral of vdu.*0060

*This integral converts into u × v is xarctan(x) - integral of vdu.*0068

*So that is x/x*^{2}+1 dx.0079

*Now we got this other interval that does not look at all like the first one.*0085

*Fortunately this one can be solved pretty easily.*0090

*We do not need to use integration by parts at all.*0093

*We can do this with a quick substitution.*0095

*Let us let u = x*^{2}+1, then our du will be 2x dx.0098

*The point of that is that we practically have du already.*0108

*We have x dx, so we just need to correct for the fact that x dx is actually 1/2 du.*0112

*This is x arctan(x) - the integral of 1/2 du all over u.*0120

*That is x arctan(x) - 1/2, and now the integral of 1/u is just ln(abs(u)).*0136

*Finally we get x arctan(x) - 1/2, ln we will substitute back, u was x*^{2}+1 and as always we attach a constant at the end.0147

*We do not have any limit values to plug in.*0168

*The only difficult part there was knowing how to get started.*0173

*The advice on that is to remember our order of functions.*0180

*First natural log, then inverse trigonometry.*0184

*We check these one by one and realize, oh, I have an inverse trigonometric function,*0186

*So, I am going to make the inverse trigonometric function be...*0193

*We have one more example for integration by parts.*0000

*We are going to try to solve the integral of sin(sqrt(x)) dx.*0005

*This is one that does not lend itself to integration by parts immediately.*0008

*What we will do is a little substitution.*0013

*Normally I use u for substitution but since I know we will be using integration by parts later on,*0016

*I am going to use a different variable here.*0023

*I will let w = sqrt(x).*0031

*Whenever you make a substitution you also have to figure out the derivative of the variable.*0032

*So dw is, if you think of sqrt(x) as x*^{1/2},0035

*Then dw is 1/2 x*^{-1/2} dx, which is 1/2 sqrt(x) dx.0041

*Which is 1/2w dx.*0059

*So, that tells you that dx is 2w dw.*0065

*We are going to make that substitution in here.*0072

*Now we have the integral of the sin(sqrt(x)) which is converted to w.*0075

*The dx is converted to 2w dw.*0080

*A lot of people forget to change the dx when they make a substitution.*0086

*That is a really important step, when you make the substitution.*0093

*Now I will pull the 2 outside, and pull the w to the other side and get w sin(w) dw.*0099

*Now this is kind of a standard integration by parts problem.*0112

*We have w sin(w).*0116

*In fact this was done as an example in my first lecture on integration by parts.*0119

*So, I will not redo it the same way.*0125

*Instead I will do the tabular integration method.*0126

*You will see another example of how to do the tabular integration to do something like this quickly.*0132

*So, I will set up w sin(w).*0137

*Remember you do derivatives on the left, so the derivative of w is 1, and the derivative of 1 = 0.*0143

*The integral of sin(w) is -cos(w), and the derivative of -cos(w) is -sin(w).*0151

*Then we make these diagonal lines with positive and negative signs on the lines, so plus, minus, plus.*0160

*Then, the answer here is what you get when you multiply along these lines.*0172

*So, -w cos(w).*0178

*The second diagonal line has 2 negatives cancelling each other.*0183

*You get + sin(w), and the third diagonal line has a 0.*0188

*That just multiplies away to nothing.*0195

*Now I will substitute back.*0198

*I get 2 × sqrt(x) cos(sqrt(x) + sin(sqrt(x)).*0202

*As always, we have to add a constant at the end.*0214

*So, the trick there was making a little substitution at the beginning.*0224

*Once we saw the sqrt(x), it looks kind of unpleasant to deal with.*0230

*So we make this substitution at the beginning that w = sqrt(x).*0235

*That allows us to convert the integral into something that is very amenable to integration by parts.*0240

*That is the end of our lecture on integration by parts.*0248

*Hi there, welcome to educator.com. This is a lecture on integration by parts.*0000

*The main equation for integration by parts is right here. *0005

*The integral of U dV is equal to UV minus the integral of V dU. *0011

*Where this comes from is the product rule in reverse. *0016

*The product rule is something you learned in Calculus 1, and is a way to take derivatives of products of functions.*0025

*This is changing around the product rule and using it as an integration formula.*0031

*The point of integration by parts is that you will be given a hard integral to solve. *0036

*What you are going to do, is take the integral that you are given and split it up into two parts, a U part and a dV part. *0044

*Then you will invoke this formula to convert it into UV minus the integral of VdU, and if you do that right, then the second integral that you get will be an easier integral. *0054

*Then, you can finish the problem by doing that easier integral. *0064

*That is the idea of integration of parts, but of course the best way to learn it is to do lots of examples. *0075

*Let us go ahead and do some examples.*0082

*Here is the first example, a very typical integration by parts problem. *0084

*We are trying to integrate X sin(X) dX.*0086

*Remember, the first part is to split this integral up into U and dV and we are going to let U be just X and dV be sin(X) dX. *0091

*You always put the dX with the dV part. *0108

*Then, we are going to figure out dU and V because those are both parts of the formula before. *0110

*dU, if U is X, dU is just dX and V if dV is sin(x), V is the integral of sin(x).*0116

*The integral of sin(X) is negative cos(X), and remember the integration by parts formula with the integral of U dV is equal to uV minus the integral of VdU. *0126

*Now, the integral that we are given, because we have converted it using our substitution, that is now the integral of U dV.*0140

*Using the integration by parts formula, that converts into UV while UV is minus X cos(x), minus the integral of VdU.*0151

*So, minus the integral of V dU, that is minus cosin(x) dX.*0167

*OK, I am going to cancel these two negative signs. *0177

*Now we have minus X cosin(x).*0180

*Now, this new integral you can see is just cosin(x).*0185

*That is a much easier integral to deal with than we started with. *0189

*The integral of cosin(x) is just sin(x) and I am going to add on a constant just because you always have a constant for an indefinite integral. *0191

*Then, we have our answer is negative X cosin(x) plus sin(x) plus a constant. *0200

*Let us try a trickier example. *0208

*X*^{2} e^{3x} dX. 0211

*Again, we are going to divide it up into a U and a dV.*0214

*We will let U equal X*^{2} and dV be e^{3x} dX.0218

*Again, we have got to figure out dU and V, so dU, if U is X*^{2} is 2x dX. 0228

*v is the integral of e*^{3x} dX. 0238

*That is one third e*^{3x}. 0239

*Then, I am going to write down the integration by parts formula again.*0246

*The integral of U dV is UV minus the integral of V dU and we have taken our example integral and we have split it up in to U dV again. *0250

*Invoking the formula again, that is UV, so that is 1/3 x*^{2} e^{3x} minus the integral of VdU.0265

*VdU is, there is a 1/3 and there is a 2, I will combine those as 2/3, and on the outside is x*^{3x} dx. 0280

*Now, what we have here is another integral. *0290

*It is easier than the first one because it has an x instead of an x squared. *0293

*However it is still not an integral that we can do directly. *0297

*What we have to do is integration by parts again. *0301

*This is a very common issue with integration by parts. *0304

*We are going to do integration by parts again on this new integral. *0307

*I will let U equal X dV equal e*^{3x} dX, and again fill in dU equals dX and V is 1/3 e^{3x} 0310

*We still have that first term minus 2/3. *0330

*Now we have the integral of U dV, so again using the integration by parts formula, that is UV, the new U and the new V. *0338

*So, 1/3 x e*^{3x} minus the integral of V dU, minus the integral of 1/3 e^{3x} dX. 0347

*I am just going to focus on the stuff on the right. *0364

*This is minus 2/9 X e*^{3x} and then the 2 minuses give you a plus 2/9. 0370

*Now the integral of e*^{3x} is 1/3 e^{3x}.0381

*If we put all of those parts together we get 1/3 x*^{2} e^{3x}. 0391

*Minus 2/9 e*^{3x} plus 2/27 e^{3x}.0398

*2/27 X e*^{3x} plus a constant, and that is the answer.0409

*The moral of that example is that sometimes you have to do integration by parts twice in the same problem.*0417

*First we had to do integration by parts to reduce the original x*^{2} down to an easier integral that just had an X in it. 0425

*But that still was not an integral that we could do directly.*0434

*We had to do integration by parts again to reduce the X, well actually to make the x go away, and give us an integral that we could do directly. *0437

*That is a pretty common story with integration by parts, that you have to do it twice. *0446

*I want to teach you a secret short cut to doing integration by parts problems. *0449

*This is just kind of a book-keeping device, but it can help you do some of these problems really quickly. *0458

*It is called tabular integration and I am going to introduce it with an example. *0465

*I am going to redo the same problem that I just did. *0469

*Remember that problem was x*^{2} e^{3x} dx. 0471

*Here is the secret shortcut. *0477

*What you do is write x*^{2} e^{3x}, and you make a little table here. 0482

*On the left hand side you write down derivatives. *0488

*The derivative of x*^{2} is 2x, the derivative of that is 2, and then the derivative of a constant is just 0. 0490

*On the right hand side, so those were all derivatives, you take integrals. *0497

*The integral of e*^{3x} is 1/3 e^{3x}. 0503

*The integral of that is 1/9 e*^{3x}. 0507

*The integral of that is 1/27 e*^{3x}. 0511

*Then this is just a clever little trick, but say this time we were doing the problem the previous way. *0517

*You draw these little diagonal lines and then you put little positive and negative signs on the diagonal lines alternating plus, minus, plus. *0523

*Then, what you do is multiply along these diagonal lines. *0533

*Those give you the terms of your answer. *0539

*So, you get x*^{2} times 1/3 e^{3x}. 0543

*Now, the next diagonal line is minus 2x/9 e*^{3x}. 0553

*The next diagonal line is plus 2/27 e*^{3x}. 0561

*You attach a constant at the end and there is your answer *0567

*That is just a clever book-keeping way of suppressing all the grunt work of going through the U and dV stuff.*0570

*It works really fast for certain kinds of problems.*0581

*If you have a polynomial like x*^{2}, times something like e^{3x} or cos(x) or sin(x) where it is easy to take integrals, then this tabular integration trick works really nicely. 0584

*I want to do a more complicated example, where we have e*^{x} cosin(x) dX. 0595

*Again, this is one where we can not do the previous shortcut, the tabular integration idea, because we do not have a polynomial.*0608

*If we took derivatives of either one of these, e*^{x} or cosin(x), we would never get down to 0.0612

*The shortcut is not available here. *0620

*But, again, we are going to start with integration by parts.*0622

*Let U equal e*^{3x}, or sorry, e^{x}, and dV equal cosin(x) dX, and fill in dU is e^{x} dX. 0625

*A lot of students leave off the dX when they are doing these problems. *0633

*It is really important to include the dX. *0643

*It makes all the notation work out and helps you to track where you are going late on, so do include the dX.*0646

*If dv is cosin(x) dX, then V is the integral of cosin(x), which is just sin(x). *0651

*This integral, again using our formula UV minus the integral of V dU, is UV. *0657

*So, e*^{x} sin(x) minus the integral of V dU. 0667

*That is e*^{x} sin(x) dX. 0677

*That gives us a new integral e*^{x} sin(x) and we are going to do integration by parts again. 0681

*We are agian going to let U equal e*^{x}. 0688

*This time dv is going to be sin(x) dx, and dU is e*^{x} dX. 0694

*V is the integral of sine which is negative cosin(x).*0700

*Invoking our integration by parts formula again, this is UV. *0709

*Negative e*^{x} cosin(x) minus the integral of V dU, that is negative cosin(x). 0718

*I will combine the negatives and that becomes positive there. *0730

*e*^{x} cosin(x) dx.0733

*I am just going to get rid of the brackets and bring the negative sign through so we get e*^{x} times sin(x) plus e^{x} cosin(x). 0741

*Minus the integral of e *^{x} cosine(x) dx.0752

*Here, we have got a strange thing happening because what we did was integration by parts twice. *0757

*If you look at this integral that we ended up with, which was supposed to be getting easier, it is exactly the same as the integral that we started with. *0764

*That seems like we are spending a lot of time to just go around in circles, because we have done a lot of work and we have come up with the integral that we started with. *0771

*In fact, we can use this to finish the problem. *0782

*The way we do that is we let i be this integral that we started with. *0789

*That means that i is equal to all these calculations that we did. *0794

*Then this integral that we get at the end is i again.*0798

*So what we can do is we can move i over to the other side of the equation. *0805

*What we get is 2I is equal to e*^{x} sin(x) plus e^{x} cosin(x). 0810

*That means we can solve for i. *0820

*I is just one half the quantity e*^{x} sin(x) plus e^{x} cosin(x).0824

*And then of course we have to add on a constant, as always.*0836

*All of a sudden, even though it looked like we were going around in circles, we have solved our integral. *0841

*That is the answer right there. *0844

*So that is another common pattern that can happen with integration by parts. *0852

*It happens a lot when you have combinations of e*^{x} and cosin(x).0857

*You have e*^{ax}, e to some number x, and then either sine, or cosine, of some number times x. 0861

*You can do this trick of doing integration by parts twice and then you get an expression that you can solve for your original integral.*0871

*I want to show you a little memory trick that can help a lot with integration by parts.*0880

*The difficult part with integration by parts is that you will be given an integral of a whole lot of stuff. *0890

*What you have to do is decide how to break up that integral into a U part, and a dV part. *0898

*That is what make integration by parts tricky, deciding what to make U, and what to make dV. *0914

*The idea is that after you go through the integration by parts formula, you want to get an integral that is easier than the one you started with, not harder. *0925

*That can often be difficult to predict ahead of time. *0932

*There is this little mnemonic that can help you remember how to split up integrals that way.*0938

*Just remember these five letters. L.I.A.T.E., lee-ah-tay. *0942

*What those stand for is the functions that you should use for U, if you have them. *0949

*Let U be the following functions if you have them. *0960

*First of all L stands for ln(x). *0966

*If you see a ln(x), that is your U. *0970

*I stands for inverse functions. *0973

*If you see an inverse trigonometric function, for example arcsine, arctan, and those are also written as sine inverse and tan inverse. *0976

*If you see those, you know that is going to be your U. *0991

*A stands for algebra. *0994

*If you see something like x or x*^{2}, that is going to be your U. 0998

*T stands for trigonometry.*1003

*Something like sin(x) or cosin(x). *1010

*E stands for exponential, so e*^{x}.1015

*Work through these functions in order, and whichever one of these you see first, that is going to be your U.*1020

*That is a very effective way of solving integration by parts problems.*1026

*As you work through your homework and try this out on different problems, keep this in mind and try it out. *1032

*I hope it works out for you. *1039

*So, that is the end of the first lecture from educator.com on integration by parts.*1040

*OK, lset us try a new example.*0000

*We have here tan*^{3}(x) × sec(x) and remember that the trick here is to look at the powers.0004

*You first look at the powers of tan(x).*0009

*Is that even?*0012

*Well 3 is odd so that does not work so we cannot use our strategy there.*0014

*The power on sec(x) is 1.*0020

*We ask, is that an odd number?*0023

*That does work, so we are going to use u = sec(x).*0026

*Then du = sec(x) tan(x) dx.*0033

*The point of that is that we can then convert this integral into something just involving sec(x).*0043

*We have tan*^{3}(x), so we will write that as tan^{2}(x) × tan(x) × sec(x) dx.0051

*Now the u is going to be sec(x), but we still need to have du.*0064

*sec(x) tan(x).*0066

*We have that right here, sec(x) tan(x).*0069

*The tan*^{2}(x), we need to convert that into secants.0073

*Since we have an even number of tangents, we can do that. *0078

*Remember our old Pythagorean Identity, tan*^{2}(x) + 1 = sec^{2}(x).0081

*The tan*^{2}(x) converts into sec^{2}(x) - 1.0090

*tan(x) sec(x) dx.*0099

*Now we are set up to put our substitution in.*0103

*Sec*^{2}(x) - 1 is u^{2} - 1.0106

*tan(x) sec(x) dx, that is just du.*0112

*All of a sudden we have a really easy integral.*0117

*We can integrate that quickly to be u*^{3}/3 - u.0121

*Then we substitute back into sec*^{3}(x)/3 - sec(x) + C.0126

*So, again, the reason that worked is because we were checking the odds and even powers on the tangent and the secant.*0143

*It worked because we had an odd power of sec(x).*0155

*There is one more example I want to work through with you.*0000

*It is a pretty tricky one.*0003

*It is sec*^{3}(x).0005

*This one is not so obvious.*0006

*What we are going to do is let u = sec(x).*0010

*But the point of that is not to make a substitution.*0017

*The point of that is to use integration by parts.*0022

*If u = sec(x), then dv = sec*^{2}(x) dx.0024

*That is the stuff left over after you use one of these secants.*0032

*That is not obvious that we should do it that way.*0036

*You might ask how did you know to break it up that way?*0041

*Well, here is the idea.*0044

*I know that sec*^{2} is the derivative of tangent. 0046

*So, by making dv be sec*^{2} v becomes tan(x).0050

*That is a relatively simple integration.*0056

*By making u = sec(x), we have du = sec(x) tan(x) dx.*0061

*That looks a little more complicated.*0070

*We will see how it works out. *0074

*Remember the formula for integration by parts.*0075

*It is always uv - the integral of vdu.*0077

*So, uv is sec(x) tan(x) - the integral of vdu.*0081

*So vdu looks a little complicated.*0086

*We have sec(x) tan(x) dx × tan(x).*0096

*So sec(x) × tan*^{2}(x) dx.0102

*This is looking a little complicated.*0108

*What I am going to do is take this tan(x) and remember the Pythagorean Identity.*0114

*tan*^{2}(x) + 1 is sec^{2}(x).0116

*I am going to write that tan(x) as sec*^{2}(x) - 1.0123

*The whole thing becomes sec(x) tan(x) - now I will distribute this secant across these two terms.*0135

*So we have - the integral of sec*^{3}(x) dx.0147

*+ the integral of sec(x) dx.*0156

*Now, it looks like we have taken what seemed like a pretty hard integral and replaced it by another hard integral.*0163

*And, an integral that is exactly the same as what we started with.*0171

*It looks like it is getting worse and worse.*0175

*Let me show you how you can resolve those two.*0178

*First of all, the sec(x) dx, there is a trick here.*0181

*Which is to multiply by sec(x) + tan(x)/sec(x) + tan(x).*0184

*You are multiplying by 1, not changing it.*0204

*But you get something very interesting when you multiply that into the integral.*0209

*The numerator will be sec(x) × (sec(x) + tan(x)). *0213

*So, sec*^{2}(x) + sec(x) tan(x).0220

*The denominator is going to be sec(x) + tan(x) all multiplied by dx.*0224

*Look at this.*0238

*If you look at the denominator, sec(x) + tan(x), the derivative of sec(x) is sec(x) tan(x).*0240

*The derivative of tan(x) is sec*^{2}(x).0250

*So the numerator is the derivative of the denominator.*0255

*So this last integral here is a derivative divided by the original function.*0259

*That last integral just integrates to ln(sec(x)) + tan(x).*0266

*That is a pretty special trick that is probably worth memorizing, just so you can do the integral of sec(x) dx.*0267

*Or else, it is just worth remembering that the integral of sec(x) dx is ln(sec(x) + tan(x)).*0287

*It is a strange and unusual enough integral that is worth memorizing for its own sake.*0296

*Let us go back to our original integral, which I am going to call i, just as we did when we did integration by parts twice.*0302

*What we have now is i has been resolved into sec(x) tan(x) - now we have i again, + that integral that we just did.*0311

*ln(sec(x)) + tan(x).*0328

*Separate that part out and so again just like we did when we had to do integration by parts twice.*0334

*We can move this i over and get 2i = sec(x) tan(x) + ln(sec(x) + tan(x).*0343

*Again, we can solve for the original integral just by dividing both sides by 2.*0364

*We get i = 1/2 of the quantity sec(x) tan(x) + ln(abs(sec(x) + tan(x).*0369

*As always, we have to add the constant at the end.*0386

*So, that is a fairly complicated answer.*0397

*That is a fairly specialized problem.*0401

*When you have a power of sec, an odd power of secant all by itself.*0405

*You want to use this integration by parts, sort of reduction formula,*0407

*To reduce it down to a lower power of secant.*0416

*If you keep reducing it using this formula, you eventually get down to just sec(x) by itself.*0420

*Remember we said that we will memorize that the integral of sec(x) = ln(sec(x) + tan(x).*0429

*That is the end of this lecture on integration of trigonometric functions.*0437

*Hello, this is educator.com and today we are going to talk about integration of trigonometric functions.*0000

*The prototypical examples of these integrals is you will have an integral and some power of sine and some power of cosine.*0006

*The important thing to focus on here is what those powers are.*0015

*In particular, which one is odd.*0022

*This is really a game of odds and evens.*0029

*You want one of those powers to be odd.*0031

*If one of them, if they are both even, then the integral gets more difficult and we will talk about that later in the lecture.*0034

*So right now the important thing is to find one of those numbers and to make it the odd or hope that it is odd.*0041

*If they are both odd, then it really does not matter, you can just pick one.*0047

*The one that is odd, what you do with it, is you are going to make a substitution and let U be the other one.*0051

*U is going to be the other one, and dU will be either + or - the one that is odd.*0076

*It seems a little strange so it would be easier to understand if we work through some examples.*0083

*You will see why this strategy works so well.*0089

*Let us take a look at some examples.*0093

*Here is the first example.*0095

*Cos(x) times sin*^{4}X, dX.0098

*Of course, cos(x) is the same as cos(x) to the 1.*0105

*So, you look at those numbers 1 and 4, and you see the odd one, that is the odd number. *0109

*What we are going to do is let U be the other one.*0114

*The other one is sin(x) and the point of doing that is dU is then cosin(x) dX.*0120

*So this integral, the cosin, and the dX, become the dU. *0128

*Those two become the dU, sin*^{4}X just becomes U^{4}. 0136

*All of a sudden you have a really easy integral, the integral of U*^{4}.0144

*Well, that is just U*^{5}/5 + C.0149

*Then you substitute back, and you get sin*^{5}X + C. 0156

*Just to reiterate there, what made that work was, we looked at which power was odd, and that was the power on cosine.*0167

*Then, we let u be the other one, sin(x)*0177

*The reason we did that is because we had that one power of cosine left over to be the dU.*0182

*So let us try that on a little more complicated example.*0188

*Cos*^{4}X sinU^{3}X dX. 0191

*Again, we look at those two powers and the odd one is 3.*0198

*We are going to let U be the other one.*0201

*U is cos(x) this time, and so dU, well the derivative of cosin is sin(x) dX *0205

*What we can do now is we can take this sinU*^{3} and we can split that up into a sin^{2}X × sin(x) dX. 0207

*The point here is that we save that sin(x) dX to be the dU.*0231

*Well, we are going to have to attach a negative sign, but that sin(x) dX is going to be the dU. *0236

*Then this sin*^{2}, we can convert that into 1 - cosin^{2}x.0243

*That is why we really had to have this odd even pattern going on.*0249

*It is because we save one sin to be the dU, or actually negative dU, and then you can convert sines into cosines two at a time.*0256

*If we save one sine to be dU, that will leave an even number of sines left over. *0266

*We can covert those all into cosines.*0273

*Our integral, after we make all of those substitutions, cos*^{4} becomes u^{4}.0276

*The sin*^{2} becomes 1 - cos^{2}, and then becomes 1 - u^{4}. 0285

*Sin(x) dX is dU, or actually negative dU. *0292

*Then we can combine these into one big polynomial. *0297

*Actually, I can bring this negative sign back inside, and make it the integral of u*^{4} × u^{2} - 1 dU.0304

*Combine those into u*^{6} - u^{4} dU. 0310

*Then, that is an easy integral because then we can just use the power rule, u*^{7}/7 - u^{5}/5.0321

*Finally, we can convert that back into u as cos(x).*0330

*So, this is cos*^{7}X/7 - cos^{5}X/5, and of course we have to attach a constant there.0337

*Again, the key thing here was recognizing those powers.*0350

*We see 4, we see 3, 3 is the odd one, so we want to let u be the other one.*0357

*The other one was cos(x).*0365

*The point of that was it gives you one extra sin(x) left over to be the dU.*0367

*Now, we know what to do if either one of the powers is odd.*0373

*What if you have both powers being even?*0382

*That makes it a trickier problem.*0384

*What we are going to use are the half angle formulas.*0386

*If both sine and cosine are even powers, then we can not use the previous strategy.*0391

*That odd strategy does not work.*0398

*We are going to use the half angle formulas.*0401

*Cos*^{2}(x) = 1 + cos(2x)/2 0403

*Sin*^{2}X = 1 - cos(2x)/2.0407

*The point of those is that those are easy to integrate.*0411

*Even if we have to multiply those together a few times, as we will see in the example, it is not too bad to integrate.*0415

*Let us try out an example of that.*0422

*Here we have sin*^{2}x cos^{2}x dX.0425

*We look at those powers, we look for the odd one, and oh no, they are both even.*0430

*So, we are going to use those half angle formulas.*0436

*Remember, sin*^{2}x is 1 - cos(2x) over 2.0438

*Cos*^{2}x is 1 + cos(2x)/2.0446

*Now I am going to pull the two halves out of the integral to get them away.*0455

*So now that is on the outside.*0460

*Now, we have the integral of 1 - cos(2x) × 1 + cos(2x).*0462

*I will multiply those two together and we will get 1 - cos*^{2}(2x).0469

*Well, the 1 is going to be easy to integrate, so I am not going to be worried about that.*0477

*But now we have a cos*^{2}(2x). 0481

*How do we integrate that?*0484

*Well, again, that is an even power of cosine.*0486

*We can use the half angle formula again and I am going to write the half angle formula over here.*0491

*But, I am going to use U this time.*0496

*1 + cos(2u)/2 because here we have cos*^{2}(2x), but if you think of that as being U, you can convert that into 1/4 the integral of 1 - cos^{2}(u).0500

*cos*^{2}(u) is 1 + cos(2u).0523

*1 + cos, now 2u if u is 2x, u is 4x.*0530

*This is 1/4 × cos(4x) dX.*0540

*That simplifies a little bit to 1/4 × 1/2 - 1/2 sin(4x) dx.*0558

*I am going to pull those halves out and combine them with the 1/4, and we get 1/8.*0573

*Times the integral of 1 - cos(4x) dX, and that is 1/8.*0579

*Well the integral of 1 is just x. *0591

*The integral of cosine is sine.*0596

*But, because it is 4x, we have to multiply on a 1/4.*0601

*Finally, we get (1/8)x - 1/8 × 1/4 = (1/32)sin(4x) and at the end of these we always add a constant.*0606

*In that one, we had to cope with the fact that there were no odd powers.*0626

*Both were even and we used the half angle formula to resolve that.*0632

*Sometimes after you use the half angle formulas, you find yourself with another even power as we did.*0634

*And so you have to use the half angle formula again to resolve that.*0642

*If you keep on using them, eventually you get down to single powers of cosine and that is something that you can integrate without too much trouble. *0647

*Let us look at another example.*0661

*This one is in terms of secants and tangents.*0663

*That is the other common pattern that you are going to see with trigonometric integrals.*0665

*That is, powers of secant(x), and tangent(x).*0671

*To do this, it is helpful to remember that the derivative of tan(x) is sec*^{2}(x).0674

*The derivative of sec(x) is sec(x)tan(x).*0685

*What we see here is again, looking at these powers, I look at the secant power and I see that that is even.*0699

*The reason that that is significant is that I can separate that out into sec*^{2}(x) tan^{2}(x) dx.0709

*And, I remember that the derivative of tangent is sec*^{2}(x).0724

*If I use the substitution u = tan(x) than dU is sec*^{2}(x) dX.0730

*The point of this is that I can convert this integral into, I have got sec*^{2}(x) dX, so that is dU, tan^{2}(x) is u^{2}.0744

*Now what about this other sec*^{2}, well remember the other pythagorean identity.0759

*Tan*^{2}(x) + 1 is sec^{2}(x).0764

*Sec*^{2}(x) can be written as tan^{2}(x) + 1 which I am going to convert directly into u^{2} + 1.0773

*This integral converts into u*^{3}, sorry, u^{4} + u^{2} dU.0786

*Again, that is a very easy integral.*0800

*That just integrates to u*^{5}/5 + u^{3}/3.0803

*Then we can substitute back, u is tan(x), so that is tan(x*^{5}/5) + tan(x^{3}/3) + C.0808

*Let us go back and look at what made that work.*0831

*What made that work again was the even power on sec(x).*0833

*If you have an even power on sec(x), then you can always use this substitution u = tan(x) dU = sec*^{2}(x).0841

*That even power of sec*^{2}(x) guarantees you 2 secants to use as the dU,0850

*and then the other secants you can convert into tangents using the pythagorean identity.*0856

*The other place you can use substitution like this is if the power on tangent is odd, then you can use the other substitution u equals sec(x).*0861

*Then your dU will be sec(x)tan(x) dX.*0892

*The point of that is that you will have an odd tangent.*0900

*Because you have an odd tangent, you will use this pythagorean identity to convert the tangents into secants two at a time.*0907

*Starting with an odd secant you will be left with exactly 1 tangent left and that tangent you can save that to be your dU.*0918

*We will see another example of that later on.*0930

*This is educator.com.*0000

*Here are a couple more examples on trigonometric substitution.*0003

*We are going to start with the integral of 6x x*^{2} dx.0006

*We have a quadratic under the sqrt.*0012

*The problem is we do not have a constant - x*^{2}.0016

*We have 6x - x*^{2}.0020

*What we have to do is complete the square on that before we go ahead with our trigonometric substitution.*0022

*Because we do not have a constant yet.*0030

*I am going to write that as sqrt, I am going to factor out the negative sign first, and I will get - x*^{2} - 6x,0031

*And now I want to complete the square here.*0043

*So, I take the middle term, the 6x, and divide it by 2 and square it.*0046

*So -6/-2 is 3, you square that and you get 9, add on 9.*0051

*Now to pay for that, well I added 9, but that was inside the negative sign right here.*0059

*So, what I really did there was I subtracted 9,*0069

*So, to balance that out I have to add 9.*0072

*We get sqrt(-x*^{2}-6x+9)+9.0079

*I will write that as sqrt(9) -, and the whole point of completing the square there is that this was (x-3)*^{2}.0090

*The first thing I am going to do here is a little substitution.*0102

*Let u = (x-3), and we always have to substitute the du as well.*0109

*The du is dx there, so it is an easy substitution but we have to do it.*0112

*What we have here is the integral of sqrt(9-u*^{2}) du.0120

*Now, this is something that looks like it is ready for a trig substitution.*0131

*We look at the trig substitution rules and we see that this one calls for a sin substitution,*0135

*Because it is 9 minus u*^{2}, the u is being subtracted.0142

*So we are going to let u = 3sin(θ).*0146

*Again, I got this 3 because it is the square root of this nine.*0154

*du is going to be 3cos(θ) dθ.*0160

*The point of doing that is the sqrt(9-u*^{2}), well, u^{2} = 9^{2}θ.0171

*If we pull a 9 out of the radical we get 3 × (1-sin*^{2}θ).0190

*1-sin*^{2}θ is cos^{2}θ, so this is just 3cos(θ).0199

*The radical turns into 3cosθ, the du turns into 3cos(θ) dθ.*0204

*This whole integral turns into 9cos*^{2}θ dθ.0219

*Again, we have a trigonometric integral in the trigonometric integral lecture, that when you see even powers of cosine,*0228

*you use the half-angle formula. *0233

*This is 9 × the integral of 1/2.*0237

*I will just pull the 1/2 outside, 9/2, times the integral of 1 + cos(2θ) dθ.*0243

*So, that is 9/2, if you integrate 1 you get θ + integral of cos is sine,*0256

*But because of that 2, we have to have a 1/2 there, and this was all being multiplied by the 9/2.*0272

*We will still have a constant.*0282

*Again, we have sin(2θ).*0285

*We have seen this earlier and the way we want to resolve sin(2θ) is 2sin(θ) cos(θ).*0290

*So, I am going to keep going on the next page here.*0301

*Let us remember the important things are that u was 3sin(θ), and in turn u was (x-3).*0307

*9 - u*^{2} was 3cos(θ).0316

*We will be using the sin(2θ) and I am just going to copy this equation on the next page and keep going.*0322

*From the previous page we have 9/2 × θ + 1/2 sin(2θ) + C.*0331

*So, 9/2, how can we sort θ out?*0347

*Well remember that u was 3sinθ.*0353

*So, θ if you solve this, u/3 is sin(θ).*0364

*θ is arcsin(u/3).*0372

*The 1/2 sin(2θ), remember sin(2θ) is 2sin(θ) cos(θ).*0378

*So 1/2 sin(2θ) is sin(2θ) cos(θ).*0390

*So we have 9/2 arcsin(u/3), sin(θ) is u/3.*0403

*Now the cos(θ), what we remember from before is that cos(θ).*0416

*Actually we cannot see it here, so maybe I will work it out again very quickly. *0424

*Cos(θ) is the sqrt(1 - sin*^{2}(θ)), which is sqrt(1) -, sin(θ) is u/3, so we know sin(θ) is u^{2}/90430

*Cos(θ) is sqrt(1 - u*^{2}/9).0445

*And, so, this turns into 9/2 arcsin, OK u/3 it is time to convert that back into, remember u is x-3.*0456

*So x-3/3 + now we have u/3 again.*0470

*That is x-3/3 again.*0475

*Now 1 - u*^{2}/9 is, 1 - u^{2} is (x-3)^{2}/9.0483

*I am just going to try to simplify this radical because I think you will recognize it after I simplify it.*0498

*That is 1 - (x-3)*^{2}, is x^{2} - 6x + 9/9.0504

*We are going to write 1 as 9/9.*0518

*The 9/9's cancel, and we get 9/2 arcsin(x-3/3) + (x-3)/3.*0524

*This radical turns into over 9 - x*^{2} + 6x.0538

*If you pull that 9 out, we get a 1/3 6x-x*^{2}, which is the same radical we started with.0547

*So, our final answer is 9/2 arcsin(x-3/3) + x-3/9 × sqrt(6x - x*^{2}) + C.0559

*That one was a pretty long example.*0581

*The key to it was recognizing this initial radical in the integral.*0589

*Recognizing that we had to complete the square on that.*0594

*Then we had to do a trigonometric sin substitution on that.*0600

*So, those were sort of the two key theoretical steps there.*0611

*The rest was just keeping track of lots of substitutions.*0617

*X into u, and u into θ*0618

*And then keeping track of a lot of different constants that were percolating throughout the whole interval. *0625

*A very long and complicated example but the basic ideas were just completing the square and doing a basic substitution.*0630

*So I want to do one more example.*0000

*This is the integral of dx/sqrt(x*^{2} + 1).0004

*Remember, when you see x*^{2} + 1, that is your sort of warning flag that you are going to need a tangent substitution.0009

*Remember a minus tells you that you are going to use either a sin or a secant substitution,*0018

*And a plus tells us that we are going to use a tangent substitution.*0025

*So we are going to use x = tan(θ), and then dx = sec*^{2}(θ) dθ.0027

*x*^{2} + 1 = tan^{2}(θ + 1), which by the trigonometric identity is sec^{2}(θ). 0043

*And, that was the whole point of making the substitution, to invoke that trigonometric identity,*0059

*That we would get tan*^{2} + 1 and could convert it into sec^{2}(θ).0064

*The sqrt(sec*^{2}(θ) is just sec(θ).0072

*When we solve this integral, or rather make this substitution into this integral, *0077

*We have dx in the numerator, so that converts into sec*^{2}(θ) dθ.0087

*In the denominator we have the sqrt(x*^{2} + 1),0093

*We already saw that that is sec(θ).*0094

*So, the secants, one of the secants cancels and we just get the integral of sec(θ) dθ.*0100

*That is the trigonometric integral that we learned in the trigonometric integral lecture.*0113

*You have to remember how to do it.*0116

*There is an old trick, where you multiply the top and bottom by sec(θ) tan(θ).*0121

*We covered that in the lecture on trigonometric integrals.*0136

*The answer for the integral of sec(θ) came out to be the ln(abs(sec(θ) + tan(θ).*0140

*That in turn, turns into, remember we have to substitute back into x.*0153

*sec(θ), I figure out what this right here, that is the sqrt of x*^{2} + 1.0156

*tan(θ) was our original substitution, that was x.*0164

*And, we have to put a constant on that.*0168

*So, that is our answer.*0175

*Again, the key step there was recognizing that we had x*^{2} + 1,0177

*And recognizing that that would give us a tangent substitution.*0181

*It was also important to keep track of the dx,*0188

*And then to work everything into θ's, and then at the end convert everything back into x's.*0190

*So that is the end of the lecture on trigonometric substitutions.*0195

*This has been educator.com.*0200

*Welcome to educator.com, this is the lecture on trigonometric substitution.*0000

*There are three main equations that you have for trigonometric substitution.*0006

*The idea is that if you see any one of these three forms in an integral, then you can do what is called a trigonometric substitution to convert your integral into a trigonometric integral.*0012

*Then hopefully you can use some of the trigonometric techniques that we learned in the previous lecture to solve the integral.*0028

*Let me show you how they work.*0035

*If you see something like the sqrt(a-bu*^{2}).0038

*Now here the a and the b are constants, and the u is the variable.*0042

*You are going to make the substitution u=sqrt(a/(bsin(θ)).*0048

*Why does that work?*0051

*Well, if you do that, then u*^{2} is a/(bsin^{2}θ), 0053

*And, so a-bu*^{2} will be a - while bu^{2} will be asin^{2}(θ)0063

*That is a × (1-sin*^{2}(θ))0079

*And 1-sin*^{2}(θ) of course is cos^{2}(θ).0088

*If you have the square root of that expression, then that will convert into cos(θ), and you will get an easier integral to deal with.*0096

*By the same idea, if you have au*^{2}-b, then you are going to have u=sqrt(b/(asec(θ)).0104

*What you are trying to do there is take advantage of the trigonometric identity tan*^{2}(θ)+1=sec^{2}(θ).0117

*Sec*^{2}(θ)-1 = tan^{2}(θ) so once you make this substitution you are going to end up with sec^{2}(θ)-1 under the square root.0131

*The last one you will have a+bu*^{2}=sqrt(a/(btan(θ))) and so again you are taking advantage of this trigonometric identity tan^{2}(θ)+1.0144

*You will end up with that under the square root, and that will convert into sec*^{2}(θ).0161

*All of these take advantage of these trigonometric identities.*0165

*One thing that is important to remember when you are making these substitutions is whenever you substitute u equals something, or x equals something, you always have to substitute in dU or dX as well.*0170

*For example, if you substitute u=sqrt(a/bsin(θ)), then you also have to substitute dU, which would be, a and b are constants so that is just sqrt(a/bcos(θ)) dθ.*0182

*You always have to make the accompanying substitution for your dU or your dX.*0204

*As with all mathematical problems it is a little hard to understand when you are just looking at mathematical formulas in general, but we will move onto examples and you will see how these work.*0211

*The first example is the integral of 4 - 9x*^{2} dx.0223

*That example matches the first pattern that we saw before, which was the square root of a minus b u*^{2}.0229

*The substitution that we learned for that is u equals the square root of a/b sin(θ).*0237

*Here, the a is 4, the b is 9, the x is taking the place of the u.*0249

*What we are going to substitute is x equals well the square root of a over b, is the square root of 4/9 sin(θ).*0259

*The square root of 4/9 is 2/3 sin(θ).*0276

*As we do that we also remember we have to substitute dX, which is going to be 2/3 cos(θ) dθ.*0279

*This is why it is really important to write the dX along with your integral.*0291

*It helps you remember that you also need to do the extra part of the substitution with the dX.*0296

*You have to convert that to the new variable as well.*0302

*So, then our integral becomes the integral of the square root of 4 minus, well 9x*^{2} is 9, x^{2} is 4/9 sin^{2}(θ) and dX is 2/3 cos(θ) dθ.0304

*Just working inside the square root for a moment here, we get the square root of 4 minus 4 sin*^{2}(θ) and we can pull the 4 out of the square root.0334

*We get 4 times the square root of 1 minus sin*^{2}θ.0343

*That is 2 times 1 minus sin*^{2}(θ) is cos^{2}(θ) so that will give us cos(θ). 0352

*Now, if we bring in the other elements of the integral, we get the integral. *0360

*I am going to collect all of the constants outside, so we have a 2, and a 2/3 that is 4/3 cos*^{2}(θ) because we have one cos here, and one cos here, dθ.0363

*We learned how to integrate cos*^{2}(θ), you use the half angle formula.0380

*This is 4/3 times the integral of 1/2 times one plus cos(2θ) dθ.*0387

*If we combine the 1/2 with the 4 we get 2/3 times now the integral of 1.*0401

*Remember we're integrating with respect to θ so that is θ plus the integral of cos(2θ).*0412

*Well the integral of cos is sin, but because of the 2 there, we have to put a 1/2 there. *0421

*This is now 2/3 and we want to convert this back to x's.*0432

*We have to solve these equations for θ in terms of x.*0448

*θ if we solve this equation in terms of x, we get 3/2 x equals sin(θ)*0449

*So θ is equal to arcsin(3/2 x)*0465

*Arcsin(3/2 x), that is the θ. Now 1/2 sin(2θ) to sort that out it helps to remember that sin(2θ) is 2 times sin(θ) times cos(θ) *0478

*1/2 2sin(θ) is sin(θ) times cos(v)*0500

*So, 2/3 arcsin(3/2 x)*0515

*Now, sin(θ) we said was 3/2 x.*0525

*Cos(θ) is the square root of 1 minus sin*^{2}(θ), so 1 minus 3/2 sin(θ) was x 9/4 x^{2}.0532

*Now we have converted everything back in terms of x.*0554

*And we have an answer.*0562

*The point here was that we started with a square root of a quadratic and we used our trigonometric substitution and we converted this integral into a trigonometric integral.*0568

*Then we learned in the other lecture how to convert trigonometric integrals, so we solved that integral in terms of θ.*0588

*Then we have to convert it back into terms of x.*0592

*So, let us try another example, this one is actually a little quicker.*0598

*We see the integral of dX over 1 + x*^{2} 0605

*The key thing to remember here is that when you see the square root of 1 + x*^{2}, or even 1 + x^{2} without a root, you want to use the tangent substitution.0611

*X equals the tan(θ)*0629

*The reason you do that is that x*^{2} + 1 is tangent^{2}(θ) + 1 but that is sec^{2}(θ)0632

*Let us make that substitution x equals tanθ.*0642

*Of course, every time you make a substitution you also have to make a substitution for dX.*0650

*dX is equal to sec*^{2}(θ) dθ0653

*If we plug that into the integral, we get dX is sec*^{2}(θ) dθ.0663

*1 + x*^{2} is sec^{2}(θ) 0672

*We kind of lucked out on this one because the sec*^{2}'s cancel and we just get the integral of θ0680

*This turns out to be a really easy one, that is just θ.*0686

*Theta, remember that x was tan(θ) we know that θ is arctan(x) + a constant.*0692

*The thing to remember about this example is that when you have a + there, you are going to go for a tangent substitution.*0706

*Whenever you have a minus, you are going to go for a secant substitution.*0720

*We are going to move to a more complicated example now.*0727

*dX/x*^{2} + 8x + 25.0729

*We have a more complicated expression in the denominator here.*0737

*What we are going to do is try to simplify it into something that admits a trigonometric substitution pretty easily.*0738

*What we are going to do is a little high school algebra on the denominator. *0748

*We are going to complete the square on that, so that is x*^{2} 0758

*Remember, the way you complete the square is you take this 8 and you divide by 2 and you square it. *0764

*So 8 divide by 2 is 4 and 4 squared is 16, so we will write 16 there.*0770

*To make this a true equation, we have to add 9 there.*0777

*Then that is x + 4, quantity squared + 9.*0780

*We are going to use that for the integral.*0791

*What we are going to do is use that for the quick substitution.*0795

*U is equal to x + 4 and again we have to convert dU, but that is easy that is just dX.*0797

*The integral just converts into dU/u*^{2} + 9.0811

*Now, to finish that integral, well, we want to make a trigonometric substitution.*0820

*We are going to let u be 3tanθ.*0829

*The way I got 3tanθ was I looked at 9 and I took the square root of that and I got 3. *0839

*The point of doing that is that u*^{2} + 9 will be 9tan^{2}(θ)+9.0844

*And, that is 9+tan*^{2}(θ)+1.0856

*Which is 9 sec*^{2}(θ).0863

*Of course with any substitution, you have to figure out what dU is.*0867

*Well if u is 3tan(θ) then dU is 3 sec*^{2}(θ) dθ.0871

*We will plug all of that in, the dU was 3 sec*^{2}θ dθ.0881

*The u*^{2} + 9 converted into 9 sec^{2}θ0890

*The three and the 9 simplify down into 1/3 and now again, the secants cancel and we just have the integral of dθ.*0900

*That is 1/3 θ.*0907

*θ is something we can figure out from this equation over here.*0914

*θ if we solve this equation u/3=tanθ. *0925

*θ is arctan(u/3).*0933

*But we are not finished with that because we still have to convert back into terms of x.*0938

*That is 1/3 arctan, now u was (x+4), remember our original substitution there, so (x+4)/3 and now we add on a constant.*0944

*Now we are done.*0960

*OK, that is the end of the first instalment of trigonometric substitution.*0966

*This has been educator.com.*0972

*OK, we are going to try a couple more examples of the partial fractions technique. *0000

*Remember that it is an algebraic technique of trying to separate a rational function,*0005

*Meaning a polynomial over a polynomial,*0013

*Into different pieces that are easier to integrate.*0016

*Remember that the very first step of partial fractions is to try to factor the denominator.*0020

*We are trying to factor x*^{2} + 4x + 5,0027

*Right away you get a problem.*0035

*It does not work.*0038

*So, even though this looks like a partial fractions problem,*0044

*It turns out that we are stuck at the quadratic level.*0048

*We cannot factor x*^{2} + 4x + 5 down into linear terms.0052

*What do we do instead?*0058

*Well, there is a two-step procedure for solving problems like this.*0060

*It is very important you do the steps in order.*0063

*The two steps are a u substitution, and then a trig substitution.*0068

*It is very important that you do those in order.*0080

*If you do not do those in the right order, you will end up giving yourself a lot of extra work.*0083

*Let us see how that plays out.*0090

*The first step here is let u = the denominator.*0094

*u = x*^{2} + 4x + 5.0098

*Then du = 2x + 4 dx.*0104

*So, what we would like to do is see if we have du in the numerator.*0114

*We do not, because we have 6x + 10, instead of 2x + 4.*0120

*So, we are going to write 6x + 10/x*^{2} + 4x + 5.0127

*We would really like to have du in the numerator.*0136

*I would really like to see du in the numerator, so I am going to write du in the numerator, 2x + 4.*0145

*That does not work, because we have a 6 over here.*0152

*To make that correct, I am going to put a 3 here.*0156

*That makes the 6's match.*0160

*We have 3 × 2x.*0163

*Unfortunately, that gives us 3 × 4 over here which is 12, which does not match the 10.*0167

*To make that match, I am going to subtract off a 2 x*^{2} + 4x + 5.0175

*So, 12 - 2 is 10, so it does match.*0185

*Now, we have 3 × 2x + 4 - 2.*0192

*We are going to attack those integral separately.*0197

*The left hand one we can use our u and du.*0200

*That turns into the integral of 3du/u.*0205

*Which is 3 × ln(u), and then we can substitute back,*0212

*And, we will be done with that one.*0217

*The right hand one is a little harder. *0220

*The right hand one is where we are going to use step 2.*0222

*We are going to use a trig substitution, but in order to do that, we have to complete the square on the denominator.*0232

*We are going to write the denominator x*^{2} + 4x + 5.0240

*Remember, completing the square, you take the middle term divide it by 2, and square it.*0252

*So, 4/2 = 2, 2*^{2} = 4, so I will write + 4. 0262

*And to make that accurate I have to write + 1 to match the 5 right there.*0268

*So we can write this as x + 2 *^{2} + 1.0277

*What I am going to do is a little substitution, u -, actually I do not want to use u because I already used u for the other integral.*0283

*So I will use w.*0290

*w = x + 2, and then dw = dx.*0294

*We have for the second integral, - the integral of 2 dw/x + what turned out to be (x + 2)*^{2}, so that is w^{2} + 1.0304

*Then, the way you solve this is by letting w = tan(θ).*0320

*We learned that on the section on trig substitution and we actually did this integral a couple times,*0327

*So, I am not going to show you all the detail again.*0331

*You might remember all the details to the pattern, but if you do not,*0334

*Go ahead and do the substitution, put in dw, and you will get the answer quickly.*0337

*So, - 2 arctan(w), which in turn converts back to - 2 arctan(x + 2).*0344

*That was just the answer to the second integral.*0363

*We still have the first integral to combine with that.*0367

*That is 3 × ln of, u was x*^{2} + 4x + 5. 0370

*Then we will add on a constant as before.*0382

*There are a couple of important points about this example.*0391

*Remember, in all partial fractions problems, you try to factor the denominator first.*0396

*In this case, the denominator did not factor.*0400

*What we do is this 2 step procedure, it must be done in order.*0406

*First we do a u substitution where u = the denominator. *0411

*Then we do a trig substitution.*0417

*A secret trick here, is that when you do a trig substitution on one of these, it is always going to be a tangent substitution.*0425

*Remember we had three types of trig substitutions, tangent, sine, and secant.*0432

*But, when you have a problem like this, you will never get sine and you will never get secant.*0440

*Here is why you will never get sine and you will never get secant.*0447

*The reason you will never get sine or secant is because you use sine when you have something like 1 - u*^{2}.0453

*You use secant when you have something like u*^{2} - 1.0460

*Now, either one of those can be factored.*0465

*u*^{2} - 1 factors as (u-1)(u+1).0469

*1- u*^{2} factors as (1-u)(1+u).0474

*So if you had either one of these trig substitutions,*0481

*It means you actually had something that would factor.*0484

*We said back here that the factoring does not work.*0489

*Essentially if you get over to the trig substitution step,*0491

*And you get a sine or a secant substitution, it means you screwed up somewhere.*0496

*So go back to the factoring step and see if you can factor it.*0503

*Check your work back there, it probably means you made a mistake.*0506

*If you did not make a mistake through any of your work, then you are guaranteed to get a tangent substitution at the end.*0514

*So you keep going through the tangent substitution and you end up with your answer.*0522

*We have got one more example on partial fractions.*0000

*It looks like the nastiest one of all.*0004

*It is a little bit more work than some of the others.*0008

*Remember, the first step is to check the degrees of the numerator and denominator.*0010

*Here we have got degree 2 in the numerator, and degree 3 in the denominator.*0013

*Remember what we are checking for.*0020

*We are checking to see if the numerator is bigger than or equal to the denominator. *0024

*Well, 2 is not bigger than 3, so it is OK, we do not have to go through long division.*0027

*So there is no long division which is a relief because these are pretty nasty polynomials.*0033

*So, we do not have to go through long division, but we do have to through and factor the denominator.*0041

*That is going to be a little bit of work, because this is a cubic.*0050

*Remember that the way you factor a cubic is you look at factors of the constant term, the 10.*0058

*Divided by factors of the leading term.*0078

*Which in this case is 1.*0085

*You look at + or - those factors and those are your possible candidates.*0089

*We have + or - factors of 10 could be 1, 2, 5, or 10.*0099

*Divided by factors of 1 is just + or - 1, so I will not even include that.*0107

*So we have 8 candidates for factors here.*0113

*I will show you how to test possible candidates.*0118

*Suppose we want x = 2.*0122

*Is that a possible candidate?*0124

*Well there is a trick you might have learned in high school algebra.*0126

*Where this is the same as checking whether x -2 is a factor of the polynomial.*0133

*What you do is you look at 2 and then you write down the coefficients, 1, 6, 13, and 10.*0139

*I am getting those from the coefficients of this polynomial.*0149

*Then you bring the one down, you multiply by 2, 2.*0154

*You add, so 6 + 2 = 8.*0160

*Multiply by 2, so that is 16, add, so that is 29.*0164

*Multiply by 2, 58, and then we add, and we get at the end 68.*0170

*So, that means that x = 2 is not a root there.*0179

*Or x-2 is not a factor.*0186

*Let us try something else, let us try x = -2.*0189

*Which means we are testing x+2 as a factor. *0192

*That means we put -2 in, we go through the same process, 1, 6, 13, and 10.*0198

*We bring the 1 down, multiply by -2, we add, so that is 4, 6 + -2 is 4.*0206

*Multiply by -2 = 8. Add and we get 5.*0218

*Multiply by -2, we get 10, and look, we got 0 at the end.*0224

*That tells us that x = -2 is a root and x+2 is a factor.*0230

*Moreover, it tells us that our polynomial factors into x+2 times, now you use these coefficients,*0239

*To get the other factor.*0258

*x*^{2} + 4x + 5.0260

*So this is all a technique that you learn in high school algebra,*0265

*But by the time that you get to Calculus 2 sometimes it is a little bit rusty.*0271

*What we are going to try to do is write, 8x*^{2} + 30x + 30/x^{3} + 6x + 13x + 10 = a/x + 2,0276

*Then it would be nice if we could factor x*^{2} + 4x + 5 down further.0298

*But notice that that does not factor anymore.*0304

*We are stuck with that as a quadratic term.*0310

*I will put that as my next denominator.*0314

*x*^{2} + 4x + 5.0316

*Because this is quadratic, I cannot just put b, I have to put bx + C.*0320

*Now, I am going to solve for a, b, and c.*0330

*If you combine these two terms over a common denominator,*0333

*The common denominator would be x + 2 × (x*^{2} + 4x + 5).0339

*It is the same as the denominator on the left.*0342

*The x*^{3} term that we started with. 0348

*What you would get is a × x*^{2} + 4x + 5.0352

*+ bx + c × x + 2.*0357

*Because you would be multiplying these terms by x + 2.*0364

*All of that is equal to 8x*^{2} + 30x + 30. 0367

*So, I am going to combine these 2 polynomials and separate them by power of x.*0376

*So, my x*^{2}, I am going to have an a here,0384

*Then from the b and the x here, we get a b x*^{2}.0389

*+, now let me see what we get for an x term, we get a 4a over here from the 4.*0395

*For an x term we get 2b + C.*0404

*The constant term is 5a + 2b, sorry not 2b.*0415

*The constant term is 2c. *0431

*That is still equal to 8x*^{2} +30x + 30.0434

*I am going to carry these equations over onto the next slide and we will see what we can do with them.*0446

*From the previous slide, we had an x*^{2} term, an x term, and a constant term.0455

*On both sides.*0465

*The x*^{2} term on one side, it was a + b. 0468

*On the other side it was 8.*0477

*The x term was 4a + 2b + C = 30.*0480

*The constant term was 5a on one side, + 2C, and on the other side, = 30.*0490

*Here you have 3 equations and 3 unknowns.*0500

*It is a little messy but it is nothing impossible from high school algebra.*0505

*You could start with a quick substitution of a = 8 - b.*0512

*Plus that into the other two equations and then you get 2 equations in b and c.*0521

*Then you get a high school algebra where you are solving for 2 equations and 2 unknowns.*0535

*It is a little messy but not too bad.*0539

*I am going to skip over the details of that and just tell you what the answers come out to be.*0542

*You can try working it out on your own.*0547

*It turns out that a = 2, b = 6, and c = 10.*0550

*What that means is that our integral, remember we had a/x+2, so that is 2/x+2 + 6x + 10/x*^{2} + 4x + 5.0557

*We have got to integrate all of that.*0584

*What we get there, this is an integral we have seen before,*0588

*This is just 2 ln(x+2) abs(x+2),*0595

*6x + 10/x*^{2} + 4x + 5.0600

*That sounds more complicated, but we saw the technique to do that in the previous example.*0604

*Remember, there was a two-step technique for that, there was a u substitution,*0611

*U = the denominator*0621

*Then the second step, after you plough through the u substitution, was a trig substitution.*0625

*Remember that the trig substitution should always come out to be a tangent substitution,*0633

*Because if it is a sine or a secant substitution, it means you really could have factored the denominator.*0639

*We said here that this is a denominator that we cannot factor. *0647

*I rigged up the numbers so that it comes out exactly the same as the previous problem.*0656

*I am not going to go through solving this again, because this is the exact same as the example we did on the previous problem.*0665

*Let me show you just what the answer came out to be.*0672

*This came out to be 3 ln(x*^{2} + 4x + 5).0675

*That was the part that came out of the u substitution.*0684

*- 2 × arctan(x + 2).*0689

*You can look this up on the previous problem or you can work it out.*0697

*That is the answer that we got.*0700

*All of this put together is the answer for the whole integral.*0703

*Let us recap what we had here.*0712

*We had the integral of a polynomial over a cubic.*0715

*What you try to do with problems like these is you try to factor the cubic.*0724

*There are sort of two ways it could factor.*0733

*It could factor into three linear terms,*0736

*If so, then you try to split it up into a over 1 linear term + b over the next.*0740

*+ C over the next.*0750

*That did not happen for this problem.*0754

*This problem factored into x over a quadratic term.*0757

*The quadratic term did not factor.*0762

*What we had to do there was write it as a over the linear term,*0767

*+, we had to go with bx + C over the quadratic term.*0775

*Solve for a, b, and c.*0783

*Then the a part, integrating that is easy.*0788

*The b part is where we go through this 2 step procedure of a u substitution and a trig substitution.*0789

*That is the end of the lecture on partial fractions.*0797

*Thank you for watching educator.com.*0799

*Hi, this is educator.com, and we are going to learn today about partial fractions.*0000

*Partial fractions is a technique used to solve integrals but it is really grounded in algebra.*0004

*We are going to be doing a lot of algebra and not that much integration.*0012

*The point of partial fractions is to solve integrals where you have something like one polynomial, or maybe just a constant, divided by another polynomial.*0018

*What you are going to do is factor the denominator polynomial.*0037

*I have an example of a really horrific one here.*0043

*In practice, the ones you get on your homework, you will not have one this bad. *0047

*But this is kind of an example of the worst it could possibly be.*0052

*The denominator is a really complicated polynomial, but we factored it down.*0055

*You will factor the denominator polynomial, and then you will do some algebra to separate this fraction into other pieces, into simpler fractions that are supposed to be easier to integrate.*0058

*As usual, it is much easier to do this using examples rather than to simply talk about it in the abstract.*0078

*Let us go ahead and try some examples.*0079

*The numerator polynomial is just 1, the denominator is x*^{2} + 5x + 6.0084

*We are going to forget about calculus for a while, and just write this as 1/(x*^{2} + 5x + 6)0092

*Just doing some algebra, the urge here is to factor the denominator.*0104

*That is an easy one to factor, that factors into (x + 2)(x + 3).*0108

*What partial fractions does, is it tries to separate that into a constant divided by (x + 2) plus another constant divided by (x + 3). *0116

*We are going to try to figure out what those constants should be. *0136

*If you imagine putting these two terms on the right back together, we would combine that back over a common denominator, (x + 2) × (x + 3).*0141

*To put them over a common denominator we would have to write a × (x + 3) + b × (x + 2). *0159

*Remember this is still supposed to be equal to 1/(x*^{2} + 5x + 6).0164

*Effectively the numerator, a × (x+3) + b × (x+2) has to be equal to the numerator 1. *0174

*I am going to expand out the left hand side as a polynomial, that is ax + bx, so that is (a+b)x + 3a + 2b = 1.*0189

*I want to think of 1 as being a polynomial, so I want to think of that as 0x + 1.*0203

*Remember that whenever you have two polynomials equal to each other, that means their coefficients have to be equal.*0212

*That is saying the coefficient of x has to be equal on both sides, so (a+b)=0 and the constant coefficients have to be equal as well. *0220

*(3a+2b)=1.*0230

*This is now a linear system that you learned how to solve in high school algebra.*0240

*There are a number of different ways of doing this.*0246

*You can do linear combinations, you can do substitution, I think the easiest way of doing this is probably substitution.*0256

*If we write a=-b, and then plug that into the second equation, we get -3b+2b=1, so -b=1 so b=-1 and a=1*0258

*That is just solving a system of variables and two unknowns just like you learned how to do in algebra 1 or 2 in high school. *0280

*So we take those numbers and we plug them back into our separation and we get that our integral is equal to two separate integrals. *0295

*a=1/(x+2) dX + b=-1/(x+3) dX*0307

*Those are both easy integrals to solve.*0317

*We can solve the first one by doing u=x+2 dU=dX, so we get the integral of dU/u, and that is equal to ln(abs(u)) which is the ln(abs((x+2))).*0319

*The second one is solved in exactly the same technique, so I will skip over the details of that.*0342

*This minus is from the negative sign above the ln(x+3) + a constant.*0353

*The integration there at the end was very easy. *0364

*The important part that we are trying to learn here is the algebra, where you start with a polynomial in the denominator, you factor it out and then you try to split this expression up into two parts with an a and a b.*0370

*Then, you combine those two parts back together, that gives you two equations for a and b.*0386

*You solve those two equations for a and b, you substitute them back into the original expression, and you get an integral that hopefully turns out to be easy to solve.*0397

*Let us try another example that gets a little more complicated.*0407

*Example 2 here, again we have a polynomial over a polynomial. *0411

*(2x*^{2} - 7x)/(x^{2}+6x+8) 0415

*There is a problem with this one, it will not work as well as the one on the previous page.*0423

*The problem is the degree of the two polynomials.*0426

*If the degree of the numerator is bigger than or equal to the degree of the denominator, then you can not go to partial fractions directly. *0436

*You have to do something else from high school algebra, which is long division of polynomials.*0460

*Let me show you how to do a long division here.*0464

*We are going to do (x*^{2}-6x+8) into (2x^{2}-7x) and I am going to stick on 0 as a constant term there.0480

*We look at x*^{2} and 2x^{2} and we divide those into each other and we get 2, and then just like long division of numbers we multiply 2 by the whole thing here.0495

*So, we get 2x*^{2}-12x+16, subtract those the 2x^{2} cancel, subtracting is the same as changing signs and adding.0505

*So, minus 7x + 12x gives us 5x - 16.*0518

*We can write (2x*^{2} - 7x)/(x^{2}-6x+8) as our answer 2 plus a remainder term (5x-16)/(x^{2}-6x+8).0527

*Now, the important thing here is that the degree of the numerator is now 1, the degree of the denominator is now 2.*0551

*The denominator now has a bigger degree and partial fractions is going to work.*0566

*Remember, partial fractions does not work if the degree of the numerator is bigger than the degree of the denominator.*0569

*That is no longer true after we do polynomial division and we can expect the partial fractions to work.*0580

*I am not going to worry about the 2 for a while, because I know that 2 is going to be easy to integrate.*0584

*I am going to do partial fractions now on 5x-16 over that denominator, and I can factor that into (x-2) × (x-4).*0598

*I am going to try to separate that out into a/(x-2) + b/(x-4).*0614

*Just like we did before, we are going to end up with a × (x-4) + b × (x-2), that comes from trying to combine these terms over a common denominator, is equal to the numerator on the other side (5x-16).*0626

*(a+b) × (x=5x and -4a-2b=-16,*0647

*so a+b = 5 and -4a -2b = 16.*0664

*Again, you get a high school algebra problem, two equations and two unknowns, and you can solve that using any technique that you prefer from algebra.*0678

*I am just going to tell you the answer now, it turns out that a=3 and b=2. *0687

*You can check that in the two equations to see that it works.*0700

*Our integral now becomes the integral of 2, plus a = 3 so that is 3/(x-2), and b was 2 so that is 2/(x-4). *0702

*All that stuff that we did so far is just algebra, we still have to integrate this problem.*0724

*However, it is a very easy integral now.*0731

*The integral of 2 is just 2x.*0733

*We saw that is easy to integrate 1/(x + or - a constant), you just get ln(abs(x-2)) + 2(ln(abs(x-4)) + a constant.*0740

*Let us reiterate, the key thing that we learned from this example is that before you start doing partial fractions, you have to check the degree of the numerator and denominator.*0761

*If they are equal, or if the denominator is smaller, then you have got to do this step of long division or the partial fractions is not going to work for you.*0782

*After you do the long division, you get a part that is easy to integrate and a remainder term and the remainder term should work out for you when you try to do partial fractions.*0796

*The next example looks pretty similar to the previous examples but it has a little twist as you will see in a moment.*0811

*Again we start out by checking the degrees.*0822

*The degree is 1 in the numerator, the degree of the denominator, the power of the highest term, is 2.*0824

*So that is ok, the numerator has a smaller degree than the denominator so we expect partial fractions to work.*0831

*We are going to go ahead and try to factor the denominator.*0843

*(x*^{2}-6x+9), that factors as (x-3)^{2} and if you are going to try to separate that out using partial fractions, your first instinct might be to try a/(x-3) + b/(x-3) the same way we did the previous examples.0845

*That does not work. *0873

*Here is why that does not work, Suppose you did that.*0875

*If you try to combine those back together you get (a+b) × (x-3) and that does not match, that cannot possibly match (3x-7)/(x-3)*^{2}. 0878

*If you naively try a/(x-3) + b/(x-3), the partial fractions is not going to work.*0897

*That would give you a big problem.*0908

*The secret to fixing this is to have a/(x-3) but b/(x-3)*^{2}0914

*That is kind of the secret to this problem, is to put in that square on (x-3).*0931

*Then if you try to find, if you try to match up the numerators, we get (3x-7)=a × (x-3).*0935

*To get the common denominator, and then + b because if you wanted to combine those two, our common denominator would be (x-3)*^{2} and b is already over that denominator, so we would not have to multiply b by anything to get that common denominator. 0948

*Then, we get ax-3a + b = (3x-7), so ax plus, let me separate out terms (3a + b)=3x-7 and again you get two equations and two unknowns except this one is very easy, you get a=3 and minus 3a+b=-7.*0970

*Again, solve that using anything you want from high school algebra and of course you get a=3 and you plug that in, it works out to b=2.*1003

*So, our integral turns into the integral of a/(x-3) + , now b=2/(x-3)*^{2}, all of that times dX.1021

*Now, we have learned already how to integrate 3/(x-3), that is ln(abs(x-3)).*1045

*2/(x-3)*^{2}, we have not seen that recently.1054

*The way you do that is you make a little substitution, u = x-3, so dU = dX.*1058

*That gives you the integral of 2dU over u*^{2}, and you think of that as being u^{-2} and the integral of that is 2u^{-1}/-1.1070

*This gives you 3ln(abs(x-3)), I am going to pull this -1 outside so we get -2.*1091

*Now, u*^{-1} = (x-3) and u^{-1} = 2/(x-3) and then I will tag on a constant. 1102

*OK, so the key observation in that way is that if you have a denominator that factors into a perfect square, you can not separate it out as you did in the previous examples.*1125

*You have to separate it out into (x-3) and an (x-3)*^{2}, and you still have an a and a b on both parts.1142

*OK, let us try some trickier examples of trigonometric substitution.*0000

*Here we have the integral of 2x - x*^{2} dx.0006

*The thing that makes that different from the basic forms,*0009

*The basic forms look like a*^{2} - u^{2}, or a^{2} + u^{2}, or u^{2} - a^{2}. 0015

*You will see in your tables of integrals, a lot of those kinds of forms.*0030

*But in all of those, the a is constant.*0036

*The problem with our integral is the 2x.*0038

*It is 2x, and not 2, so that makes it a bit harder.*0042

*However, if you look towards the end of the section of integral tables,*0049

*There is one that covers our situation and I am going to write down the integral formula for you.*0057

*It is number 113 in Stewart's book, it might be a different number in your book.*0060

*But it says that the integral of sqrt(2au - u*^{2}) du = u - a/2 × sqrt(2au - u^{2}) + a^{2}/2, arccos(a - u)/a + C.0066

*So, we are going to figure out how to invoke that formula.*0100

*The x is acting just like the u in the formula.*0108

*So, we have 2x, and in the formula we have 2au.*0113

*That tells us that the a must be 1 and the u is acting like the x.*0120

*So, we can invoke the formula directly.*0126

*U - a = x-1/2.*0131

*sqrt(2au - u*^{2}) = 2x - x^{2} + now a^{2}/2, a = 1, so that is 1/2 arcos(a-u), is 1-x over a is just 1.0137

*That can be simplified down to 1-x, plus a constant.*0162

*That one was a complicated formula, but we were actually able to find it directly in the book.*0172

*So that one went a little more quickly than some of the more tricky examples.*0183

*Finally, I would like to mix an exponential function and a trigonometric function.*0000

*The integral we have here is e*^{3x} × cos(5x) dx.0007

*Again, I am going to use Stewart's book, and he has a section called exponential and logarithmic forms.*0014

*I will show the formula that I am going to be using.*0026

*It is formula number 99.*0030

*So, 99 says the integral of e*^{au} × cos(bu) du = e^{au}/a^{2} + b^{2} × acos(bu) + bsin(bu).0034

*Again, plus a constant.*0071

*So, let us try to figure out how to reconcile the integral we are given with this formula.*0078

*Well, obviously the a=3, and b=5.*0085

*We can just leave dex/dx as the same as being u.*0092

*Then we can read off the answer from this formula.*0103

*We have e*^{au}, OK, that is e^{3x}.0105

*Now a*^{2} + b^{2}, that is 3^{2} + 5^{2},0111

*3*^{2} is 9 and 5^{2} is 25.0116

*9 + 25 = 34.*0121

*a = 3 so, 3cosine, b=5, so 5x, + b is 5, 5sin(5x) + C.*0127

*That is our answer there.*0146

*That is the last example on using integration tables.*0159

*I want to emphasize here something that a lot of students do not realize.*0160

*All of these formulas in the integration tables were derived using the techniques we learned in the lectures here. *0170

*So, these formulas should not come as sort of mysterious, magical formulas that you just sort of invoke blindly.*0175

*All of these formulas come from the techniques we have been learning.*0184

*This one that we just used here for example is actually coming from integration by parts.*0190

*We did a problem in the integration by parts lecture where we did integration by parts twice.*0195

*We essentially derived this formula.*0200

*If you are ever stuck without a table of integrals, you can figure out all of these formulas using what you have learned in the other lectures.*0205

*What the role these formulas play is if you do not want to go through the steps for integration by parts twice, *0216

*If you have done that enough times where you understand the procedure,*0220

*And you have to do the integral again,*0225

*You can quickly look it up in an integration table in the back of the book, or on the web.*0230

*You can jump essentially straight to the answer here.*0234

*It is the same answer that you would have gotten had you gotten integration by parts twice, and gone through a little extra arithmetic there.*0237

*This concludes the lecture on integration tables.*0245

*This has been educator.com.*0250

*This is educator.com, and today we are going to talk about how to use integration tables to solve integrals.*0000

*If you are taking a calculus class, you have a calculus book, and if you look inside the back cover of your calculus book, you are going to find several pages of integration tables.*0008

*You can also find these tables on the web. *0026

*We are going to learn to use these tables here today.*0032

*The key thing is that you want the pattern in the integral you are given to match the pattern in the integration table exactly.*0034

*Often that means you are going to have to do a little bit of work with the integral that you are given to get it into a form that matches the integration table. *0045

*We are going to see some examples where you can practice doing that work.*0050

*You get it into a form where you can use the integration table. *0056

*A big part of that is that the integration table will always have a dU.*0063

*You have to make your integral match that, including the dU.*0066

*You will see when we go through the examples.*0074

*So, the first example here, we have the sqrt(x*^{2}+2x+5).0077

*The calculus book I am using is James Stewart's Essential Calculus, and I am look back at the integration tables there, *0085

*His formula 21 says that if you have the integral of a*^{2}+u^{2} dU. 0094

*Then that converts, or that solves to, (u/2) × sqrt(a*^{2}+u^{2})+a^{2}/2 × ln(u+sqrt(a^{2}+u^{2})+C.0105

*That is an easy formula to use if we can get our integral to look like that pattern.*0133

*It does not quite look like that pattern yet.*0143

*What we have to do is complete the square on what is underneath the radical sign. *0145

*So we are going to write that as x*^{2}+2x, and then we are going to complete the square on that. 0153

*Remember, completing the square you look at the x term and you divide it by 2 and square it. *0160

*So, 2/2=1 and 1*^{2}=1, so I am going to write 1.0164

*But actually, what we had here was x*^{2}+2x+5 so to make that equal I am going to write + 4.0173

*The point of completing the square was so that we could write this as (x+1)x*^{2}+4.0185

*We are going to let u=x+1, and then dU=dX.*0190

*You always have to substitute for the dX as well, and so our integral converts into the integral of ux*^{2}+4 dU.0200

*Now, that matches this formula from the integration table.*0213

*The a*^{2}=4, a=2 and the u=x+10220

*Now we can just read off the answer from the integration table.*0230

*Our answer is u/2, that is (x+1)/2 × the sqrt(a*^{2}+u^{2}).0232

*Well a*^{2}+u^{2} was u^{2}+4 which came from x^{2}+2x+5.0244

*+a*^{2}/2, that is 4/2 = 2, times ln(u), u is x + 1, + that square root again, a^{2}+u^{2}, came from u^{2}/4 came from x^{2}+2x+5.0255

*Then we finish it off with a constant.*0280

*The key step there was looking at the integral, identifying which integral formula from the book was going to be useful,*0295

*Or which integral formula from the web if you are looking at the web,*0305

*But then matching the integral we are given to the pattern in the formula in the book.*0309

*The integral we were given does not exactly match, so we had to do this step of completing the square to make it match the formula exactly and then we could invoke the answer.*0315

*So, let us try another example of that. *0327

*We have here arcsin3x dX. Remember, inverse sin is the same as arcsin, not sine*^{-1}.0330

*If we have a look at the table of integrals, and there is a whole section on inverse trigonometric formulas. *0344

*Formula number 87 in Stewart's book, it might be a different number in your book, is the integral of arcsinu dU = u × arcsindU+sqrt(1-u*^{2})+C.0361

*Again, our interval does not exactly match the formula from Stewart's book because we have a 3x and Stewart has a u. *0386

*So what we are going to do, is let that u=3x, but in order to make that substitution we also have to convert the dX. *0401

*We have dU=3dX, so dX=1/3dU*0410

*Thus, our integral becomes, I will pull the 1/3 outside, 1/3 × integral of arcsin(3x) and the dX=1/3, we already put the 1/3 outside so now we can say dU.*0420

*Now, it matches the formula from the book perfectly. *0437

*In order to make that work we had to go through the substitution with the 3 and the 1/3, so that 1/3 outside is very important.*0441

*Now that it matches, we can just invoke the formula from the book, so u=3x, arcsin(3x)+sqrt(1-u*^{2}) is 1-9x^{2} + C.0454

*I should have had parentheses around the whole thing there. *0482

*We can bring the 1/3 in there and get your final answer.*0487

*x × arcsin(3x)+1/3sqrt(1-9x*^{2})+C.0493

*Again, the key step there is identifying which formula you want to use and then making the integral match that formula exactly.*0515

*Making the integral match that formula exactly, which sometimes involves some substitution along the way. *0521

*Let us try a slightly tricker one.*0534

*This is a trigonometric integral, tan*^{2}x × secx dX.0536

*I am going to look at the section in the book under trigonometric forms, and I do not see this integral in there exactly. *0539

*But what I do see is an integral, this is formula 71, that says the integral of sec*^{3}u dU = 1/2 secu × tanu + 1/2 ln(secu + tanu) + C.0552

*So, I do not have exactly the integral of sec*^{3}u, but I know that sec^{2}, I know that I can write that sec^{2}x-1. 0585

*I can write this integral as the integral of sec*^{2}(x-1) × secx dX, which in turn is the integral of sec^{3}x dX - the integral of sec(x) dx).0603

*Now, I do have sec*^{3}x dX, which is the same as sec^{3}u dU, except that you are just substituting x for u. 0629

*I can write this as 1/2 secx × tanx + 1/2 ln(secx + tanx).*0640

*I will add the constant at the end of this when I am all done.*0657

*In the mean time, I still have to cope with this integral of sec(x), *0661

*But we learned in the lecture on trigonometric integrals that we can solve that one by multiplying top and bottom by tan(x) × sec(x),*0664

*The point of that was that it solved the whole integral into the ln(secx+tanx). *0674

*This actually simplifies a little bit, 1/2 secx × tanx, now we have 1/2 ln of all that minus ln of all of that.*0699

*We can combine those into 1-ln(secx+tanx).*0712

*Now I will add on the constant because we are done with the whole integral.*0721

*Just to reiterate here, you might not be able to invoke the formulas from integral tables immediately.*0733

*Sometimes you have to take the integral you are given and do a little bit of manipulation on it the way we did with the tan*^{2}, 0740

*And convert it into some form where you can use an integral table.*0745

*You do that, you may get extra terms, like we had the extra term of the integral(secx), *0752

*You still had to remember how to integrate that from what we learned in the trigonometric integral lecture.*0758

*We are going to do a couple more examples here.*0000

*I want to get some more practice with the midpoint rule.*0003

*We are going to estimate the integral from 1 to 2 of x × ln(x).*0006

*Now, here b = 2, a = 1.*0011

*So, δx, which is b-a/n, is 2-1/4, so that is 1/4.*0020

*The midpoint rule says the integral is approximately equal to 1/4 × f(the midpoint of these 4 intervals).*0030

*So if we take the interval from 1 to 2 and split it into 4 pieces, that is 1 and 1/4, 1 and 1/2, 1 and 3/4.*0042

*Now we want the midpoints of those 4 integrals.*0050

*The midpoint of the first one is 1 and 1/8.*0056

*The midpoint of the second one between 1 and 1/4 and 1 and 1/2 is 1 and 3/8.*0064

*The midpoint of the third one is 1 and 5/8.*0072

*The last one is 1 and 7/8.*0076

*That is 1/4.*0081

*Now the function here, we get from the integral. *0084

*That is x ln(x), so that is 1 and 1/8, we want to plug that in there.*0088

*1 and 1/8 is 1.125 × ln(1.125).*0095

*We plug each one of these values in there and I will not write them all down, but the last one here is 1 and 7/8.*0102

*Plug that in, and we get 1.875 × ln(1.875).*0110

*That is an expression now that you could plug into your calculator.*0123

*You just take these numbers and these expressions and plug them all into your calculator.*0128

*What you come up with is .634493.*0135

*So, that tells us that our integral is approximately equal to .634493.*0143

*Notice there that we never actually solve the integral as we would have using some of our earlier techniques in Calculus 2.*0154

*We just picked different points and plugged them into the function and got an approximation of the area.*0160

*So the next example I would like to do is using the right endpoint rule.*0000

*We are going to find the integral of sin(x) from 1 to 2.*0007

*It says to use n = 4, so we will split that into 4 pieces.*0013

*δx is b-a/n, so that is 2/1/4, which is 1/4.*0019

*Then, the right endpoint rule says you take the width of these rectangles, which is 1/4, that is the δx...*0028

*... times, now you plug in for the height, you plug in the right endpoint.*0037

*So we are not going to look at the left endpoint of the first interval.*0040

*We are going to look at the right endpoint of those four intervals.*0047

*So we look at f(1 and 1/4) + f(1 and 1/2) + f(1 and 3/4) +f(2).*0050

*f(2) represents the right endpoint of the last interval there.*0065

*Now the function f here is sin(x), so we will be doing sin(1.25) + sin(1.5) + sin(1.75) + sin(2).*0073

*Again, this is something you can plug into your calculator.*0093

*When I plugged it into my calculator and simplified it down,*0100

*I got .959941 as an answer.*0103

*So, that is the approximate area under sin(x) using the right endpoint rule without actually doing any actual integration.*0112

*That is the end of the lecture on approximate integration.*0126

*We covered the trapezoid rule, the midpoint rule, and the left and right endpoint rules.*0128

*Those were all brought to you by educator.com.*0138

*This is educator.com, and today we are going to discuss three methods of integration approximation, the trapezoidal rule, the midpoint rule, and the left and right end-point rules.*0000

*The idea here is that you are trying to approximate the area under a curve*0014

*The function here is f(x) and we are trying to approximate that from x=a to x=b and we are trying to find that area.*0027

*What you have done so far in your calculus class, is you just take the integral of f and then you plug in the endpoints. *0044

*The point is that there are a lot of functions that you will not be able to take the integral of directly.*0050

*So, what we are going to try to do is find approximation techniques that do not rely on us being able to take the integral.*0055

*The idea for all of these techniques is that you start out by dividing the region between a to b into n equal partitions.*0062

*Then we are going to look at the area on each one of those.*0082

*That is the first part of the formula here.*0086

*Each one of these partitions is δx y, and δx comes from b - a, that is the total width, divided by n because there are n of these segments.*0093

*Now on the trapezoidal rule, what we are going to do is label each one of these points on the x axis.*0108

*This is x*_{0}, x_{1}, x_{2} all the way up to x_{n} is b.0115

*That is where the next part of this formula comes from.*0124

*x*_{0} is a, x_{1} is a + δX because it is a and then you go over δx, x_{2} is a + 2 δX all the way up to x_{n} is a + n δx.0127

*But, of course a + n δx is a + b - a/n. *0147

*So, that is a + b -a which is b, so xn is the same as b*0156

*We have labelled these points on the x axis, and what we are trying to do is approximate the area.*0162

*What we do to approximate the area is we are going to use several different rules.*0172

*The first one is called the trapezoidal rule.*0179

*The trapezoidal rule means that you draw little trapezoids on each of these segments, and then you find the area of these trapezoids.*0183

*The area of a trapezoid is = 1/2(base*_{1} × base_{2} × height).0200

*That area of the trapezoid is reflected in this formula right here.*0218

*The 1/2 gives you that 2 right there, the height of a trapezoid, that is the height, and that is the width of one of those trapezoids, and that is δX*0225

*Then you have base*_{1} + base_{2} is, I am going to show this in red, base_{1} + base_{2}, that is for the first trapezoid.0243

*For the next trapezoid, base*_{1} + base_{2}, and so on, up to the last trapezoid, base_{1} + base_{2}.0260

*What you are doing is you are plugging each of these x*_{0}, x_{1}, x_{2} into f to get these heights 0279

*But you only have 1 of the end one and 2 of each of the middle ones.*0287

*That is why you get 1 here, and two of each of the middle ones, and one of the end one. *0297

*So, that is where the formula for the trapezoidal rule comes from.*0305

*Let us try it out on an example.*0308

*Example 1 is we are going to try to estimate the integral from 1 to 2 of sin(x) dX.*0313

*Here is 1, and here is 2, and we are going to try to estimate that using n = 4.*0322

*That means we are going to divide the region from 1 to 2 into 4 pieces.*0330

*Using the formula for the trapezoid rule, we have δx/2.*0336

*Well δx is (b - a)/n, so that is 1/4.*0341

*δx/2 is 1/8, so we are going to have 1/8 times f(x*_{0}) + 2f(x_{1}) + 2f(x_{2}) + 2f(x_{3}) + only 1 of f(x_{4}). 0352

*These x*_{0}'s are the division points in between 1 and 2, so this x_{0} is 1, so that is sin(1) + 2, now that is 1, x_{1} × δx is 1/4.0385

*x*_{2} can go over another unit of δx, so that is 1 and 1/2, 1 and 3/4 and finally, sin(2).0410

*We are going to take all of that, and multiply it by 1/8.*0430

*At this point it is simply a matter of plugging all of these values into a calculator.*0433

*I have a TI calculator here, and I am going to plug in 1/8 sin(1) is 0.01745 + 2sin(1.25), which is 0.0281, and so on.*0437

*You can plug the rest of the values into a calculator.*0468

*What you get at the end simplifies down to 0.951462.*0476

*So, we say that the integral from 1 to 2 of sin(x) dX is approximately = 0.951462*0486

*Next rule we are going to learn is the midpoint rule.*0503

*It is the same idea, where you have a function that you want to integrate from a to b, and you break the region up into partitions.*0508

*So, you have x*_{0} = a, x_{n} = b and a bunch of partitions in between, each partition is δx y.0522

*Except, in each partition, instead of building trapezoids, we are going to build rectangles.*0533

*We are going to build rectangles on the height of the middle of the partition. *0542

*Here, we are going to look at the middle of the partition, *0548

*We see how tall the function is at the middle of the partition and we build a rectangle that is that height.*0555

*We do that on every rectangle.*0564

*The formula we get in total is δx, that is the width of the rectangles, times the height of these rectangles, *0575

*I have labelled the midpoints of those rectangles x*_{1}* and x_{2}* and x_{n}*.0596

*Those represent these midpoints, so that is x*_{1}*, there is x_{2}*, and so on.0616

*Those are the midpoints so x*_{1}* is just x_{0} + x_{1}/2, x_{2}* is just x_{2} + x_{2}/2, 0622

*And so on and those just represent the midpoints of each of these intervals.*0632

*We plug those midpoints in to find the heights of the rectangles and estimate the area.*0640

*What we are going to do for the last rule today, is we are going to use instead of the midpoints, we will use the left endpoints of each rectangle.*0650

*Instead of having to find the midpoints, the x*_{1}× and the x_{2}× and so on will be the left endpoint of each interval.0663

*We will use those to get the heights.*0677

*We will see those in the second. *0680

*First we will do an example with the midpoint rule using the same integral as before.*0682

*Sin(x), there is 1, there is 2, again we are using n = 4 so we will break it up into 4 partitions. *0686

*Except we are going to use a slightly different formula to solve it.*0698

*Again, δx = 1/4, because that is the width of each of these rectangles, but now we are going to look at the midpoints of those 4 rectangles to find the heights.*0700

*The midpoints are, well, this is 1 right here, that is 1 and 1/4, the midpoint there is 1 and 1/8.*0715

*The next midpoint is halfway between 1 and 1/4 and 1 and 1/2 and that is 1 and 3/8.*0731

*The formula that we get is δX(f(1 + 1/8) + f(1 + 3/8) + f(1 + 5/8) + f(1 + 7/8) and that is just 1/4.*0740

*Now the function here is sin(x) so we will be doing sin(1 + 1/8), that is 1.125, + sin(1 + 3/8), that is 1.375, sin(1 + 5/8) is 1.625, and sin(1 + 7/8) is 1.875.*0771

*Now it is just an expression that we can plug into our calculator.*0800

*I worked this out ahead of time, I got 0.958944*0805

*That is our best approximation for the integral using the midpoint rule.*0827

*The next formula we want to learn is the right and left endpoint rule.*0835

*We will talk about the left endpoint rule first.*0840

*It is pretty much the same as the midpoint rule.*0845

*Again, you are drawing these rectangles except instead of using the midpoint to find the height of the rectangle, you are using the left endpoint.*0849

*So that means you start out with the exact same formula, *0862

*Except that the star points that you choose to plug in to find the heights are just the left endpoints, x*_{0}, x_{1}, up to x_{n-1}.0865

*You do not go up to x*_{n}.0876

*For the right endpoint rule it is the same formula except you use the right endpoints.*0879

*The right endpoint would be x*_{1}, x_{2}, up to x_{n}.0885

*You do not have x*_{0} anymore because that is the first endpoint of the left formula.0890

*Let me draw these in different colors here.*0896

*The left endpoints give you x*_{0}, x_{1}, up to x_{n-1}, so that is the left endpoint rule.0902

*The right endpoint, I will do that one in blue, is x*_{1}, x_{2}, up to x_{n-1}, and x_{n}.0912

*You are using the right endpoints so that is the right endpoint rule. *0928

*We will do another example.*0932

*Again we are going to figure out the integral, or estimate the integral of sin(x) from 1 to 2.*0934

*We are going to use the left endpoint rule so that means the key points that we plug in for the heights are 1, 1 + 1/4, 1 + 1/2, and 1 + 3/4.*0947

*Again δx = 1/4 and our formula says δx × f(left endpoint).*0964

*So, f(1) + f(1 + 1/4) + f(1 + 1/2) + f(1 + 3/4).*0975

*We do not go to 2 because that was the right endpoint of the left interval.*0990

*The integral is approximately equal to that.*1000

*That is sin(1) + sin(1 + 1/4) + sin(1 + 1/2) + sin(1 + 3/4).*1010

*That is something that you can plug into your calculator, and when I did that ahead of time I got 0.942984.*1022

*That is our estimation of that integral using the left endpoint rule.*1038

*We are going to work out a couple more examples on Simpson's Rule.*0000

*We are going to start with the example of the integral from 1 to 2 of x log(x) using Simpson's Rule with n = 4.*0004

*We are integrating from 1 to 2, and we are uisng n=4.*0014

*Remember, for Simpson's Rule, n always has to be even.*0022

*So we break the interval from 1 to 2 into 4 equal pieces.*0026

*So, that means that our breakpoints are 1, 5/4, 3/2, 7/4, and 2.*0030

*Then, that means that δx which is the width of the intervals, is 1/4.*0040

*We are going to invoke Simpson's Rule formula.*0047

*It starts out with x/3, so that is 1/12.*0050

*Then you plug these points into the function you are integrating, in this case x (log(x)).*0059

*Then you attach these coefficients, remember that funny pattern, 1, 4, 2, 4, 2, 4, as many times as you need until the last is 4 and then 1.*0066

*The first point is 1.*0083

*So, 1 × ln(1) + 4 × 5/4 ln(5/4) + 2 × 3/2 ln(3/2) + 4 × 7/4 ln(7/4) + 1 × 2 × ln(2).*0086

*This can be a little simplified, because ln(1) is 0.*0124

*The other values are not common values we know easily.*0139

*This would be something you would plug into a calculator.*0144

*I have already worked it out on my own calculator.*0148

*Here is what you get, 0.6723348708*0150

*To reiterate there, we are estimating an integral as the area under a curve using the Simpson's Rule formula.*0160

*So, that is the answer.*0176

*In our last example today we are going to compare the accuracy of using the midpoint rule and Simpson's Rule,*0000

*Both with n = 4, on the integral from 1 to 2 of x ln(x) dx.*0006

*We just worked out what the answer was using Simpson's Rule.*0015

*The Simpson's Rule gave us the answer 0.6723348708.*0023

*Th midpoint rule is something we did earlier in another lecture.*0037

*I am going to remember what the answer was there.*0046

*That was done as an example in the previous lecture on the trapezoidal rule, the midpoint rule, and the left and right endpoint rules.*0051

*The answer of this integral using n=4 was 0.6344928081.*0056

*Those are the two answers we get using the different estimation rules.*0071

*You can also do this integral directly, the integral from 1 to 2 of x ln(x) dx.*0079

*I am not going to show the details of this but I will tell you how you can figure it out.*0091

*You can use integration by parts,*0095

*And the way you would break it down using integration by parts,*0100

*If you remember the LIATE rule, that was the order in which you look at the functions to use integration by parts.*0106

*LIATE stands for ln, inverse functions, algebraic functions, trigonometric functions, exponential functions.*0110

*Very first on that list is natural log.*0120

*Remember LIATE means you try to make whatever function comes first be the u.*0124

*For integration by parts on this, we see a ln(x), so a good choice is to make u be ln(x), and dv be x dx.*0130

*Then you can work out this integral using integration by parts.*0144

*It takes a couple steps so I will not show the details.*0147

*What you get as the final answer is -x*^{2}/4 + x^{2}/2 × ln(x).0151

*All of that evaluated from x = 1 to x = 2.*0164

*Then if you plug in those values, you get -3/4 + ln(4).*0170

*If you plug that into a calculator, you get 0.6362943611.*0182

*That is as close as we can come to the true value of the integral.*0196

*Let us see how accurate our two approximation techniques were.*0202

*Let us start with our midpoint approximation.*0205

*If we look at the midpoint - the true value of the integral, remember this was the true value over here.*0210

*If we do the midpoint approximation - the true value, we get - 0.001801553.*0221

*That looks very accurate, because our error here, if we subtract the true value from the midpoint rule approximation.*0235

*Our error is just 0.001.*0245

*Let us look at Simpson's Rule though.*0252

*If we do the Simpson's Rule approximation minus the true value, we get 0.000015469.*0257

*That error is really tiny, very small error using Simpson's Rule.*0272

*In fact, when we calculated the integral using Simpson's Rule, we did not have to do anymore work than the midpoint rule.*0283

*We still were just plugging in those values at 1, 5/4, 3/2, 7/4 and 2.*0292

*So, it was about the same amount of work, but Simpson's Rule is much more accurate.*0300

*Using about the same amount of work.*0310

*So, the moral of the story there is that we have various approximation techniques.*0325

*The trapezoid rule, the midpoint rule, the left and right endpoint rules, and the Simpson's Rule.*0337

*However, by far the most powerful of all of these is the Simpson's Rule.*0340

*It gives us extremely accurate results.*0346

*So that is the end of this lecture.*0350

*Today we are going to talk about Simpson's Rule.*0000

*Simpson's Rule is a way of estimating the value of an integral when you cannot solve it by traditional integration techniques.*0005

*We already learned a few of those, the trapezoidal rule, the midpoint rule, and the left and right endpoint rules.*0011

*Simpson's Rule is based on the same kinds of ideas but it is a little more sophisticated.*0017

*We will see later on that the answers it gives are a lot more accurate.*0023

*The idea here is that we are trying to estimate the area under a curve, y = f(x).*0029

*Just as we did with the previous rules, what we are going to do is divide the input into partitions.*0039

*An important difference between Simpson's Rule and the previous ones is that we have to use an even number of partitions.*0050

*Just as before, we have an x*_{0}, x_{1}, x_{2}, up to x_{n}, but remember n has to be even now.0058

*The principle of Simpson's Rule is that you look at those partitions two at a time.*0072

*Let me draw an expanded version of that.*0073

*So you look at two partitions here, and you have got the function above there.*0078

*What you do is look at the three values on the boundaries of those partitions and you draw a parabola through it, through those three values.*0084

*Then you find the area under that parabola and you add up those areas.*0109

*So, you find this area and you add all those up and that gives you your approximation of the value of the integral.*0116

*Finding the formulas to calculate that area would be a little bit tedious if you had to do it every time, so I will just give you the answer.*0124

*The answer turns out to be, for the area under one parabola, δx/3, remember δx is the width of one of these three partitions,*0133

*Times the value of f(x) at these three points multiplied by certain coefficients.*0145

*The coefficients are 1, 4, 1, which is kind of strange but that is just the way the math works out.*0153

*And then you go through, and you do this over each set of partitions, over the entire area.*0160

*What you end up doing is you end up adding 1, 4, 1, and then 1, 4, 1, for the next one but that overlaps here *0168

*Because remember the last endpoint of one set of partitions is the first endpoint of the next set.*0179

*And then 1, 4, 1 and so on for however many double partitions you have.*0186

*Then you add up those coefficients and you got 1, 4, 2, 4, 2, 4, 1.*0193

*You get this funny pattern 1, 4, 2, 4, and that keeps alternating until the final coefficient is 1.*0202

*So that is where we get the final formula for Simpson's Rule. *0210

*This δx/3 and then the coefficients 1, 4, 2, 4, 2, 4, 1, for however many partitions you have.*0215

*Let us try that out with some examples and you will see how it works.*0225

*The first example we are going to do is the integral of sin(x) as x goes from 1 to 2.*0229

*That is an integral that you could do without using Simpson's Rule.*0236

*Certainly we know what the integral of sin(x) is, but it is an easy function to practice using Simpson's Rule on.*0239

*What we are going to do is look at the interval from 1 to 2, and it says to use n = 4, so we are going to divide that into 4 sub-intervals.*0248

*That means that δx is (b - a)/4, so δx is 1/4*0265

*Our points that we are going to check are x*_{0} = 1, x_{1} = 5/4 because we are going over by 1/4 each time, x_{2} = 3/2, x_{3} = 7/4, and x_{4} = 2.0276

*The Simpson's Rule formula, so δx/3, so that is 1/12, times f(left endpoint), *0302

*So that is sin(1) + 4sin(5/4) + 2sin(3/2) + 4sin(7/4) and remember that the last coefficient is 1, so we get + sin(2).*0310

*That is our approximation for the area.*0340

*Now it is just a matter of plugging those values into a calculator.*0347

*Take your calculator and plug these values in and I have already actually worked it out on my calculator, and it turned out to be 0.9564700541.*0352

*That is our guess for this area using Simpson's Rule*0383

*So let us try that out on a slightly more complicated example.*0391

*It is the same integral, but now we are going to use n = 8, so we have a much finer partition there.*0393

*Now δx is 1/8 and our partition points are x*_{0} = 1, x_{1} = 9/8, x_{2} = 5/4, all the way up to x_{8} = 2.0406

*So, our Simpson's rule formula says we do δx/3, *0434

*So, that is 1/8/3, so 1/24 × (sin(1) + 4sin(9/8) + 2sin(5/4) + 4sin(11/8) + 2sin(3/2) + 4sin(13/8) + 2sin(7/4) + 4sin(15/8) + 1sin(2)).*0436

*Again, that is a complicated expression but it is easy enough if you just plug it into your calculator.*0511

*I already have worked it out and what I got was 0.9564504421.*0519

*That is what we get using Simpson's Rule, using more partitions, n=8 instead of n=4.*0536

*Remember n has to be even every time in order for this formula to work.*0543

*Our next example is to look back at the two previous examples and compare the accuracy of using n = 4 and n = 8.*0548

*Now, in order to do this, to say how accurate they are, we really need to know the true value of this integral.*0558

*This is an easy integral to solve using calculus 1 techniques.*0564

*The true value is the integral from 1 to 2 of sin(x) dx, which is, well the integral of sin is -cos.*0569

*Evaluated from x = 1 to x = 2, so -cos(2) + cos(1)*0587

*Again if you plug that into a calculator, you will get 0.9564491424.*0600

*That is the true value of the integral, but we are interested in figuring out how accurate our approximations using Simpson's Rules were.*0617

*Let us look at the n = 4 values, remember that was 0.9564700541, and if we subtract the true value from that, what you get is 0.000020912.*0624

*That is extremely accurate.*0661

*We get 4 decimal places of accuracy using n = 4.*0665

*But let us try n = 8.*0671

*The value we had was 0.9564504421, and if we subtract the true value from that, what you end up with is 0.000001300, *0674

*Which is much smaller than what we got with n = 4.*0706

*The point of that is that using n = 8 gave us a much more accurate result than using n = 4 did.*0715

*The conclusion there is that n = 8 is much more accurate.*0723

*OK, that finishes this section on Simpson's Rule.*0739

*In this case, we are given the integral from 0 to infinity of 8 e*^{-x}.0000

*If you graph that, e*^{-x} has the basic shape of e^{x}.0006

*Except it is flipped around the y-axis.*0015

*The 8 multiplies it up and makes it a little bigger but it is not going to change the same basic shape.*0021

*What we are really trying to do is find that area and determine whether it is finite or infinite.*0025

*The problem actually asks a little more.*0031

*If it diverges we have to determine whether it is positive or negative infinity or neither.*0035

*If it converges, we have to find out exactly what that area is.*0039

*Let us set it up as a limit.*0044

*We take the integral from 0 to t of 8 e*^{-x}.0047

*That is a pretty easy integral to do, if you make the substitution u = -x,*0061

*You get -8e*^{-x} from x=0 to x=t.0065

*If you plug that in, that is -8e*^{-t} - 8e^{0}.0073

*Sorry, subtracting a negative sign, so + 8e*^{0}.0088

*Then we take the limit as t goes to infinity of this.*0093

*So this is minus 8e*^{-infinity} + 8e^{0} = 1.0101

*Well e*^{-infinity}, this is 1/e^{infinity}, well actually I am not looking at the -8 part, just the e part which goes to 0.0111

*And our answer for the whole thing there is 8.*0128

*So we say that this converges to 8.*0141

*So just to recap there, what we did was we handled the impropriety of letting x go to infinity by replacing infinity with t.*0148

*We worked through the integral and then took the limit as t goes to infinity.*0161

*We got the answer 8, and what that represents is that this area,*0166

*Even though it goes on infinitely far, the total area is only 8 square units.*0170

*So our last example is the integral from 2 to infinity of dx/x+sin(x).*0000

*Again, because we have an impropriety at infinity, we are going to integrate from 2 to t of dx/x+sin(x).*0007

*We would like just to do the integral as the next step.*0021

*The problem is that the integral of 1/x + sin(x) is something very difficult, something we do not know how to do.*0025

*How do we handle this?*0033

*The way you want to think about this is kind of ask yourself,*0035

*What does 1/x+sin(x) look like?*0037

*Now I am not asking for a detailed graph, but asking you to think about how big 1/x+sin(x) can be.*0045

*Well, x+sin(x), something I know here is that sin(x) is always between -1 and 1.*0056

*So x + sin(x) is always less than or equal to x+1.*0068

*So 1/x+sin(x), when you take 1 over something, it reverses the inequality.*0075

*That is bigger than 1/x+1.*0082

*So this integral, whatever it is, is bigger than the integral from 2 to t of dx/x+1.*0087

*That is an easy integral to do.*0096

*We can u = x+1.*0099

*And we get the ln(x+1) evaluated from x=2 to x=t.*0100

*That is the ln(t+1) - ln(2+1).*0115

*We want the limit as t goes to infinity, so this is the ln(infinity), which is infinity - ln(3),*0126

*Which is just a finite number.*0144

*So this diverges to infinity.*0148

*However, that was the integral of dx/x+1.*0152

*We are actually trying to look at the integral of dx/x+sin(x).*0158

*What we have learned though is that the integral from 2 to infinity of dx/x+sin(x),*0161

*Is bigger than or equal to the integral from 2 to infinity of dx/x+1.*0174

*That in turn is infinity, so the area under this function is infinite, this is even bigger.*0182

*So, our integral if it is even bigger than our infinite value, it diverges to infinity too.*0197

*There are a couple points I want to make about this.*0216

*One is that in order for this kind of argument to work, the inequality has to go the right way.*0219

*If this inequality had turned out to go the other way,*0225

*If this had been a greater than or equal to,*0230

*I will write this in red so that you can see this is not the actual value.*0235

*So what if that had been greater than or equal to?*0240

*That would have made this inequality less than or equal to. *0243

*Then in turn this inequality would have been less than or equal to, and this inequality would have been less than or equal to.*0247

*Then we would have something less than or equal to an infinite area.*0253

*That is not something that we can say is finite or infinite.*0258

*If something is less than infinity, well it might be finite, or it could still be infinite.*0264

*So what is very important for this kind of argument to work, is that the inequality must go the correct way.*0267

*The second point I wanted to make about this is that when you use this kind of argument,*0288

*You are not figuring out the exact value of the integral.*0301

*This kind of argument will never tell you the exact value of the integral,*0305

*Even when you get an integral that converges.*0307

*This never tells you the exact value of the integral.*0311

*Now, in this problem, we got kind of lucky because we were not asked what the exact value of the integral was.*0329

*We were only asked to determine whether the integral converges or diverges.*0336

*In that sense, that is sometimes a tip-off that you want to use this comparison idea.*0342

*If the problem does not ask you what the exact value is,*0347

*If it just asks you whether it converges or diverges,*0353

*That is a little hint that you might want to use this idea of comparison.*0359

*Today we are going to talk about improper integration.*0000

*The idea of improper integration is the function that you are trying to integrate has either a horizontal or a vertical asymptote.*0005

*For example, if you have a horizontal asymptote, that means something like this, *0016

*Where the curve goes on forever and gets closer and closer to a horizontal line.*0020

*What you might be trying to do is find the area under the curve but this is an area that goes on forever.*0027

*The formal notation for this is to take the integral from x=a and we say the integral from a to infinity of f(x) dX.*0038

*The way you really do that since we do not really know what it means to plug in infinity for a function is we cut off this integral at a certain point which we call T.*0048

*Then we just find the integral from x=a to x=t.*0065

*So, we are finding the area under the curve from x=a to x=t, and then we let t get bigger and bigger to go off to infinity.*0070

*We take the limit as t goes to infinity.*0081

*What can happen is it can turn out that that can approach a finite limit or it can diverge to infinity.*0085

*The same way that the limits that you learned about in Calculus 1 can approach a finite limit or they can diverge to infinity, or negative infinity.*0094

*When we do some examples here, we will be asking whether these integrals converge to a finite limit or if so what the limit is,*0104

*Or whether they diverge to either negative infinity, or positive infinity, or neither.*0112

*The other kind of integral we will be looking at is a vertical asymptote.*0117

*That means that the function you are trying to integrate maybe goes up to infinity, in other words it approaches a vertical line.*0129

*If it has a vertical asymptote at b, we might want to try to find the total area under that curve between a and b.*0134

*We try to find that area but the reason the function probably goes up to infinity at b, *0148

*Is because we are dividing by 0 at b and it is not legitimate to just plug in a value at b for the function.*0155

*We cut off the function a little bit short of the vertical asymptote, so if there is b, and there is a, *0168

*What we do is we cut off the function a little bit short.*0179

*We call that value t.*0183

*We cut it somewhere short of the vertical asymptote where the function is still defined and it does not blow up to infinity and we find the area between a and t.*0185

*Then what we do is we gradually move t closer and closer to b, *0196

*That this area approaches the area under the entire curve that we are looking at.*0200

*Again, this area can turn out to be finite or infinite, *0213

*So, we will talk about the integral converging to a finite number or diverging to infinity or negative infinity, or neither.*0217

*Again, we handle it with limits, we take the limit as t goes to b.*0226

*Since t is coming towards b from the left hand side, *0230

*That is why we put a little negative sign down there to show that t is coming to b from the negative side.*0234

*We can also talk about functions that approach vertical asymptotes from the right hand side.*0242

*We can ask ourselves what is the area under that function between a and b where a is the vertical asymptote.*0254

*Then what we would say is that the integral from a to b of f(x) dX,*0261

*We would look at the integral and now we would have t be just a little bit to the right of a, *0270

*And t would approach a from the right, and we would integrate from t to b of f(x) dX.*0277

*Then we would take the limit as T gets closer and closer to a, but this time t is on the right-hand side of a.*0285

*I am going to put a little plus sign there to show that t is approaching a from the positive side.*0294

*Let us try some examples and see how it works out.*0299

*Our first example is to look at the integral from 0 to infinity of 1/(x+1).*0304

*The key thing here is that if you graph 1/(x+1), well, at 0 it is 1, and as x approaches infinity it goes down to 0.*0310

*We are trying to find that area there and see whether that area is finite or infinite.*0325

*The way we set this up is we take the integral from 0 to t, of 1/(x+1) dX.*0331

*We will go ahead and do that integral and then we will take the limit as t goes to infinity.*0341

*We are cutting this integral off at some value of t that will then let the t get bigger and bigger.*0348

*The integral of 1/(x+1) is a pretty easy integral if you let u = x+1 and put the substitution through there, then it turns into ln(x+1).*0356

*Evaluated from x=0 to x=t, and so we get ln(t+1) - ln(0+1), so ln(1) but that goes away to 0.*0367

*We get ln(t+1) and we want to take the limit of that as t goes to infinity.*0390

*If you remember, ln(x), y = ln(x).*0401

*As you plug in bigger and bigger numbers, ln(x) goes to infinity.*0411

*If we plug in bigger and bigger values for t, this limit goes to infinity.*0417

*What we say about this integral is that it diverges to positive infinity.*0424

*This represents the fact that there is actually infinite area under that curve.*0435

*Again, the way you handle these improper integrals is you convert the infinity into a t and then you do the integration*0444

*Then you take the limit as t goes to infinity and you see whether it turns out to be a finite or an infinite area.*0453

*Let us try another example.*0463

*Again, the same function actually, 1/(x+1).*0464

*But, this time we are taking the integral from -1 to 0 so the point here is that if you plug in x=-1 to this function, *0471

*It blows up because you get 0 in the denominator and because this function has a vertical asymptote at x=-1.*0488

*We are trying to figure out whether the area under that function as it approaches that vertical asymptote is finite or infinite.*0500

*Again, we handle this with limits.*0509

*We will take a value t that is just to the right of -1, *0512

*And we will integrate from t to 0, instead of from -1 to 0 because you cannot plug in -1 to that function.*0515

*We will look at dX/(x+1) and then we will take the limit as t gets closer and closer to -1.*0524

*I will put a little positive sign to indicate that t is approaching -1 from the positive side.*0535

*Again, we will take the integral and we get ln(abs(x+1)) evaluated from x=t up to x=0.*0543

*That is ln, if you plug in x=0, you get ln(1) - ln(t+1), so ln(1) goes to 0.*0555

*This leaves us with -ln(t+1).*0573

*Now, let us see what happens when we take values of t that are closer and closer to -1.*0581

*Well, if t goes to -1, that means that t+1 is going to 0, so what we are trying to do is take the ln, I should keep my absolute values here *0589

*Take the ln of something that is going to 0 from the positive side.*0610

*If you remember what the graph of ln(x) looks like, when you go to 0 from the positive side, ln(x) goes down to -infinity.*0615

*This is minus -infinity so we say the whole integral diverges to infinity.*0631

*Which represents the fact that the area under that curve is infinite.*0649

*Here we had a vertical asymptote and again the way to handle the vertical asymptote was to replace the problem value of x.*0659

*The problem value of x was -1, with t.*0668

*Go ahead and work through the integral and then take the limit as t approaches -1, *0672

*And I should have put a little plus in here to show that it is approaching -1 from the positive side.*0679

*We take that limit and we see whether it comes out to be a finite or infinite area.*0685

*Our next example is a little different.*0694

*We are going to look at cos(x).*0697

*We are going to look at the integral from 0 to infinity of cos(x).*0699

*For this one, just like the others, it is very helpful to draw a graph, so I will draw a graph of cos(x).*0703

*It starts at 1 and then it goes down to 0, -1, up to 1, down to -1, and so on.*0714

*So that is y=cos(x).*0722

*Now what we are really doing here is we are finding the area under the curve of cos(x), *0728

*Where we count area above the x axis as positive and below the x axis as negative.*0735

*We are really trying to add up all of those areas where this counts positive and this counts negative, positive, negative, and so on.*0743

*You can see easily just by looking at the graph here that there is going to be infinite positive area and also infinite negative area.*0758

*What we are really trying to do with this integral is add up infinitely many positive things and infinitely many negative things.*0784

*So, a positive infinite and a negative infinity.*0792

*A common question at this point is to ask whether the positive infinity and negative infinities cancel each other.*0797

*The answer is that there is no mathematically meaningful way to give rules for positive and negative infinities to cancel each other.*0807

*What you do in a situation like this is you say that the integral diverges but not to infinity or negative infinity.*0815

*You might be tempted to say that all of the positives and the negative infinities cancel each other and the whole integral is equal to 0.*0844

*Or you might be tempted to say that, well, these two areas cancel each other and this area cancels with the next one, so all of these areas cancel*0852

*Then you are just left with this area and you can find the value of that area.*0862

*There are all sorts of ways that you might think you could resolve this.*0868

*Because those all contradict each other, we do not sort of put the stamp of legitimacy on any of those.*0873

*We just say the whole thing diverges but not to positive infinity or negative infinity.*0881

*We just say it diverges. *0886

*You could also do this one by actually calculating the integral from 0 to 1 of cos(x) dx.*0888

*Which would give you sin(x) evaluated from 0 to 1.*0896

*That in turn gives you sin(t) - sin(0), which gives you sin(t) and then you are finding your limit as t goes to infinity of sin(t).*0905

*Again, we know that since sin oscillates forever, this does not exist.*0923

*It is not positive infinity, it is not negative infinity, it is not 0, we just say the whole thing does not exist.*0927

*It diverges.*0943

*OK, let us do another example.*0950

*Again this is one that if we look at the graph it is something we can resolve quickly.*0953

*We are looking at the integral from negative infinity to infinity of x*^{2} dx.0958

*In this case we actually have improprieties, places where the intervals are improper at both ends.*0963

*If we graph this one, y = x*^{2}, it looks like that.0971

*What we are trying to do is find the area under this curve as we go to infinity in both the positive and negative directions.*0979

*Now, clearly if you look at this area, the area is infinite.*0991

*Both on the positive and the negative sides.*0998

*Here is another common mistake that Calculus 2 students make*1000

*They will look at this area over here and say wait, is that not negative area?*1004

*That is a little mistake because remember, negative area is when the area is below the x axis.*1012

*This area is not below the x axis, it is to the left of the y axis, but it is still above the x axis.*1020

*That is not negative area.*1029

*It is still positive area because it is above the x axis.*1035

*If you simply look at the graph here, we are looking at positive infinite area + another batch of positive infinite area.*1041

*If you put those together we have two positive infinities.*1064

*Remember we learned in the last example that you cannot cancel a positive and a negative infinity, but what about 2 positive infinities?*1066

*You can add those up and say that the answer will be a positive infinity.*1074

*You can say that it diverges to positive infinity.*1081

*You do not just say this one diverges, you say that it diverges to positive infinity.*1088

*Because, if you look at these two areas we definitely have positive infinite area, positive infinite area, we put them together and we have got positive infinite area.*1095

*There are a few integrals that come up very, very often when you are looking at improper integrals.*1109

*The most important ones are the integral from 0 to 1 of dX/(x*^{p}).1115

*The reason that is improper is because when x=0, you have 0 in the denominator here so that blows up that x=0, so x=0 is improper there.*1124

*The other one that comes up very often is the integral from 1 to infinity of dx/(x*^{p}), again that is improper because of the infinity there.1139

*These integrals come up so often that it is worth working them out once and then probably memorizing the answers.*1153

*Especially this one, the integral of 1 to infinity of dx/x*^{p} is worth remembering.1162

*It is worth remembering because it comes up later on in Calculus 2.*1177

*When you start looking at infinite series.*1183

*When you start looking at infinite series we are going to be using something called the integral test to determine whether infinite series converge or diverge.*1188

*The integral test says instead of looking at a series, you look at an integral.*1202

*The series that we are going to start with is 1/n*^{p}.1210

*We will convert that into the integral of 1/x*^{p} dx.1215

*Then we will remember what this integral was in order to determine whether the series is convergent or divergent.*1222

*It is worth remembering these. *1230

*Each of these integrals, you can evaluate yourself.*1233

*I am not going to work through the details there.*1236

*But you can evaluate yourself and it turns out that it depends on the values of p.*1239

*What values of p determine whether these integrals converge or diverge.*1245

*You want to remember that the first integral, if p < 1, it converges.*1250

*If p = 1, it diverges.*1258

*If p > 1, it also diverges.*1261

*The second integral, if p < 1 it diverges.*1265

*Equal to 1, it diverges, and p > 1 it converges.*1269

*That seems like a lot to remember.*1272

*You kind of remember, if p = 1, they both diverge, but if p < 1 or p > 1, they kind of flip flop back and forth there.*1276

*We will use these on the next example to help us determine whether an interval converges or diverges.*1286

*Here we have been given the integral of from 6 to 11 of 11/(x-6)x*^{1/3}.1297

*We have been asked to determine whether it converges or diverges.*1305

*The first thing to look at with this is to look at it and see if we can make a substitution to make it simpler.*1309

*We can.*1316

*Let us do u = x - 6. *1317

*Then, whenever you make a substitution, you also have to change the differential. *1320

*So, du, if u is just x - 6, that is very easy.*1329

*Let us also change the limits.*1331

*If x = 1, then u will be equal to 5 because that is 11 - 6.*1334

*If x = 6, then u = 0.*1340

*This integral converts into the integral from 0 = u to u = 5 of 11.*1347

*I am just going to write the 11 on the outside.*1356

*1/u*^{1/3} du.1360

*Now it is a little more obvious why this is an improper integral, because u = 0, you will get 0 in the denominator.*1363

*That is obviously an impropriety there.*1374

*Actually you might have been able to notice that from the original integral there.*1378

*When x = 6, you might plug in x = 6 to the denominator you get 0 in the denominator and that is clearly improper.*1380

*I am not going to worry about the 11 for the time being because it is not going to affect whether the thing converges or diverges.*1391

*What we are going to do is instead of 0, I am going to change the 0 to t.*1400

*We are going to integrate from t to 5, of 1 over, now u to the cube root = u*^{1/3} dU.1407

*Then we take the limit of that as t goes to 0 from the positive side.*1420

*Now I can split this integral up from t to 1 of 1/u*^{1/3} du.1428

*Plus the integral from 1 to 5 of 1/u*^{1/3} du.1445

*Now the point about that is that this is not an improper integral.*1455

*This completely proper and it will not determine whether the whole thing converges or diverges.*1460

*This will give us a finite number.*1469

*This is the improper part right here.*1473

*What we have here is the integral of du/u*^{1/3}.1476

*That is exactly the integral that we were looking at on the previous page.*1482

*I will just write it as 0*^{1} for now, with a p value of 1/3.1492

*On the previous page, we learned that the integral from 0 to 1 of dx over x*^{p},1498

*If p < 1, converges.*1505

*That tells us that this whole integral converges, or at least this part of it converges.*1514

*Then, the second part with some finite number,*1523

*I have not really incorporated the 11 but we can have an 11 times the whole thing.*1531

*An 11 multiplied times a whole number is not going to make it more likely to go to infinity or less likely.*1536

*So, the whole thing goes to it converges to a finite number.*1545

*In this case, the problem only asked us whether the integral converges or diverges.*1558

*At this point, we are finished.*1562

*In fact, this integral is completely feasible.*1566

*You could if you wanted, go back and evaluate the integral.*1572

*What you should get is 3/2 × 5*^{3/2}.1584

*That is what you get if you plug in the whole integral and then you multiply the whole thing by 11. *1593

*That is what you get if you evaluate the whole integral.*1600

*The important thing here is that is a finite integral, so we would say the integral converges.*1603

*There is a very common sort of ambush in Calculus 2, where they will give you a sort of innocent-seeming problem.*1613

*It turns out what the problem has is hidden discontinuities.*1622

*Let me give you an example of that.*1628

*The integral from 1 to 4 of dx/x*^{2} - 5x + 6.1630

*This is an example that is not that hard to integrate using the techniques we learned in earlier lectures.*1636

*In particular, if you use partial fractions.*1645

*If you use partial fractions on this integral, you get 1/x*^{2} - 5x + 6.1656

*Well, remember what we learned in partial fractions was to factor the denominator.*1671

*So, 1/(x-2) × (x-3).*1675

*If you do the partial fractions work, I will not spell out the details now, but we learned how to do that in a previous lecture.*1683

*You get 1/(x - 3) - 1/(x - 2).*1688

*So it is very tempting to do that partial fractions work and then integrate it.*1697

*You integrate it and you get ln(x-3) - ln(x-2).*1703

*Then you think, OK, I can plug in my bounds, x=1 and x=4.*1713

*You can plug in those numbers just fine.*1717

*And, you get a nice numerical answer.*1721

*That is the temptation in this kind of problem.*1724

*That is exactly the trap that your Calculus 2 teacher is trying to make you fall into.*1734

*You work all the way through this problem and you get a nice numerical answer and you present it as your answer.*1741

*The fact is that that is flat wrong.*1747

*What is wrong with that?*1752

*Well, what is wrong with that is that if you look at the function we had to integrate there, 1/(x-2) × (x-3).*1754

*That blows up at x=2 and x=3.*1765

*Because if you plug in x=2 or x=3, you get 0 in the denominator and the thing explodes.*1779

*So, this integral, which looked very tame and safe and looked like a fairly easy Calculus 2 integral actually is an improper integral.*1786

*Since we are talking about it in the section in improper integrals.*1795

*Maybe it is obvious to look for that, but if this comes in a swarm of other integrals where you are using all kinds of other techniques,*1799

*It is very easy to overlook things like this.*1809

*Instead of just solving it, using the generic partial fractions idea,*1812

*What you have to do is you have to split this up into pieces at the discontinuities.*1819

*So, you split it up at the integral from 1 to 2 of dx over the denominator + the integral from 2 to 3 of dx over the denominator + the integral from 3 to 4 of dx over the denominator.*1828

*You split it up at these two places, at 2 and 3, because those are the two discontinuities.*1851

*In fact, this middle integral, is now discontinuous at both ends because 1 end is 2 and 1 end is 3.*1860

*This middle integral is discontinuous in two different place.*1868

*We are going to split that up from 2 to 2.5 of dx over the denominator + the integral from 2.5 to 3 of dx over the denominator.*1872

*So, then we look at these 4 different improper integrals.*1886

*Four improper integrals.*1899

*You solve all four of them.*1907

*If any one of them diverges.*1909

*Or if more than one diverges, then we say the whole original integral from 1 to 4 diverges.*1920

*Now, each one of these 4 integrals you could work on using partial fractions.*1939

*So, using that technique that I outlined up here,*1947

*Solve each one using partial fractions.*1953

*It turns out that all 4 of those integrals diverge.*1972

*So, we say the original integral from 1 to 4 diverges, as our answer.*1987

*This is really quite dangerous because this depends on your noticing that there are these hidden discontinuities between the bounds of integration.*2008

*In order not to fall into this trap, what you have to do is look at the thing being integrated and ask yourself when does that blow up.*2023

*In this case that blows up at x=2 and x=3.*2031

*2 and 3 would be in the bounds of integration.*2035

*They are between 1 and 4 and so that is why we have this problem.*2038

*If we were asked to integrate the same function from 4 to 6 of dx/x*^{2} - 5x + 6,2044

*We would have no problems here and would not have to split it up. *2059

*It would not be improper.*2064

*We could go back and use this regular technique of partial fractions and just plug in the bounds and we would not have to worry about limits at all.*2067

*The reason there is that the place where it blows up, x=2 and x=3, is not in between 4 and 6.*2077

*There would be no discontinuities in our region of interactions.*2084

*That is a very dangerous kind of integral. *2088

*You kind of have to watch out for when you see a denominator*2090

*Ask yourself, where is the denominator 0, where does that make you function blow up.*2095

*If that is inside your region of integration, then you have to split up the integral and work on each part.*2100

*If any one of them diverges. *2107

*Then you say the whole integral diverges.*2109

*We will do some more examples later on.*2114

*This has been Will Murray for educator.com.*2116

*We are here to look at a couple more examples of arc length problems.*0000

*The first one I have set up here is y = x*^{4}/8 + 1/4x^{2}.0005

*So remember with the arc length, you do not integrate it directly.*0012

*That is a common mistake that Calculus 2 students make is just integrating the function they are given.*0015

*When you want to find the arc length, you have got to find the derivative f'.*0019

*So here, the derivative of x*^{4} is 4x^{3}/8 is just x^{3}/2.0026

*Now, if you think of x*^{2} in the denominator as x^{-2},0037

*The derivative of that is x*^{-2} x^{-3}.0042

*That is why I have a negative here - 2x*^{-3},0048

*Then because of that 4 in the denominator, the two and the four cancel and we get a 2 in the denominator.*0052

*x*^{-3} gives you an x^{3} in the denominator.0060

*So, that is what we get for the derivative.*0066

*We need to square that, f'(x)*^{2} is x^{6}/4 + 1/4x^{6}, squaring out both terms.0069

*Then, minus 2ab, so minus 2 × 1/2 × 1/2 is 1/2.*0085

*Then the x*^{3} cancel, and this term is what we get by doing the -2ab of the a - b^{2} formula.0094

*So, that is where that - 1/2 comes in.*0107

*1 + f'(x)*^{2} is x^{6}/4, still + 1/4x^{6}.0109

*Now, - 1/2 + 1 gives us + 1/2.*0122

*That is almost the same function that we had before except that the minus has turned into a plus.*0130

*What started out as a-b*^{2}, this is now going to turn into a+b^{2}.0137

*So this is x*^{3}/2 + 1/2x^{3} quantity squared.0144

*Again, this is a very common pattern in arc length problems.*0154

*They often kind of rig up these examples so that the answer you get for 1 + f'(x)*^{2} factors nicely into a perfect square.0158

*It is sometimes not so obvious but if you look for it, it is often there.*0168

*That makes it nice because 1 + f'(x)*^{2}, when we take the square root will just simplify back down to x^{3}/2 + 1/2x^{3}.0173

*So, the arc length is the integral of that function.*0186

*So that is the integral from x=1 to x=2, getting those values from the stem of the problem.*0191

*Of x*^{3}/2 + 1/2x^{3} dx.0199

*If we integrate x*^{3} that is x^{4}/4, so this is x^{4}/8.0210

*Now, the integral of x*^{-3} is x^{-2}/-2, using the power rule.0217

*So this is -, because the -2 in the denominator, 1/4x*^{2}, that is -2x^{-2}.0229

*Then there is another 2 coming from the 2 there.*0240

*So, we want a value like this from x=1 to x=2.*0244

*So, what we get is 2*^{4}/8, that is 16/8, so that is just 2.0251

*Minus 1/4 × 2*^{2}, so that is 1/16.0257

*Minus, now if you plug in x=1, you get 1/8.*0266

*Plus 1/4, so if we think of everything in terms of 16ths there,*0270

*That is 32 - 1 - 2 + 4/16.*0277

*So that simplifies down to 33/16.*0287

*As our final answer for the arc length*0294

*So the point there is that we take the function we are given, find its derivative, run it through this Pythagorean formula,*0297

*And we got lucky here in that the perfect square and the square roots cancel.*0308

*Then integrate the thing you get there.*0315

*Let us do 1 more example here, we want to find the arc length of y=ln(x)/2-x*^{2}/4.0000

*So we do not integrate the function directly, we look at its derivative.*0010

*f' is, well, the derivative of ln(x) is 1/x so this is 1/2x - the derivative of x*^{2} is 2x.0016

*So the 2 and 4 cancel, so we get x/2.*0025

*So then we are going to find f'(x)*^{2},0029

*So that is a*^{2}, so 1/4x^{2},0035

*+ b*^{2} + x^{2}/4.0041

*Now, - 2ab, so -2 × 1/2 × 1/2 is 1/2.*0046

*Then the x's cancel, this term right here came from -2ab.*0056

*In the formula for the difference of squares.*0064

*Again, we do 1 + f'(x)*^{2}, so that is 1/4x^{2},0066

*+ x*^{2}/4, now - 1/2 + 1 gives us + 1/2.0073

*Again, that converts a - 2ab formula into a + 2ab formula.*0083

*So this factors as a perfect square, 1/2x + x/2 quantity squared.*0093

*So if you look at the square root of 1 + f'(x)*^{2},0103

*Remember that is what we have to integrate to find the arc length, we get 1/2x + x/2.*0110

*So we set up our arc length, and it is the integral from x=1 to x=e.*0116

*Of, 1/2x + x/2 dx.*0123

*Now that integrates to ln(x)/2,*0132

*X/2 integrates to x*^{2}/4 and we evaluate that whole thing from x=1 to x=e.0136

*So we get ln(e)/2 + e*^{2}/4 - ln(1/2) - 1/4.0148

*This simplifies a bit, of course ln(e) is just 1 and the ln(1) is 0.*0161

*So, we get 1/2 + e*^{2}/4 - 1/4.0168

*So that is 1/4, because that is 1/2 - 1/4, + e*^{2}/4.0176

*So just to reiterate there, the arc length formula, we take the function we are given,*0187

*We take its derivative, we run it through this Pythagorean Formula, and then you integrate what you get from that.*0196

*That is the end of our arc length lecture.*0204

*Today we are going to talk about arc length.*0000

*So there is one main formula for arc length that you need to know*0005

*And that is if you are trying to find the length of a curve y=f(x)*0008

*And you are trying to find the length from a to b.*0016

*The way I have drawn it kind of looks like we are looking for the area under the curve.*0020

*That is not what we are looking for today, not the area, but the length of that curve right there.*0023

*The way you work it out is this integral formula*0035

*The integral from x=a to x=b of the sqrt(1+x'(x)) that is the derivate of x*^{2} dx.0039

*Where this formula comes from is it comes from the pythagorean theorem.*0048

*This is the length of the hypotenuse of a triangle where the base has side length 1, the height is f'(x).*0053

*So the length of the diagonal is 1 + f'(x)*^{2} and then take square root of all of that.0066

*That is where the formula comes from.*0077

*It can help you remember that.*0079

*It is a very common mistake for Calculus 2 students to make to integrate the original function. *0081

*You will get an arc length problem, and then what you will try to do is integrate the original function.*0090

*That is very tempting. *0098

*There is nothing inherent in the problem that will tell you you are making a mistake there. *0101

*You may well go ahead an integrate that and get an answer.*0105

*That is a big no no in Calculus because what you actually computed there was the area, not the arc length.*0110

*You use this pythagorean formula with the sqrt and so you do not make this mistake of calculating the area instead.*0123

*Let us try this out with some examples.*0131

*The first example is the length of the curve x*^{2}/8 - ln(x) from x = 1 to x = e. 0133

*We are going to use the formula but first we are going to figure out f'(x).*0145

*Derivate of x*^{2}/8, derivative of x^{2} is just 2x so this would be x/4.0149

*Minus the derivative of ln(x) is 1/x.*0156

*So if we square that, f'(x)*^{2}0160

*It is x*^{2}/16 + 1/x^{2} - remember that (a+b)^{2} or (a-b)^{2} is a^{2} + b^{2} - 2ab.0170

*So minus 2(x/4) × 1/x.*0181

*If we combine those, the x's cancel and we just get - 1/2.*0190

*Remember the formulas that we have to look at 1 + f'(x)*^{2}.0196

*That is x*^{2}/16 + 1/x^{2} now we had minus 1/2 before.0203

*If we add - 1/2 + 1, that is + 1/2.*0212

*The nice thing about that is it factors again as a perfect square.*0219

*That is (x/4 + 1/x)*^{2}.0225

*Because, if you squared this out, you would get,*0230

*Well, x*^{2}/16 + 1/x^{2} + 2 × 1/x × x/4 would give you exactly 1/2 again.0234

*This factors as a perfect square and is a very common feature of arc length problems.*0248

*We will see that again in some of our problems.*0252

*You will probably see that as you work through your homework problems that you are sort of rigged up to make this work out.*0255

*Remember that what we are supposed to do is integrate the sqrt(1+f'(x)*^{2}) and that just cancels off the prefect square we had.0264

*That is x/4 + 1/x.*0276

*We are going to calculate the arc length, is the integral of that from x=1 to x=e.*0280

*x/4 + 1/x dx, and so x/4 integrates back to x*^{2}/8 and 1/x integrates back to ln(x).0288

*We evaluate that whole thing from x=e down to x=1.*0307

*We get e*^{2}/8 + ln(e) - 1/8 - ln(1).0316

*Of course ln(e) is just e, ln(1) is just 0.*0327

*So, we get e*^{2}/8 + 1 - 1/8, so that is + 7/8.0337

*That is our answer for the length of that curve.*0355

*Just to review what we have to do for this problem,*0360

*We are given a function here, and we do not integrate it directly.*0364

*First we calculate its derivative.*0368

*Then we plug it into this formula, sqrt(f'(x*^{2}))0374

*Then we integrate that and that gives us our answer.*0381

*Let us try another one.*0388

*This time we have to find the length of the curve y = 4x*^{3/2} + 1.0390

*Again, we have to look at f'(x), so we have to take the derivative of that.*0397

*The derivative of x*^{3/2} is 3/2 x^{1/2}0401

*So this is 4 × 3/2x*^{1/2}, derivative of 1 is 0.0406

*This just simplifies into 6x*^{1/2}.0416

*f'(x)*^{2} would be 36, (x^{1/2})^{2} would just be x, 0421

*if we add 1 to both sides we get 36x + 1.*0430

*What we are really integrating is just the sqrt(f'(x)*^{2})0435

*Which would be the sqrt(36x +1).*0445

*We are going to calculate the integral from x=0 to x=1 of the sqrt of 36x + 1 dx.*0450

*The way we want to do this integral, and this is common if you have the square root of something linear,*0465

*This is a common technique, we are going to use u =36x + 1.*0470

*Remember that whenever you use a substitution, you also have to work out du.*0476

*du is 36 dx, and so dx is du/36.*0480

*This turns into the integral, I am going to write x=0 to x=1, so that we remember those variables refer to the x values.*0492

*The square root of u and then dx is du/36, I will just put the 1/36 on the outside.*0502

*So I get 1/36, now this square root of u is the same as u*^{1/2}0509

*So the integral of u*^{1/2} is u^{3/2}, and then we have to divide by 3/2.0523

*Then we evaluate that from x=0 to x=1, but of course we cannot plug that in right away until we convert back to x's.*0530

*This is 1/36, 3/2 in the denominator, I am going to flip that up to be 2/3.*0541

*Then we have u*^{3/2} but u was 36x + 1.0548

*36x+1, so (36x+1)*^{3/2}, all that evaluated from x=0 to x=1.0555

*Now, 1/36 × 2/3 the, 2 and the 36 can cancel.*0571

*That is 1/18, and that is 1/54.*0578

*Now, 36x + 1, if we plug in x=1, that gives us 37*^{3/2} - x=0, just gives us 1^{3/2}.0582

*This reduces down to 1/54 × 37*^{3/2} is the same 37 times the square root of 37 - 1^{3/2} is just 1.0604

*That is as far as we can simplify that one.*0620

*Again, the critical step there was looking at the function, y = 4x*^{3/2} + 1. 0624

*Remembering that for arc length you do not integrate it directly.*0632

*You take its derivative and then you run it through the square root formula.*0637

*Then you integrate the thing that you get there.*0641

*From the on, it is really an integration problem.*0646

*OK, we are going to look at another example problem.*0652

*y = ln(cos(x)) from x=0 to x=pi/3.*0655

*Again, we wanted to find the derivative so f'(x).*0660

*The derivative of ln is 1/x, so the derivative of ln(cos) is 1/cos(x) × the derivative of cos by the chain rule.*0667

*Well the derivative of cos, by the chain rule, is the -sin(x).*0676

*This simplifies down to - tan(x).*0680

*f'(x)*^{2} is tan^{2}(x)0685

*So, (1 + f'(x))*^{2} = 1 + tan^{2}(x).0691

*But, if you remember the pythagorean identity, that is sec*^{2}(x)0702

*So, the square root of (1+f'(x))*^{2} is just sec(x).0707

*That is what we have to integrate.*0715

*We set up the integral from x=0 to x=pi/3 of sec(x) dx.*0716

*Now, sec(x) is one of those integrals that we learned how to solve in our section on trigonometric integrals.*0728

*It was one of those tricky ones that you just kind of have to remember.*0736

*The trick was to multiply top and bottom by sec(x) + tan(x).*0739

*That is probably worth just remembering because it is not something that you are that likely to figure out in a pinch.*0746

*When you do that, the integral is the ln(sec(x)) + tan(x)).*0756

*That is what we need to evaluate from x=0 to x=pi/3.*0765

*Now, that turns into the ln(sec(x) = 1/cos(x) + tan(x).*0775

*We are evaluating that from x=0 to x=pi/3*0789

*The cos(pi/3) is 1/2 tan(pi/3) is the sin(pi/3)/cos(pi/3). *0794

*That is sqrt(3)/2 divided by 1/2, so that is sqrt(3).*0812

*That is what we got by evaluating x = pi/3.*0815

*Minus 1/cos(0), the cos(0) is 1, + tan(0) is 0.*0822

*This right hand term gives us the ln(1) which goes away to 0.*0837

*This left hand term gives us the ln of 1/1/2 is 2 + s1rt(3). *0843

*I have absolute values signs here but I do not really need them because 2 + sqrt(3) is going to be positive.*0850

*So, that is our answer.*0856

*Ln(2 + sqrt(3)).*0858

*Again, what we did there was look at the function y = ln(cos(x)).*0863

*We did not integrate directly, instead we took its derivative and plugged it into this pythagorean formula and we integrated that.*0868

*From there on, it was an integration problem.*0874

*I am going to do a couple more examples of surface area of revolution.*0000

*We are going to start out by rotating the graph of 2x*^{3} from x=0 to x=1 around the x axis.0004

*So here, our f(x) is 2x*^{3}, we want to find f'(x), which is 6x^{2},0012

*f'(x)*^{2} is 36 x^{4}.0025

*1 + that is 36x*^{4} + 1.0034

*The square root of 1 + f'(x)*^{2} is just the square root of 36x^{4} + 1.0041

*The point of doing that was that you want to use your surface area formula. *0054

*That involves this big radical expression.*0057

*Our surface area is equal to the integral from x=0 to x=1, of 2pi × f(x), which is 2x*^{3}.0060

*× the radical expression, 36x*^{4} + 1 × dx.0079

*Now, I am going to pull the 4pi outside.*0092

*x=0 to x=1.*0097

*Key thing to notice here, this 36x*^{4} looks pretty nasty, but its derivative is exactly x^{3}.0101

*Well not exactly x*^{3}, but x^{3} times a constant.0110

*That suggests the substitution, u = 36x*^{4} + 1.0115

*Then du is 4 × 36 x*^{3} dx.0123

*We kind of have the du here when we have x*^{3} dx.0131

*The only thing that is not quite right is the 4 × 36.*0134

*We can correct for that by dividing on the outside by 4 × 36.*0139

*Then we will have the x*^{3} dx gives you du.0145

*We now have the square root of 36x*^{3} + 1, that is u.0149

*So this is pi/36.*0156

*Integral of sqrt(u) is not too bad, you think of that as u*^{1/2}.0161

*So the integral is u*^{3/2} and divide by 3/2, which is the same as multiplying by 2/3.0169

*We want to evaluate this from x=0 to x=1.*0180

*We cannot do that directly because everything is still in terms of u.*0184

*I am going to write this as pi, I guess we can simplify the 2 and 36 to be 1 and 18.*0190

*Then 18 × 3 is 54.*0198

*Now u*^{3/2}, (36x^{4} + 1)^{3/2}, and we want to evaluate that from x=0 to x=1.0202

*That is pi/54, now if we plug in x=1 to 36x*^{4} + 1, we will get 37 ^{3/2}.0218

*- x=0 in there just gives you 1*^{3/2}.0235

*This can be slightly simplified to pi/54 ×, 37*^{3/2} is the same as saying 37×sqrt(37) 0243

*- 1/2*^{3/2} is just 1.0253

*So we get our final answer there.*0258

*So, the key part of that problem is identifying y=f(x).*0266

*Then working through the formula, finding f'(x) and figuring out what the square root of 1 +f'*^{2} is.0270

*Then plugging the whole thing into the surface area formula that we learned at the beginning of the lecture.*0280

*Then it looks like a tricky integral but the key thing is to notice that if we let u = 36 x*^{4} + 1,0286

*We basically have our du set up for us with 2x*^{3}, we just need to correct for the constant there.0294

*For our last example, I would like to find the surface area by rotating the graph of y=sqrt(2x) around the x axis.*0000

*Here, our f(x) is sqrt(2x), and f'(x),*0009

*If you think of f(x) is 2x*^{1/2}, then f'(x) is 1/2 × 2x^{-1/2} × the derivative of 2x,0015

*So those 2's cancel and you get 1/sqrt(2x).*0031

*Now f'(x)*^{2} is just 1/2x, and if we add 1 to that, we get,0038

*If we put those over a common denominator we get 2x+1/2x.*0053

*Finally the sqrt(1+f'(x)*^{2}) is sqrt(2x+1/2x).0061

*Now we are ready to invoke our surface area of revolution formula.*0070

*The surface area of revolution is the integral on the bounds we are given are x=0 to x=1, of 2pi × f(x),*0075

*Which is the sqrt(2x).*0088

*× this big square root formula, 2x+1/2x.*0091

*All integrated with respect to dx.*0100

*So, this is actually pretty nice because the sqrt(2x) here, cancels with sqrt(2x) in the denominator there.*0106

*Now we have the integral of sqrt(2x+1), and a natural thing to do there is to let u=2x+1.*0115

*Then a du is just 2dx.*0130

*I am going to pull a 2pi outside, we get 2pi × integral from x=0 to x=1.*0135

*Of the sqrt(u) and dx converts into du, except that dx is 1/2 du.*0145

*I will put that 1/2 outside.*0154

*Sorry the integral from x=0 to x=1, of sqrt(u) du.*0157

*So, the point of that was that sqrt(u) is an easy integral.*0172

*You think of that as u*^{1/2}, and the integral of u^{1/2} is u^{3/2}/3/2 0180

*Which is the same as multiplying by 2/3.*0193

*And we are going to evaluate that from x=0 to x=1.*0196

*But you cannot evaluate that yet because we have the thing in terms of u, and we need to convert back into x's.*0198

*We remember that u=2x+1, so we get 2pi/3, 2x+1*^{3/2}, evaluated from x=0 to x=1.0206

*So this is 2pi/3, now if you plug in x=1 to 2x+1, that turns into two times 1+1, so 3*^{3/2},0223

*- x=0 in there, into 2x+1, just gives you 1*^{3/2}.0240

*Again it simplifies down to 2pi/3.*0247

*3*^{3/2} is the same as 3sqrt(3) - 1^{3/2} is just 1.0250

*And we get our final answer.*0262

*To recap there, we identified f(x), that was y=sqrt(2x),*0265

*We plugged that into the formula to sqrt(1+f'(x)*^{2}),0270

*Then we plugged our answer into the whole surface area formula right here.*0277

*Luckily that simplified down a little bit, then to get the integral we just needed to substitute u = 2x+1.*0283

*That finishes off that example and our lecture on surface area formulas is finished.*0292

*Today we are going to learn how to find the surface area of revolution.*0000

*What that means is that we are going to take a function y=f(x)*0006

*We are going to revolve this function around the y axis.*0016

*We are going to take this curve and spin it around the y axis and create a surface.*0022

*Then, what we are asking today is if you were going to paint that surface, how much paint would it require.*0039

*In other words, what is the surface area of the surface that you would obtain.*0045

*We have a nice formula that tells us the answer here.*0052

*The formula is the integral from x=a to x=b of 2pi × f(x) × sqrt(1 + f'(x)*^{2}) dx.0054

*You might recognize part of this formula as something we saw in a previous lecture on arc length.*0072

*Indeed, that is not a coincidence.*0078

*The surface area formula comes from looking at a small piece of the curve,*0080

*Calculating its arc length, and then calculating what the surface area would be if we rotated that around the x axis.*0085

*Of course, that is where you get the 2pi/f(x) part of the formula.*0097

*Let us try this out with some examples.*0104

*The first example we are going to be finding the surface area of the cone*0107

*If we take the graph of y = 3x and we rotate that around the x axis from x=0 to x=2.*0112

*We would get that cone and we are trying to figure out what the surface area is.*0128

*Remember our formula is the integral of 2pi f(x) × sqrt(1 + f'(x)*^{2}) dx.0134

*In this case, our f(x) is 3x.*0147

*So f'(x) is just 3.*0150

*1 + f'(x)*^{2} is 1 + 3^{2} which is 10.0156

*So this part of the formula, the square root part, is sqrt of 10.*0162

*We are integrating from a to b is 0 and 2, from x=0 to x=2 of 2pi × f(x) is 3x.*0165

*× the square root of 10, dx.*0178

*That is our surface area formula.*0183

*I am going to pull the 2pi × 3 to the outside.*0188

*That gives us 6pi.*0192

*I will pull the sqrt(10) outside as well, that is the sqrt(10).*0197

*The integral of x is x*^{2}/2,0203

*And we want to evaluate that from x=0 to x=2.*0205

*That gives us 6pi × sqrt(10) × if you plug in x=2 there, you get 4 - 0.*0212

*Sorry, 4/2 - 0.*0228

*That is 6pi × sqrt(10).*0232

*4/2 is just 2 so 2 - 0.*0236

*The whole answer is 12pi × sqrt(10).*0239

*Key points here, we are rotating something around the x axis.*0247

*We look at the f(x) that we are given and we calculate this square root formula by doing f'(x) and then 1 + f'*^{2}.0253

*We plug it all in to the surface area formula and then do the integration to finish that off.*0262

*Let us try another example.*0270

*This time we are rotating the graph of y = x*^{3} around the x axis.0272

*Our f'(x) is 3x*^{2}.0279

*Our f'*^{2} is 9x^{4}0288

*1 + f'*^{2} is 1 + 9x^{4}.0297

*And, the square root of that is sqrt(9x*^{3} + 1).0304

*So, our surface area is the integral from x=0 to x=1 of 2pi f(x).*0313

*So, I put 2 pi, now f(x) is x*^{3}, so 2pix^{3} × integral of 9x^{4} + 1 dx.0326

*Now, this looks like a rough integral but it is actually not so bad.*0342

*If you notice, you have 9x*^{4} + 1 under the radical.0348

*The derivative of that would 36x*^{3}, and we have an x^{3} outside.0352

*What we can do is let u = 9x*^{4} + 1.0358

*Then du is 36x*^{3} dx.0368

*So, what this converts into is, I guess I can write dx or x*^{3} dx, is 1/36 du.0376

*That takes care of the dx and the x*^{3}0390

*I will put the 1/36 outside.*0394

*I am also going to put the 2pi outside.*0398

*We still have the integral from x=0 to x=1.*0403

*Now, 9x*^{4} + 1 just turned into u, and then we have du.0408

*That is really a very simple integral now.*0416

*This is now pi/18, u*^{1/2} is the same as sqrt(u).0420

*If we integrate that, that integrates to u*^{3/2}/3/2 or, 2/3u^{3/2}.0429

*We are evaluating this not using u but using x=0, not x=1.*0442

*So let us keep going with that.*0450

*We still have a pi.*0452

*2/3 I can cancel that with the 18 to get 9 here and a 1 there.*0455

*Pi over 27.*0461

*Now u was 9x*^{4} + 1, and we are evaluating that from x=0 to x=1.0468

*We get pi/27.*0480

*Now if you plug in x=1, I am sorry I left out my 3/2 there.*0485

*That 3/2 came from that right there.*0490

*Then we have 9x*^{4} + 1.0495

*If we plug in 1 to 9x*^{4} + 1, that is 10^{3/2}0500

*X=0, if you plug it in just gives you 1 - 1*^{3/2}.0507

*This gives us pi/27 × 10*^{3/2}, is the same as saying 10 sqrt(10), - 1.0516

*That is the answer for the surface area of this graph we rotated around the x axis.*0535

*Again, the key point there is identifying your f(x)*0543

*Running it through this formula, plugging it into the general surface area formula*0548

*Then, we had a pretty tricky integral there.*0554

*The key thing there was observing that the derivative of 9x*^{4} + 1 was more or less x^{3},0556

*So we could make this substitution that gave us a very nice integral to solve.*0568

*Let us try another example*0575

*This time we are rotating the graph of y = 2x*^{2} + 1 from x=0 to x=1.0578

*There is a very important difference in this example which is we are rotating around the y axis.*0584

*Remember that our surface area formula that we learned before used the x axis.*0591

*We had the x axis before, that means we need to convert our surface area formula*0605

*To adapt to the fact that we are now rotating around the y-axis.*0611

*That means we need to look at the surface area formula and sort of switch everything from x's to y's.*0615

*Let me write that down.*0621

*The surface area is now the integral from y =, I will not write a and b, c to y = d, of 2pi f(y).*0625

*× sqrt(1+f'(y)) dy.*0640

*That means we need to convert everything here into functions and terms of y.*0648

*Including our function and the limits.*0654

*We were given this as if y were a function of x.*0659

*Instead we need to convert everything to x as a function of y.*0663

*Let us go ahead and convert that.*0667

*Looking at the function first, we get y-1 = 2x*^{2} 0669

*We can divide both sides by 2 there, so x = sqrt(y-1/2).*0679

*If you plug in x=0, into the function,*0688

*That would convert into y = 1.*0693

*If you plug x=1 again into the function, *0700

*That would convert into y=3.*0704

*Now we have everything in terms of y.*0710

*Again, let us try to figure out what f'(y) and what this square root formula turns into.*0713

*f'(y), well we have got a square root of something*0722

*So its derivative is 1/2*0728

*Because we think of its square root as all of that stuff to the 1/2.*0732

*1/2 times all of that stuff to the -1/2*0736

*I will write it down here in the denominator, y-1/2*0739

*× the derivative of that inside stuff by the chain rule*0745

*The derivative of y-1/2 is 1/2.*0750

*Let me try to simplify this.*0754

*I am going to put the two 1/2's together and put 1/4.*0757

*Now this denominator, I am going to flip it and bring it up to the numerator*0760

*That is 2/y-1.*0765

*That is f'*0770

*We need to find f'*^{2}.0772

*f'*^{2} is 1/16 × 2/y-1.0776

*We can simplify that into 1/8 × y-1.*0782

*1 + f'(y)*^{2} is, 8 × y-1, 8y-8+1/8y-1.0790

*What I did there is I wrote that 1 as 8y-8/the denominator.*0809

*8y-8, because I wanted to combine everything over a common denominator.*0818

*I can simplify that down a little bit into 8y-7/8(y-1).*0826

*Now we need to take the square root of that.*0837

*What we have is the square root of 1 + f'(y)*^{2}.0840

*Is the sqrt(8y-7/8(y-1)).*0853

*I am going to put all of these pieces together into a big integral. *0863

*The integral is the surface area from y=1 to y=3.*0867

*Got those from here and here.*0874

*2pi f(y), I got that from here, that is the sqrt(y-1/2) × sqrt(1+f'(y*^{2})),0878

*That is 8-7/8(y-1) × dy.*0894

*That looks pretty messy but it does simplify.*0905

*The y-1's cancel and then we are left with a 2 and an 8 in the denominator, that is 16.*0906

*We can pull that out of the square root and that is a 4 in the denominator,*0913

*And so this simplifies down a little bit.*0919

*2/4 gives us 1/2, I will pull the pi outside.*0922

*So, we get the integral from y=1 to y=3 of square root of, I think the only thing that is left there is 8y-7 dy.*0930

*Now we can use u = 8y-7 so du = 8 dy.*0945

*dy is 1/8 du.*0955

*What we get now is pi/2 × the 1/8 that we got from the du here.*0962

*The integral from y=1 to y=3 of u*^{1/2} du.0974

*That is pi/2 × 1/8 u*^{1/2}0986

*The integral of u*^{1/2} is u^{3/2}/3/2, which is the same as multiplying by 2/3.0993

*And, we want to evaluate that from y=1 to y=3.*1004

*I can simplify a little bit.*1008

*My two's cancel, and we get pi, combine the 3 and the 8 to give 24.*1010

*That is pi/24*1019

*Now, u was 8y-7*^{3/2}1022

*Finally, we get pi/24 × *1034

*Now, if we plug in y=3 to 8y-7.*1044

*That is 24-7, which is 17*^{3/2}1047

*If we plug in y=1 to 8y-7 which is 8-7, so -1*^{3/2}1055

*We can clean that up a little bit into pi/24 × 17 sqrt(17).*1063

*1*^{3/2} is just 1.1075

*Our final answer is pi/24 × 17 sqrt(17) -1.*1080

*Probably what made that problem a little difficult besides it being a little bit complicated on the algebraic side.*1085

*Was the fact that we were given the y axis instead of the x axis*1093

*Which means you have to take your original x formula and translate everything into terms of y.*1098

*In turn, you have to translate things from y as a function of x to x as a function of y.*1106

*The x values that you are given have to be converted into y values.*1112

*Once you do that, you walk through the process of calculating this radical 1 + x'(y)*^{2}.1122

*Then it is still kind of a messy integral but the square roots sort of cancel nicely *1133

*And, you end up with something that is not too complicated at the end.*1136

*We are working on some more examples of calculating the force due to hydrostatic pressure.*0000

*In this example, we are using the same situation that we had in an earlier example.*0006

*Let us recall that there we had a semicircular plate of radius 1.*0013

*We worked out last time that a depth of y, then the width there,*0024

*Would be l(y) = 2 × sqrt(1-y*^{2}).0034

*Well we are changing things up in this example.*0042

*In this example, the plate is now submerged so that its diameter is below the surface of water.*0043

*Instead of being flush to the surface of water, we have the water up here.*0053

*The plate is d meters below the surface of the water.*0058

*Remember the whole point of force due to hydrostatic pressure is that,*0063

*When you submerge something deeper in the water, there is more force deeper down in the water.*0070

*Because you have got more water piled up on top of it.*0075

*There is more pressure at lower depths.*0079

*So, let us try to figure out how having the thing deeper in the water changes what our answer is going to be.*0082

*Again, we are going to set up our integral formula, which remember is w × integral from our 1 value of y to another,*0090

*of l(y) × d(y) dy.*0103

*So again, we are going to run our y from y=0 to y=1.*0111

*But our l(y) is still 2 × 1 - y*^{2}, because the width of the thing has not changed.0122

*What has changed is the depth, the d(y).*0130

*What is new in this problem is the d(y),*0134

*Remember before was just y,*0139

*But now, it is y + d.*0142

*So I am going to set up the integral, just like before, except that we have y + d instead of y.*0146

*Our w, remember that was the weight density of water.*0153

*We got that by multiplying the density of water × the acceleration due to gravity.*0157

*The w was the density of water × acceleration due to gravity.*0163

*That is all the same.*0170

*That is 1000 × 9.8, that is 9800.*0172

*What we get here is the force is w, 9800, × the integral still from y=0, y=1.*0176

*Of 2 ×, instead of y, y + d.*0188

*× the sqrt(1-y*^{2}) dy.0194

*Now if you look at this integral, it is quite similar to what we had before, in particular the y term is the same,*0204

*As in the earlier example.*0212

*We worked that out before.*0216

*We worked that out to be 19,600/3 Newtons.*0219

*That is what the y term gives you.*0226

*But we still have the rest of the integral to solve.*0228

*9800 × the integral from y=0 to y=1 of 2d × sqrt(1-y*^{2}) dy.0233

*Now, this integral unfortunately is not so pleasant.*0245

*We have to integrate sqrt(1-y*^{2}).0252

*How do we handle that?*0255

*We now learned several different techniques where how to do that integral.*0258

*You can use a trig substitution on that.*0267

*Where you would use y = sin(θ).*0274

*We have learned how to do that on the chapter on trig substitution.*0279

*You can also look up the answer in an integration table in the back of your textbook.*0284

*You will see in the back of your textbook how to find the integral of a*^{2} - u^{2} du.0291

*And, it will give you a formula to do that.*0299

*You can use either one of these techniques, trig substitution, or an integration table.*0302

*To do that integral.*0308

*What it converts into is,*0310

*I am just going to show the integration part,*0317

*The first part the nice 19,600/3, that is going to stay the same.*0321

*I am going to pull the 2d and then 9800 out of the integral and combine them.*0324

*So, we get 19,600d.*0332

*I am not going to go through the process of looking up this integral in a table,*0340

*Or going through the trig substitution, because we have already worked through those in previous lectures.*0349

*I am just going to go ahead and say what the answer is according to a table in a calculus book.*0354

*The answer for the integral of sqrt(1-y*^{2}),0360

*Is y/2 × sqrt(1-y*^{2}) + 1/2 arcsin(y).0362

*Remember we have to integrate this from y=0 to y=1.*0380

*So, we get 19,600d.*0388

*If we plug in y=1 to the first term, that sqrt(1-y*^{2}) goes away and gives you 0.0395

*The second term gives us 1/2 arcsin(1),*0404

*So, 1/2 arcsin(1).*0410

*We have to remember what arcsin(1) is.*0415

*If we plug in y=0, that first term, the y/2 part of it, gives you 0.*0419

*Then we have -1/2 arcsin(0).*0425

*arcsin(0) is just 0, because sin(0) is 0.*0436

*arcsin(1), you have to think ok, what angle has sin(1).*0442

*Well, the answer is pi/2.*0446

*So this turns into pi/2.*0448

*So we get 19,600 d × pi and we have still got this 1/2 × pi/2.*0454

*So 1/2 × pi/2 gives us pi/4.*0465

*Let me now bring in this other term that we had at the beginning.*0468

*We get 19,600/3 + 19,600/4 is 4900 d × pi.*0473

*This whole answer is in terms of Newtons.*0491

*Let us review what made that example work.*0505

*First we drew a picture of the plate that was submerged in the liquid.*0508

*That was very important.*0513

*We figured out what the depth function was, that was y + d, because it made sense to run this from y=0 to y=1.*0515

*So our depth function was y + d.*0524

*We figured out what the width function was,*0529

*Actually we brought that over from the previous problem.*0532

*But that was something we figured out in the previous problem,*0536

*So we figured out the depth function, the width function, *0539

*Then we figured out the weight density of water by taking the density of water × the acceleration due to gravity.*0545

*We plugged all of those into our formula for force due to hydrostatic pressure,*0549

*Then we worked through the integral,*0555

*Which was not too bad once we are armed with all of the integration techniques we learned earlier.*0556

*Then we got our answer.*0564

*Density is 300 kg/m*^{3}.0000

*It says the plate is oriented vertically with 1 corner at the top flush with the surface of the fluid.*0004

*Again the first thing to do here is draw a picture and try to figure what is going on.*0012

*So, let me draw a picture.*0017

*Here is the top of the fluid, so the surface of the fluid.*0020

*The triangular plate is 2m on each side and it has a corner at the top there.*0025

*So there is our plate.*0029

*Again, we are going to use our formula for force due to hydrostatic pressure.*0034

*We can use the depth again to be y.*0044

*So d(y) = y.*0050

*For the width, we have to figure out what the width is at any particular value of y.*0053

*What we can do there is say well if the width is x, *0061

*Then we can use our formulas for triangles.*0066

*What we know here is that x/2, if we look at that 30-60-90 triangle,*0074

*This is x/2 and that is y,*0080

*x/2 × sqrt(3) = y, using what we remember about relationships between sides in 30-60-90 triangles.*0083

*So, the width is x = 2y/sqrt(3).*0092

*So, the width, l(y) is equal to, if I rationalize that, I get 2×sqrt(3y/3).*0103

*Remember the w is the weight density, and that is the density of the fluid × acceleration due to gravity.*0118

*That is 300 × 9.8, which comes out to be 2940.*0128

*So, our integral is 2940 × the integral as y goes from, *0133

*OK that is y=0,*0144

*Now, this bottom part is.*0151

*Ok, again we have to remember our formulas for triangles.*0157

*So if that is 2 meters, then half of that is 1 meter, and so the total depth is sqrt(3).*0163

*So y = sqrt(3).*0176

*Then the integral of l(y) × d(y), so 2 sqrt(3) y*^{2}/3 dy.0183

*If we square the 2 sqrt(3)/3 and the 2940, we get 1960sqrt(3).*0198

*Times the integral from 0 to sqrt(3) of y*^{2} dy.0209

*That is 1960 × sqrt(3)/3y*^{3}.0214

*The integral of y*^{2} is y^{3}/3.0222

*Evaluate that from y=0 to sqrt(3).*0226

*We get 1960 × 3sqrt(3) × sqrt(3)/3.*0230

*That simplifies down to 1960 × 3, so 5880 Newtons.*0243

*I did the integration at the end there a little bit quickly.*0257

*The key part there is setting up the formula, finding expressions for d(y) and l(y),*0260

*Plugging them into the formula,*0267

*Then it just turns into an integration problem*0270

*Today we are going to learn about an important application of integration. *0000

*We are going to use integration to calculate the force due to hydrostatic pressure.*0004

*Let me explain the situation there.*0009

*The idea there is that we have some kind of think plate submerged in a liquid.*0012

*This could be a thin plate or it could be the wall of a dam, or anything that is enclosing some fluid.*0030

*We are trying to calculate how much force the fluid puts on that thin plate.*0038

*The key point here is that the fluid does not put so much force on it, but when it is deeper down, *0044

*There is more fluid piling up and pushing against that plate or against that wall and so there will be a greater force.*0053

*We are going to calculate this using this integral formula but I have to explain what each of these terms means.*0062

*This W outside represents the density of the fluid, which is something you would measure in kg/m*^{3} × the acceleration due to gravity.0069

*That of course is always the same, and we will use the value here, we will round that to 9.8 m/s*^{2}.0100

*We are using metric units but of course you could also use English units.*0113

*That W when you multiply them together, what you get is something that has units of kg/m*^{2}×s^{2}, 0115

*That is called the weight density of the fluid.*0126

*That is what that W represents, and that will be a constant.*0137

*What this d(y) and l(y) represent are, if this is plate that is submerged in the fluid, what the d(y) and l(y) represents are at any given point, *0141

*D(y) represents the depth and l(y) represents the width of the plate at that particular depth.*0160

*So l(y) is the width, and d(y) is the depth.*0194

*We will use this formula to calculate the total force due to hydrostatic pressure on a plate.*0197

*Let us try an example.*0207

*We are given a semi-circular plate of 1 m submerged in water so that its diameter is level with the surface to the water.*0208

*Let me start by graphing that.*0217

*Here is the surface of the water.*0221

*It is a semi-circular plate whose plate is level with the surface of the water and we know that the radius is 1m.*0223

*We want to find the force due to hydrostatic pressure on the plate and we are given that water has a density of 1000 kg/m*^{3}, 0233

*And of course we know the acceleration due to gravity.*0245

*What we want to do is find the width and depth of the plate at different depths.*0249

*Let me assume that we are at depth y, so we are measuring y from the surface of the water down.*0264

*I want to try to find the width of the plate at that particular depth.*0271

*The way I can figure that out is to remember that this is a semi-circular plate that has radius 1, so that radius is 1.*0275

*That width there, well, that little area just halfway across the width is going to be the sqrt(1-y*^{2}).0285

*The total width is l(y), which is twice that, which is 2×sqrt(1-y*^{2}).0298

*The depth, which is d(y), is just y itself.*0305

*So, let us fill in our formula for hydrostatic pressure.*0310

*Remember, W is the density of the fluid, which in this case is water × the acceleration due to gravity.*0317

*In this case the density is 1000, the acceleration due to gravity is 9.8, and so the whole term for W is 9800.*0329

*The total force is W, so 9800, times the integral from y = 0 to y = 1,*0347

*Because those are the smallest and largest values of y that we are going to see in this plate.*0363

*So integrate from y=0 to y=1, of l(y) which is 2sqrt(1-y*^{2}) × d(y), which is just y dY.0370

*Now we need to integrate that.*0389

*That is not such a bad integral because we can notice that if we let u be 1 - y*^{2}, then dU is -2y dY, 0395

*Which we almost have except for the negative sign.*0407

*We just about have 2y dY there.*0409

*I will just pull that negative sign outside, and we get nevative 9800 × the integral from y=0 to y=1 of the sqrt(u)×dU.*0413

*This is -9800, now the integral of the sqrt(u), you think of that as u*^{1/2}.0440

*The integral of that is u*^{3/2}, divided by 3/2, which is the same as multiplying by 2/3 so put a 2/3 in there.0450

*Then we want to integrate this from y=0 to y=1.*0468

*We want to be careful about keeping the u's and y's straight.*0474

*To make it easy I will substitute everything back into y's.*0477

*So, we get -9800 × 2/3.*0481

*Now u was (1-y*^{2})^{1} evaluated from y=0 to y=1.0486

*This is then -9800 × 2/3.*0500

*Now if you plug in y=1, you just get 0. if you plug in y=0 then you just get -1.*0509

*But we are subtracting that because that was the lower limit, so we get 0 - (-1), or sorry, *0520

*If you plug in y=0 to 1-y*^{2}, you just get 1 so that gets subtracted.0530

*So, we get 0-1, and then that negative sign cancels out with the negative sign we had on the outside.*0539

*What we finally get is 9800 × 2/3.*0545

*9800 × 2 = 19600/3, and remember our units here were kg/m*^{2} per s^{2} which are Newtons, 0552

*And so our answer here is in Newtons.*0567

*The trick there is to make a graph of the situation that you are looking at.*0579

*In this case we had semi-circular plates, so you draw the semi-circular plates, We figure out what our W is, that is density always a constant.*0586

*Times gravity, which is always 9.8 assuming you are using metric units.*0595

*Then we figure out d(y) is y and l(y) is the width, that was a bit trickier, *0600

*We had to do a bit of work with the graph to figure out that was 2 × sqrt(1-y*^{2})0607

*Then we plug the whole thing into our integral formula for the force due to hydrostatic pressure.*0614

*Then we work it through and we get our answer.*0620

*A couple of notes here, we will be using this same example in another example later on.*0624

*Hang on to this answer, 19600/3 Newtons, and we will come back to it later.*0629

*In the meantime, this has been Will Murray for educator.com and we will try a couple more examples later on.*0636

*We are here to do a couple more example on centroids.*0000

*Our example 4 is to find the centroid of the region under the graph y = sin(x) from x=0 to x=π.*0004

*It is always good to draw a graph of these things.*0016

*There is the graph of y = sin(x).*0021

*There is x=0, and there is x=π.*0025

*We are trying to find the centroid of that region.*0030

*First of all, remember that the centroid formulas have an area in them.*0034

*We first need to find the area of that region.*0040

*The area is just the integral from 0 to π, of sin(x) dx.*0045

*The integral of sin is cos(x).*0055

*Evaluate that from x=0 to x=π.*0060

*Sorry, the integral of sin is -cos(x).*0066

*That gives us -, -1 -1, which gives us 2 for the area.*0069

*Now we want to find the two coordinates, the x and y coordinates of the centroid.*0077

*The x coordinate, if you will look at this region, it is completely symmetric around x = π/2.*0082

*That has to be the balance point in the x direction.*0093

*We know that the x coordinate, of the centroid must be π/2, just by symmetry.*0096

*Now we could work out the integral formula, but that would be a lot more work.*0107

*I am going to skip that.*0111

*Instead, I am going to look at the y coordinate of the centroid.*0113

*The formula that we have is 1/2 × the area, integral from a to b, of f(x)*^{2} dx.0117

*So that is 1/2 × the area.*0127

*Well the area we figured out was 2, so this is 1/4.*0133

*× integral from 0 to π of sin*^{2}(x) dx.0138

*Remember what we learned in the section on trigonometric integrals.*0147

*When you have an even power of sin or cosine, you want to use the half angle formula.*0153

*When you convert this sin*^{2}(x) into 1/2 × 1 - cosine(2x),0155

*Then that is something much easier to integrate.*0167

*If I pulled that 1/2 out and combine it with that 1/4, I get 1/8.*0171

*Then, if I integrate 1, I get x.*0177

*Integral of cosine(2x) is 1/2sin(2x).*0181

*I am evaluating this from x=0 to x=π.*0189

*This is 1/8.*0195

*Now if you plug in x=π, you get π.*0198

*sin(2π) = 0, so that is just - 0.*0204

*Plug in x=0 and you get - 0, and then the sin(0) is 0 again.*0208

*So we just get π/8.*0216

*That was the y coordinate of the centroid.*0219

*If we package those two answers together,*0225

*x-bar, y-bar, we get (π/2, π/8) is the balance point of this region.*0230

*Finally, I would like to do an example with a triangular region.*0000

*The coordinates of the corners are (-1,0), (1,0), and (0,6).*0004

*Like the other examples, it is very useful to graph this region before we start.*0012

*So, (-1,0) is there.*0020

*(1,0) is there.*0024

*(0,6) is up there.*0030

*So, we are looking at this triangle.*0035

*We are trying to find the centroid of that region.*0043

*Again, we can exploit symmetry on this.*0046

*This region is symmetric in the x direction.*0053

*The x coordinate of the centroid must be exactly halfway in between the left and right extremes.*0058

*The x coordinate of the centroid must be 0, by symmetry.*0065

*The y coordinate is not so obvious, we actually have to do some calculus for that.*0075

*I think what I am going to do is find the centroid of the right hand triangle there.*0081

*That is the one I am outlining in red.*0090

*The point there is that if I can find the point where that balances. *0092

*Then that will be the same as where the entire triangle balances in the y direction.*0099

*If I just find the centroid of this triangle on the right here, *0106

*It will make the calculations a little bit easier and avoid some ugly negative numbers.*0110

*Let us just assume we are looking at this right hand triangle, *0116

*And find the y coordinate of the centroid for this.*0122

*So, as usual, we need to find the area of that region.*0127

*Well the area of a triangle is 1/2 base × height.*0133

*The base is 1, and the height is 6, so the area is just 3.*0139

*Then, our formula for the coordinates of the centroid tells us that y-bar is 1/2 × the area *0144

*× integral from a to b of f(x)*^{2} dx.0156

*That means I have to figure out what f(x) is.*0164

*f(x) is that line right there.*0167

*We need to find the equation of that line.*0170

*That is not so hard, because that line has slope -6, because it is dropping down 6 units for 1 unit it goes over.*0174

*Its intercept is given by y = 6 right there.*0183

*So, the equation for that line is just y=-6x + 6, or 6-6x.*0194

*That is the f(x) we will be using in our formula.*0202

*Then the f values are x=0 to x=1.*0206

*Because we are only looking at the right hand triangle now.*0212

*Let us plug everything in here.*0217

*1/2a, our a was 3 so this is 1/6.*0218

*Our little a and b are 0 and 1.*0224

*Out f(x)*^{2} was 6-6x, so this is (6-6x)^{2} dx.0229

*The numbers are going to get a little bit if we leave those 6's in there.*0240

*So I think what I will do is pull a 6 out of that square.*0244

*When you pull that out, that will give us a 36.*0249

*I will pull that all the way out of the integral, that gives us 36/6 × integral from 0 to 1,*0250

*Of now there is just (1-x)*^{2} dx.0258

*So, that in turn becomes 6 × integral from 0 to 1 of (1-x)*^{2}, 0264

*is x*^{2} - 2x + 1 dx.0270

*That is a very tractable integral.*0279

*That is 6 × x*^{3}/3 - integral of 2x is x^{2} + integral of 1 is just x.0282

*We want to evaluate that from x=0 to x=1.*0293

*So that is 6 × 1/3 - 1 + 1.*0299

*Then if you plug in x=0, you just get a bunch of 0's.*0304

*These ones cancel and so you get 6 × 1/3, so that is just 2, as the y coordinate of the centroid.*0310

*So the centroid here is the x coordinate and the y coordinate packaged together into a point.*0328

*The x coordinate was 0 and the y coordinate was 2.*0335

*The balance point of that whole triangle is (0,2).*0343

*Again, what we did there was we looked at the question. *0351

*We immediately drew a graph of it because you do not really understand it until you draw a graph of it.*0356

*Just by looking at the graph we were able to realize the x coordinate of the centroid is 0 by symmetry.*0362

*Because the right and left sides obviously have the same mass so the x coordinate is going to be 0.*0371

*To find the y coordinate we used our formula for y-bar.*0377

*Which in turn required us to know the area.*0381

*Since the thing was symmetric, we just found the y coordinate for the triangle on the right,*0385

*But that would give us the same y coordinate for the entire triangle.*0394

*Once we plugged that in, we got a very easy integral to find the y coordinate.*0398

*So, we found the center of mass there.*0403

*That concludes the lecture on centroids and centers of mass.*0406

*Today we are going to learn how to use integration to calculate the center of mass of a region.*0000

*The idea here is that we will have a function y = f(x).*0010

*We will look at the region underneath it from x = a to x = b.*0017

*We are going to imagine that we cut out a thin plate that fills that region.*0025

*We want to figure out exactly where the center of mass is.*0030

*In other words, if we were going to balance this region on a particular point, where would it balance.*0035

*We want an x coordinate and a y coordinate of the center of mass.*0043

*We are going to call those coordinates x-bar and y-bar.*0052

*We want to figure out how to find those.*0057

*The center of mass is also known as the centroid.*0060

*On some of the examples you will see the word centroid.*0066

*We have 2 equations that both involve integrals that tell us how to find those 2 coordinates.*0070

*The x-bar is given by this integral equation*0077

*The y-bar is given by this integral equation.*0081

*The one thing that may not be clear here is that the a is equal to the area*0085

*Of course you can find the area by integrating from small a to small b of f(x) dx.*0092

*Let us try that out on some examples and see how that works out.*0100

*The first one we are given is to find the centroid of the region inside a semi-circle of radius 1.*0105

*We are definitely going to need to start with a graph there.*0113

*There is a semi-circle of radius 1 and we want to figure out where the centroid is.*0119

*We are trying to find the region inside it.*0126

*We want to find the centroid of that region.*0131

*Now one thing is obvious by symmetry, both sides of this semi-circle, or sorry this half disc, are going to weigh the same amount.*0132

*The x coordinate of the centroid is certainly going to be 0.*0143

*That is just by noticing the symmetry of the region.*0150

*The y coordinate is not going to be so easy.*0157

*We actually have to do some calculus for that.*0159

*First of all, remember that the area of the region, remember we need that for the formula.*0165

*Since it is a circle of radius 1, well the area of a circle of radius 1 is pi, but we just have half of that.*0171

*Then the y coordinate of the centroid is given by our integral formula.*0178

*I will copy that down here, 1/2a × the integral from a to b of f(x)*^{2} dx.0183

*Now, we need to figure out what the function is for the circle.*0195

*That function is, well remember the function, or the defining equation for a circle is x*^{2} + y^{2} = 1.0199

*If we solve that for y in terms of x.*0207

*We get y = sqrt(1-x*^{2})0210

*In this case, if we plug that into our formula, we get 1/2 × area*0215

*Which is pi/2.*0222

*× integral our bounds on x are x=-1 and x=1 here, so the integral from -1 to 1.*0226

*f(x)*^{2} well since f(x) was a square root, this is just 1 - x^{2} dx0241

*This in turn becomes 1/pi × *0250

*Now, if we integrate 1/x*^{2}, that is easy, that is just 1-x^{3}/3.0255

*Evaluate that from x = -1 to x = 1.*0265

*This gives us 1/pi × 1 - 1/3 - (-1), so + 1, - (-1/3).*0270

*Sorry, minus, then there is another minus, then there is a third minus, so this whole thing is negative 1/3.*0288

*That gives us 1/pi × 2 - 2/3, which is 4/3.*0300

*That gives us 4/3 pi.*0311

*Remember that was just the y coordinate of the centroid*0315

*So, the centroid is the point located at x = 0 and y = 4 over 3pi.*0319

*So, that is the point at which this object would balance if you tried to balance it on a point.*0333

*Let us review there.*0341

*We were given a two dimensional shape and we want to use our integral formulas to find the centroid.*0344

*The x=0, that just came from the fact that both the left and the right hand sides of the circle look the same.*0353

*The area pi/2 just came from the formula for the area of a circle,*0360

*And then we use the integral formula to find the y coordinate of the centroid.*0364

*That turned out to be not too bad of an integral.*0369

*Our second example is to find the centroid of the region under the graph of y = 1/x from x = 1 to x = e,*0375

*Let me draw that, that should not cross the x axis, it is asymptotic to the x axis.*0387

*There is x=1 and x=e.*0392

*We are looking at that region and we want to find the center of mass of that region.*0401

*Let us start out by finding the area, that is just the integral from 1 to 3 of f(x).*0408

*That is 1/x dx*0413

*The integral of 1/x is ln(x).*0417

*Integrate that from x=1 to x=e*0420

*We get the ln(e) - ln(1), but the ln(e) is 1 and the ln(1) is just 0.*0426

*That is the first ingredient we needed for our formulas.*0440

*Now let us find the centroid.*0441

*x-bar, the x coordinate of the formula is 1/area × the integral from a to b of x(f(x)) dx.*0445

*The area is just 1 so that just turns out to be 1*0460

*Now we want the integral from a to b, that is the integral from 1 to e.*0465

*f(x) is 1/x so f(f(x)) is just 1 dx.*0470

*You integrate that and you get x evaluated from 1 to e, so you get e-1.*0480

*So, that was the x coordinate.*0486

*The y coordinate is a little bit harder but not too much.*0490

*The y coordinate, again our main formula tells us it is 1/2 × the area, integral from a to b of f(x)*^{2} dx0493

*The area was 1, so this is 1/2 integral from 1 to e *0507

*f(x)*^{2} is 1/x^{2} dx.0514

*So, this is 1/2, now the integral of x*^{2}, you want to think about it as x^{-2}0520

*Its integral is x*^{-1}/-1, or -1/x.0526

*Evaluated from x=1 to x=e.*0535

*That is 1/2 -1/e + 1 *0542

*That could be written as 1/2 - 1/2e*0550

*That was the y coordinate of the centroid.*0555

*you put those together and you get the centroid, the balance point, is*0558

*Put the x and the y coordinate together to get an ordered pair*0565

*You get (e-1, 1/2 - 1/2e)*0570

*The point of that problem is that we had to figure out the area first*0582

*Then we plug that into the formula for the x coordinate of the centroid.*0587

*Work that out, and then plug that into the formula for the y coordinate of the centroid.*0592

*Then we sort of package them together to give us the coordinates of the centroid.*0597

*Let us do another example of that.*0603

*We want to find the centroid now of the region inside the unit circle, inside the first quadrant.*0607

*If we graph that, it is always good to start with a graph.*0613

*That is that part of the unit circle and we are trying to find the centroid of that region.*0617

*The first thing to notice here is that we can exploit some of our earlier work.*0627

*Remember in example 1 that we found the centroid of the semi-circle, or half disc.*0632

*The y coordinate of the centroid should be the same as the y coordinate of the centroid we had before.*0646

*Wherever this thing balance in the y direction is the same as where the half disc balances in the y direction.*0659

*Let us recall that that was y = 4/3pi.*0663

*Now the x coordinate of the centroid, *0669

*Again our formula is 1/area × the integral from a to b of x × f(x) dx.*0673

*We figured out before that the equation here is y = sqrt(1-x*^{2})0684

*We are now going from x=0 to x=1.*0693

*The area there, that is 1/4 of a circle.*0697

*A circle of radius 1 would have area pi.*0701

*The area is pi/4.*0704

*1/pi/4 integral from 0 to 1 of x × f(x) is 1 - x*^{2} dx.0708

*Clean it up a little bit, we can flip over the pi/4 and get 4/pi × the integral from 0 to 1 of x × 1-x*^{2} dx.0721

*At this point there are two ways you can proceed.*0733

*Actually you could have proceeded a different way earlier but I wanted to show how you can set up this method.*0735

*You can go ahead and work out this integral*0740

*It is not that bad.*0745

*What you can do is say u = 1 - x*^{2},0748

*Then du is -2x, dx*0752

*That is going to work pretty nicely because you already have an x there that provides your dx.*0757

*That would not be such a bad integral if you use a u substitution there.*0767

*On the other hand, you could also notice that this region is symmetric about the line y=x.*0774

*The x coordinate of the centroid should be the same as the y coordinate of the centroid*0795

*This region is symmetric in the x and y directions so it should balance at the same point in the x and y directions.*0804

*What you should be able to figure out just be exploiting the symmetry is that the x coordinate is also going to be 4/3pi.*0812

*If you had not noticed that, you could certainly work out this integral.*0822

*You should get 4/3pi by the integral formula as well.*0827

*Either way, you will get the same answer.*0831

*It is a little easier if you notice this, but if not you can still do the integral formula.*0834

*You end up with the two coordinate of the centroid being (4/3pi, 4/3pi).*0840

*Again here, you can use the formulas that we learned to calculate the coordinates of the centroid.*0857

*Or you can exploit the symmetry.*0862

*I guess you could not have exploited the symmetry if you did not already know the y coordinate.*0868

*We did use the formulas to calculate the y coordinate.*0870

*So you will have to do a bit of calculus at some point.*0874

*Having done it in the previous problem we can make our lives simpler in this problem and exploit that symmetry.*0876

*We will do some more problems later.*0886

*Welcome back and we are trying some more examples of arc lengths.*0000

*And other problems involving parametric equations.*0005

*We have here the length of a curve given by x(t) = cos*^{2}(t) and y(t) = sin^{2}(t) as t goes from 0 to π/2.0010

*Remember our arc length formula is x'*^{2} + y'^{2}, take the square root of that and integrate it.0023

*Let us calculate x'.*0029

*X is cos*^{2}(t), so that would be 2 × cos(t) × -sin(t) using the chain rule there,0034

*The - sign comes from the derivative of cosine.*0044

*y'(t) is 2 × sin(t) × derivative of sin which is cos(t).*0049

*If we square each one of those, x'(t)*^{2} is 4cos^{2}(t) sin^{2}(t).0059

*y'(t)*^{2} is 4sin^{2}(t) cos^{2}(t).0070

*If we add those up, x'*^{2} + y'^{2},0082

*Well those are the same thing.*0091

*We just get the sqrt(8sin*^{2}(t)cos^{2}(t)).0094

*The sqrt(8) is 2×sqrt(2).*0102

*The square root of sin*^{2} and cos^{2}, since sin and cos are positive when t is between 0 and π/2,0105

*This is sin(t) × cos(t).*0117

*So, that is what we want to integrate.*0122

*The integral from t=0, to t=π/2.*0125

*I will write the 2×sqrt(2) on the outside.*0130

*sin(t) cos(t) dt.*0138

*This integral is not too bad.*0143

*We can use the substitution u = sin(t).*0145

*The reason that works so well is because du = cos(t) dt.*0149

*So, what we have here is the integral of u du.*0156

*That gives us u*^{2}/2.0160

*Now to convert things back into t.*0169

*2×sqrt(2) × sin*^{2}(t)/2.0172

*Evaluated from t=0 to t=π/2.*0180

*Those 2's cancel each other so we get sqrt(2) × sin(π/2)*^{2},0188

*That is just one.*0195

*Minus sin(0)*^{2}, which is just 0.0197

*So our arc length is just sqrt(2).*0200

*Again there, the calculus worked nicely.*0206

*What we did was find x' and y' and plugged them into this Pythagorean Formula.*0208

*Then we integrated to get the answer.*0215

*There is actually another way to see how this problem works.*0218

*We can see through this problem geometrically.*0220

*If you actually try graph the path that these equations are describing, *0226

*Notice that cos*^{2}(t) + sin^{2}(t) = 1.0228

*So, cos*^{2}(t) = x, sin^{2}(t) = y.0240

*So this path is actually taking place on the line y+x = 1.*0245

*So there is that line,*0255

*And if you plug in t=0, then x=1 and y=0.*0257

*There is t=0.*0264

*At the point (1,0).*0266

*If you plug in t - π/2, then y = 1 and x=0.*0269

*So, there is t=π/2 at the point (0,1).*0277

*So, what this path is really doing is just following a straight line from (1,0) to (0,1).*0286

*Of course the length of that line is sqrt(2).*0293

*That is kind of a geometric check on the calculus we just did.*0297

*Let us try one more example.*0000

*We want to find the length of the curve given by x(t) = 7 + 2t,*0002

*And y(t) = e*^{t} + e^{-t}, where t goes from 0 to 1.0006

*Again we want to find x' and y', square each one, add them up and take the square root.*0013

*x'(t) = 2, y'(t), y = e*^{t} + e^{-t}, so the derivative of e^{t} is just e^{t}.0020

*The derivative of e*^{-t} is e^{-t} × the derivative of -t, which is -1.0037

*So we get -e*^{-t}.0047

*So x'(t)*^{2} + y'(t)^{2}.0050

*Well, 2*^{2} is just 4. 0058

*Now, if we square y'(t), then we are going to follow the formula (a + b)*^{2},0060

*Is a*^{2} + 2ab + b^{2}.0074

*Here a is e*^{t}, so (e^{t})^{2} is e^{2t}, b is -e^{-t},0080

*So this is -2ab, well ab is e*^{t} × e^{-t},0090

*Then + b*^{2} is e^{-t}, so that is e^{-2t},0102

*But look at this e*^{t} × e^{-t}, is e^{0} which is 1.0111

*This is just -2.*0116

*We also have a 4 here, so what we get is e*^{2t} - 2 + 4, is just + 2.0120

*+ e*^{-2t}.0131

*The clever thing to do here is to write that 2 as 2e*^{t} × e^{-t} again + e^{-2t}.0134

*This is (e*^{t} + e^{-t})^{2}.0145

*The square root of all that, x'(t)*^{2} + y'(t)^{2}.0155

*The square root cancels that for a fixed square,*0164

*So we get e*^{t} + e^{-t}.0168

*That is what we want to integrate to find the arc length.*0173

*We take the integral from t=0 to t=1.*0175

*We are getting those bounds from the original problem of e*^{t} + e^{-t} dt.0181

*That integral is not too bad.*0192

*The integral of e*^{t}, is just e^{t} itself.0194

*The integral of e*^{-t}, is e^{-t}/the derivative of -t, which is -1.0197

*That is the same as multiplying by -1.*0207

*So this is e*^{t} - e^{-t}.0210

*Then we want to evaluate that from t=0 to t=1.*0213

*That is e*^{1} - e^{-1} - e^{0} + e^{0}.0221

*The e*^{0}'s are both 1 so those cancel each other out.0232

*This is e - e*^{-1}.0235

*I will write this as 1/e, and that is our answer for the arc length.*0240

*Again, what we did there, we looked at the x and y that we were given,*0246

*We took the derivative of each one, squared them, added them up, took their square root.*0252

*And integrated to get our answer for the arc length.*0257

*Thanks for watching, this has been educator.com.*0261

*Hi this is educator.com and we are here to talk about parametric curves.*0000

*The idea about parametric curves is that you are given the equation x(t) and y(t) and those define how a point is moving around in the plane.*0007

*The x(t) gives you the x coordinate at a particular time and the y(t) gives you the y coordinate at a particular time.*0019

*There are basically two calculus problems associated with parametric equations.*0026

*One is to find the tangent line to occur at a particular point.*0030

*The way you are going to do that is we will fine the slope of the tangent line by looking at d(y) dt.*0039

*So, just looking at the derivative of the y equation, and dividing by d(x) d(t), which is the derivative of the x equation.*0048

*That will give us the slope of the tangent line and we will also know one point on the tangent line.*0052

*We can use the point slope equation to find the slope of the tangent line. *0059

*The other equation that you use with parametric equations a lot is *0063

*You find the length of a curve and essentially the formula from that comes from the Pythagorean distance formula.*0075

*You look at x'(t*^{2} + y'(t^{2}) and then you find the square root of that.0076

*That is a unit of arc length representing the length travelled in a very small amount of time.*0084

*Then you integrate that from your starting time to your ending time and that formula represents the total length of the curve.*0089

*Let us try this out with some examples*0097

*The first examples is the equations are x(t) = t+1, y(t) = t*^{2}.0100

*Maybe I will just graph a couple of points there.*0109

*If t = 0 then x = 1 and y = 0.*0111

*If t = 1 then x = 2 and y = 1.*0118

*If t = 2 then x = 3 and y = 4.*0125

*This point is travelling along a parabolic path here.*0133

*What we are asked to do is find the tangent line at t=1.*0138

*At t=1, remember, x = 2 and y = 1.*0143

*We are trying to find the tangent line at that point right there.*0149

*We need to find the slope but our slope is d(y) dt/d(x) dt.*0156

*Now dy dt, since y = t*^{2} is 2t.0170

*Dx dt, since x = t+1 is just 1.*0174

*That is 2t, and when we plug in t = 1, that gives us the slope of 2.*0180

*Now we have the slope and we have a point and it is just an algebra problem to find the equation of a line.*0189

*We use the point slope formula, y - y*_{0}, which is 1 here, is equal to the slope × x - x_{0} which is 2.0195

*This is 2x - 4.*0207

*We get the equation of the tangent line, y = 2x-3.*0213

*To recap there, what we are trying to do is to find the tangent line to a curve that is defined by parametric equations.*0220

*What we did was we plugged in the time value we were given, t =1 to get a point on the tangent line.*0230

*Then we used our equation for the slope dy dt/dx dt.*0240

*We figured those out using the equations for x and y that we were given.*0245

*We plugged in the same value of t and we got our slope and then we had a point on our slope,*0250

*And we could use the old point-slope formula to find the equation for our tangent line.*0259

*Let us find another tangent line.*0263

*This time the curve is x(t) = cos(t) y(t) = sin(t).*0265

*You will hopefully recognize that as the equations for a circle because cos*^{2}(t) + sin^{2}(t) = 1.0272

*Those are the equations that define a point moving around in a circle.*0282

*We want to find the tangent line at the point, (sqrt(3/2), 1/2).*0288

*That is about right there on the circle*0294

*The difference between this one and the previous one is we have not been given a t value.*0297

*We have to figure out what value of t gives us the point (sqrt(3/2), 1/2).*0302

*What value of t if you plug it in to cos(t) sin(t) gives us the sqrt(3/2)?*0308

*The answer is pi/6 because the cos(pi/6) is sqrt(3/2) and the sin(pi/6) is 1/2.*0318

*So, we know that t is pi/6.*0328

*Again, to find the slope, we use dy dt/dx dt.*0330

*Now y = sin(t) so the derivative there is cos(t).*0340

*X = cos(t), the derivative of that is -sin(t).*0345

*Then we plug in the value t = pi/6 to get a number for the slope so the cos(pi/6) = sqrt(3/2)*0354

*The sin(pi/6) = 1/2, and we still have our negative there.*0364

*The two's cancel so we get our slope is -sqrt(3).*0370

*That gives us the slope and we also have a point, so I will plug those into the point slope formula.*0376

*y - 1/2 = -sqrt(3) x - sqrt(3/2).*0380

*Those values are coming from plugging the points in there.*0394

*We can simplify this a bit.*0396

*This is = sqrt(3)x.*0398

*The minuses cancel each other out plus sqrt(3) × sqrt(3) = 3.*0400

*Then we can bring this half over to the other side so we get y = -sqrt(3)x + 3/2 + 1/2, is just 2.*0410

*We get our tangent line to be y = -sqrt(3)x + 2.*0429

*The way that this problem was different from the previous one was that we were not given a value of t.*0435

*We had to look at the point we were given and we had to figure out what the value of t should be.*0440

*From then on we used the same formulas to find out the slope and the equation of the tangent line.*0450

*By the way, this problem you can also check the answer geometrically if you draw that tangent line.*0454

*It forms a 30-60-90 triangle with the x and y axis.*0464

*That is 30 degrees and that is 60 degrees.*0470

*We know what those angles are because the tangent line is perpendicular to the radius of the circle there,*0476

*We also know that the radius of the circle is 1 so we have a smaller 30-60-90 triangle in there.*0484

*With a short side of length 1, so the long side has length 2.*0501

*That confirms that the y intercept of this tangent line = 2.*0506

*That is a little check on our work using some trigonometry and no calculus at all*0511

*Let us try an example of arc length now.*0519

*We are given the curve x(t) = 6 - 2t*^{3}, y(t) = 8 + 3t^{2}.0520

*We want to find the length of that curve from t = 0 to t = 2.*0530

*Remember our arc length formula says you want to take the integral of sqrt of x'(t)*^{2} + y'(t)^{2} dt. 0536

*x'(t), the derivative of x if x is 6 - 2t*^{3},0549

*You take its derivative and the 6 just goes away, so the derivative is -6t*^{2}.0556

*y'(t), if y' is 8 + 3t*^{2}, again the 8 does not have any effect.0568

*The 3t*^{2} you take the derivative and you get 6t.0575

*What we want then is sqrt(x')*^{2}, well that is (-6t)^{2}, so that is 36t^{4}0584

*+ y'*^{2} is 36t^{2}.0596

*That can simplify because we can pull a 36 out and just get 6, *0600

*we can pull a t*^{2} out and get t.0610

*Under the radical, what we have left is just t*^{2} + 1.0615

*The arc length = integral from t=0 to t=2 of 6t × sqrt(t*^{2}+1) dt.0620

*Now, this integral is not too bad because what we can do is make a little substitution.*0645

*u = t*^{2} + 1.0650

*Then, du = 2t, dt.*0654

*The reason that works so nicely is that we already have the t and the dt, so we basically have du.*0658

*In fact, 6t dt, is just 3 du.*0666

*Then, we still have the integral from t = 0 to t = 2 of now the sqrt(u) du.*0674

*You can think of the square root as u*^{1/2}.0683

*To integrate that we get u*^{3/2}/3/2 which is the same as multiplying by 2/3.0687

*Then we still have that 3 on the outside.*0698

*This is evaluated from t=0 to t=2.*0701

*I have to convert the u's back into t's*0706

*These 3's cancel, so we get 2 × u was (t*^{2} + 1)^{3/2} evaluated from t=0 to t=2.0709

*This gives us 2 now if we plug in t=2, we get 2*^{2} + 1, so that is 5^{3/2}.0727

*- 0*^{2} + 1, so that is 1^{3/2}.0737

*We can simplify that to a little bit.*0744

*5*^{3/2} is the same as 5 × 5^{1/2}.0746

*1*^{3/2} is just 1.0749

*We get our final answer for the arc length there.*0758

*What made this problem work is having this formula that came from the pythagorean theorem.*0762

*We just take the x(t) and the y(t) that we are given and we plug them into this formula which involves computing a couple derivatives,*0768

*Simplifying a little bit if we can, then we integrate and we get out answer.*0775

*We will try a couple more examples later.*0782

*OK, let us try another area problem.*0000

*We want to find the area inside the graph of r = 8sin(θ) and outside the graph of r = 4.*0003

*This one is a little tricky because we have not been given limits of integration,*0011

*And we have been given two different functions.*0016

*Somehow, we have to figure out what those functions look like.*0018

*Also, figure out the limits of integration ourselves.*0022

*r = 4 is probably the easier one, so I will start with that.*0027

*That is just all points at radius 4 from the origin.*0031

*That is just a circle.*0036

*r = 8sin(θ) is a little trickier.*0039

*If you graph r = 8sin(x), you get an exaggerated sin curve.*0042

*That goes up to 8 and down to -8.*0053

*Comes back to 0 at π, and again at 2π.*0061

*If you graph r = 8 sin(θ), *0066

*That starts out when θ = 0, it starts out at 0.*0072

*When θ = π/2, it goes up to 8.*0084

*That is 8.*0092

*When θ = π, it comes back to 0.*0094

*When θ is 3π/2, it is gone to -8.*0101

*It is just, let me show this in blue, retracing itself again.*0110

*When θ gets to 2π, it has come back to 0.*0116

*That is what the graph of r = 8sin(θ) is.*0131

*Its intercept on the π/2 axis is 8.*0140

*I want to show those 2 graphs together.*0144

*Remember r = 4 had a radius of 4.*0146

*Let me put that graph on there.*0150

*We are actually looking at two circles here.*0156

*We want to find the area inside the graph of r = 8sin(θ)*0162

*Outside the graph of 3=4.*0167

*Let me try to color that area in, in red.*0170

*That is the area that we are trying to find.*0175

*This means we have to figure out where those two graphs intersect.*0177

*Then integrate from 1 intersection point to the other.*0185

*To figure out where those two graphs intersect, I am going to set them equal to each other.*0193

*4 = 8sin(θ).*0197

*That means 1/2 = sin(θ).*0203

*There are two angles that have sin = 1/2.*0207

*That means θ = π/6, and 5π/6.*0210

*You can kind of see that in the picture.*0216

*This is π/6.*0218

*This is 5π/6.*0221

*What we are going to do is integrate using our area formula.*0225

*I think what I am going to do is just find 1/2 of this area.*0235

*I am just going to find that segment.*0240

*I will color that in black.*0243

*I will find that area and then multiply it by 2, then integrate it from π/6 to π/2.*0245

*Now I want to find the outside area - the inside area.*0256

*Well our formula for area is f(θ)*^{2}/2.0262

*Here the outside area is 8sin(θ).*0267

*The outside curve is 8sin(θ),*0269

*So I am going to find 8sin(θ)*^{2}/2, that is the outside area,0272

*- the inside area which is just 4*^{2}/2.0284

*That is what I am going to integrate.*0291

*That is nice because the 2's cancel right away.*0294

*So we get the integral from π/6 to π/2 of 8 sin(θ)*^{2},0298

*So that is 64sin*^{2}(θ) - 4^{2} is 16.0303

*Remember the 2's cancel on the outside d(θ).*0313

*That is the integral of sin*^{2}(θ),0319

*We are going to use a half-angle formula to deal with that.*0323

*Remember sin*^{2} is 1 - cos(2θ)/2.0324

*So, 64sin*^{2}(θ) is 32 × 1 - cos(2θ).0330

*We still have - 16 dθ.*0340

*Here we have 32 - 16.*0347

*I will put those together and get a 16 - 32 cos(2θ), just 16(θ).*0352

*Now the integral of cos(2θ) is sin(2θ)/2.*0362

*The integral of 32cos(2θ) is 16sin(2θ).*0370

*We said we wanted to integrate this from θ = π/6 to θ = π/2.*0378

*We will plug those bounds in, and we get 16 × π/2 minus,*0388

*Now, if we plug π/2 into sin(2θ),*0397

*That is giving us sin(π), and that is just 0, - 16/π/6.*0401

*Which is 16 × π/6 for now + 16 × sin(2) × π/6.*0413

*Which is the sin(π/3).*0423

*We can simplify this.*0428

*π/2 - π/6 = π/3.*0430

*If we combine those two terms, we get 16 π/3.*0433

*Now, the sin(π/3) is sqrt(3)/2.*0441

*So 16 × sin(π/3) = 8×sqrt(3).*0447

*We get our answer is 16 π/3 + 8×sqrt(3).*0460

*This problem was pretty tricky but it was really the geometry that made it tricky and not the calculus.*0465

*We were given two curves and we were not really told how to handle them.*0470

*The good thing to do is to start out by graphing the two curves, putting them together on the same graph,*0475

*Then trying to find these intersection points.*0480

*We actually found the intersection point algebraically by setting the two equation equal to each other.*0486

*Once we found the intersection points, those gave us the bounds on the integrals.*0493

*Then we used this area formula.*0498

*Remember the area was f(θ)*^{2}/2 dθ.0500

*Since we had to find the area between two curves, we looked at the area of the outside one - the area of the inside one.*0505

*Then we worked out the integration and we got our answer.*0514

*So our last example is another length of the polar curve r = 3+3cos(θ).*0000

*Again, this is one where we are not given limits of integration.*0008

*We have to figure it out by ourselves.*0011

*The way we are going to figure it out is by graphing the thing.*0014

*So, the way we can graph it is we can plug in some easy values of θ.*0028

*When θ = 0, cos(θ) = 1 so this is 6.*0033

*When θ = π/2, cos(θ) has gone down to 0, so we are just talking about r = 3.*0043

*By the time θ = π, cos(θ) = -1, so that has gone down to 0.*0054

*At 3π/2, cos(θ) is back up to 0, so r has come back up to 3.*0063

*At 2π, cos(θ) is back up to 1.*0073

*So 3 + 3cos(θ) is back up to 6. *0084

*That is the curve we are looking at.*0088

*We are trying to find the length of that curve.*0091

*I think it will be easier if we just find the length of the top part of this curve, the part I am showing in red right now.*0096

*So, we are actually going to integrate from 0 to π, and then we will multiply our final answer by 2.*0100

*We are going to do 2 × the integral from 0 to π.*0110

*Now, remember we have to use the arc length formula, which is this Pythagorean Formula.*0120

*We take sqrt(f'(θ)*^{2} +f(θ)^{2}).0126

*f'(θ) is very easy.*0139

*The first 3 goes away because it is just a constant.*0143

*The derivative of 3cos(θ) is just -3sin(θ).*0147

*If we square that out we get 9sin*^{2}(θ).0156

*f(θ)*^{2} is a little bit more messy.0162

*We have to square out 3 + 3cos(θ).*0164

*That is 9 + 3 × 3 × 2 is 18 cos(θ) + 9cos*^{2}(θ) dθ.0168

*That looks like it is going to be a pretty messy integral,*0187

*But good things happen here because we have 9sin*^{2}(θ) and 9cos^{2}(θ) 0190

*Those combine just to be 9.*0196

*We get 2 × integral from 0 to π of square root of,*0198

*Well the cos*^{2} and the sin^{2} give us 9, 0206

*And we have another 9 right there,*0209

*So that is just 18 + 18cos(θ) dθ.*0212

*Now, I can simplify this a little bit if I pull out the sqrt(18), so 2 sqrt(18) × integral from 0 to π.*0223

*Of 1 + cos(θ) dθ.*0236

*This is kind of a tricky integral.*0242

*There is a trick for doing integrals like this and it may not be completely obvious.*0246

*The trick is to remember the half angle formula.*0252

*Let me write that over here.*0255

*cos*^{2}(u), remember that that is 1 + cos(2u)/2.0258

*If I solve that for cos(u), that says that cos(u) = sqrt(1+cos(2u)/2).*0268

*That is pretty nice because we have the sqrt of 1 + cos(θ).*0281

*We almost have this half angle formula set up for us.*0286

*All we have to do is let the θ be 2u.*0290

*dθ = 2du.*0297

*2 sqrt(18), what I would like to do is create this formula exactly.*0310

*I am going to put a sqrt(2) in here.*0322

*Then, multiply on a sqrt(2) on the outside.*0325

*So we get the sqrt(18) × sqrt(2).*0330

*Now the integral of 0 to π, of 1 + cos,*0336

*Now θ is 2u so this is 2u/2.*0340

*dθ is 2 du.*0347

*Let me simplify this a bit.*0355

*I am going to combine this 2 and this 2 and get a 4.*0357

*sqrt(18) × sqrt(2) is sqrt(36).*0363

*Now we have the integral from,*0365

*This 0 and π, those were values for θ*0374

*If θ = π, then u, since θ = 2u, u will be π/2.*0379

*If θ = 0, then 0 will also be 0.*0388

*In terms of u, this is the integral from u = 0 to u = π/2.*0394

*Now, sqrt(1+cos(2u)/2), the whole point of making this substitution was that that would convert into cos(u).*0400

*This is just cos(u).*0409

*This 2 I already moved to the outside du.*0413

*That took what looked like a very difficult integral, and converted it into a very easy one.*0419

*The sqrt(36) is just 6, so this is 24 × integral of cos(u) is sin(u).*0423

*We are evaluating this from u=0 to u=π/2.*0434

*That just gives 24 × sin(π/2) = 1 - sin(0) = 0.*0440

*Our final answer there is 24.*0452

*Again this qualifies as probably a pretty tricky polar coordinate problem.*0457

*What we are given there is just a polar curve.*0460

*And we were not given anything about any boundaries.*0465

*We had to first draw the graph of the curve and figure out what we were dealing with.*0470

*Once we draw the graph of the curve, it is not too hard to find out which part of the curve we want to integrate,*0479

*And multiply by an appropriate factor to set up our integral.*0485

*Then we used the formula for arc length, which is this Pythagorean Formula,*0492

*With sqrt(f'(θ)*^{2} + f(θ)^{2}).0500

*We plugged that into the integral, simplified it a little,*0505

*And then we discovered that what looked like a very difficult integral, sqrt(1+cos(θ),*0511

*But in order to handle that, we remembered this half angle formula, cos*^{2}(u) = 1 + cos(2u)/2.0516

*We did a little manipulation to get the integral to match the half angle formula,*0527

*And once we did, it turned into a fairly easy integral in terms of u.*0533

*Then it was easy to finish and get the entire length of the curve.*0540

*Hi, this is educator.com and we are here to talk about polar coordinates.*0000

*The idea of polar coordinates is that we are not going to keep track of things in terms of x's and y's anymore.*0006

*Instead, we are going to keep track of points in terms of the radius r and the angle θ.*0014

*Every point now will have coordinates in terms of r and θ.*0025

*We will talk about functions r = f(θ).*0030

*There are sort of two places that calculus comes in in polar coordinates.*0037

*If you have a function, here I am talking about plugging in different values of θ and getting different values of r as your output.*0042

*There are two things you might be interested in calculating.*0054

*One is the area, before we calculate under, here it makes more sense to talk about calculating the area inside a curve.*0060

*The equation we have for that is the integral from, these are values of θ, θ = a to θ = b.*0070

*Of f(θ*^{2}/2).0085

*If you are given f(θ), this f(θ) is just whatever the r is, you square that and divide by 2.*0090

*Then you take the integral with respect to θ.*0098

*The second thing we are interested in calculating is the arc length.*0103

*What is the length of that curve?*0111

*The way you figure that out is you find f(θ*^{2}), f'(θ^{2}).0114

*Then you use the Pythagorean theorem, you add them, take their square root,*0122

*And integrate from θ = a to θ = b.*0127

*Let us try those out using some examples.*0135

*The first example is the area inside the graph of r = θ, from θ going to 0 to pi/2.*0138

*If r = θ, then when θ = 0, r is just 0 but as θ increase to pi/2, r gradually increases.*0145

*We are trying to find that area there.*0160

*Let us work it out.*0165

*Our formula for the area is f(θ*^{2}/2) dθ0170

*In this case, our f(θ) is just θ itself, so this is the integral of θ = 0 to pi/2.*0180

*Of θ/2 dθ.*0192

*The integral of θ*^{2} is θ^{3}/30194

*This whole thing is theta*^{3}/3/2, so over 6, evaluated from θ = 0 to θ = pi/2.0196

*That is just pi/2*^{3}/6.0211

*Now, 2*^{3} is 8 and 8 × 6 is 48 so our answer is pi^{3}/48.0220

*That represents that area inside that curve.*0226

*To recap here, what we did was look at the function we were given,*0235

*R = f(θ), and then we just plugged it into this integral formula f(θ*^{2})/2.0242

*Then we worked out the integral.*0249

*Let us try that out with a slightly harder example.*0252

*We want to find the area inside one loop of the graph of r = cos(2θ).*0255

*Perhaps the first thing that makes this example difficult is we have not been told what the boundaries of θ are.*0260

*We really need to look at a graph to figure this out.*0266

*What does a graph of cos(2θ) look like.*0273

*Well if we graph cos(x) to warm up, it starts at 1, it goes down to -1, then it comes back to 1 at 2pi.*0276

*If we graph y = cos(2x), we get a graph with the same basic shape, but it oscillates twice as fast.*0290

*If that is pi, and that is 2pi, it does a complete period in the space of pi, and then another complete period by 2pi.*0305

*If we graph r = cos(2θ), we take that graph and we wrap it around a circle.*0320

*In the sense that when θ = 0, we start out at 1 at radius 1.*0330

*By the time θ gets to be pi/4, it has gone to radius 0.*0341

*This goes down to radius 0.*0350

*At pi/2, r = -1 so you know that if pi/2 is up here, the radius has gone down to -1.*0353

*It comes back to 0.*0365

*Now we are graphing this part of the graph.*0368

*By the time we get up to pi it has come up to 1 again, so here is pi in this direction.*0370

*Then between pi and 5pi/4 it goes back down to 0.*0379

*That is graphing that part of the graph and if we graph this part of the graph going from pi/4 to 3pi/2,*0387

*It is negative so the graph ends up here and then it spirals back down to 0.*0400

*And back to 1 again when it comes back to 2pi.*0412

*We get this interesting 4-leaf clover and we are trying to find the area inside one of those loops.*0417

*A good way to do it might be to find that area inside there, inside half of one of those loops.*0425

*That is between θ = 0 and θ = pi/4.*0432

*Then we will multiply our answer by pi/2.*0441

*Our area is 2 × the integral from 0 to pi/4.*0445

*Remember f(θ*^{2})/2 so our f(θ) = cos(2θ) this is cos^{2}(2θ)/2 dθ0454

*Our 2's will cancel, that is convenient, so we get the integral from 0 to pi/4.*0468

*Remember how to integrate cos*^{2} of something.0475

*You write that as 1+cos(2 × that thing), but the thing is 2θ, so this is actually 1 + cos(4θ). *0478

*All of that over 2.*0484

*So this 2 over here was not the 2 above, that cancelled, but this came from the half angle formula.*0492

*Now we integrate this with respect to θ.*0497

*I am going to pull the 1/2 outside now, we get 1/2.*0500

*Now we have to integrate 1 + cos(4θ).*0505

*The integral of 1 is just θ + the integral of cos(4θ) is just sin(4θ)/4.*0509

*Then we evaluate that from θ = 0 to θ = pi/4.*0525

*Then we get 1/2 of pi/4.*0540

*Plugging in θ = pi/4 + 1/4 sin(4θ).*0545

*Sin(4θ) is sin(pi) which is just 0.*0553

*- plug in θ = 0, we get 0 - sin(4×0) is just 0 again.*0557

*Our final answer is just pi/8.*0566

*The tricky part there is that we were not given the boundaries of integration.*0570

*We really had to look at the graph of r = cos(2θ)*0580

*Then interpret from the graph what useful boundaries of integration we could use to find the area.*0584

*Once we found the boundaries of integration, we just plugged it into the formula f(θ*^{2})/2.0590

*Then we worked out the integral and it was not too bad.*0600

*We will try some more examples later, this is educator.com.*0602

*Another example is the length of the polar curve, r = e*^{5θ} as θ goes from 0 to 2pi.0610

*Fortunately we have been given the limits of integration.*0618

*We are going to set up our arc length formula.*0623

*Which remember is f'(θ*^{2}) + f(θ)^{2} dθ.0629

*Our f'(θ) is well, r = e*^{5θ}.0640

*f'(θ) would be (5e*^{5θ})^{2} + just f(θ^{2}) so that is (e^{5θ})^{2}.0650

*We want to square root that and integrate it.*0665

*5 squared is 25, e*^{5θ})^{2} is e^{10θ}.0670

*Square root that and integrate it.*0685

*This is sqrt(26e*^{10θ}).0690

*I am going to pull the sqrt of 26 all the way out of the integral.*0704

*Then we have the sqrt(e*^{10θ}) which is just e^{5θ}.0710

*We want to integrate this from θ = 0 to 2pi.*0718

*Now the integral of e*^{5θ} is just e^{5θ}/50728

*We want to evaluate that from θ = 0 to 2pi.*0742

*I can write this as sqrt(26/5) × *0750

*If you plug 2π into e*^{5θ} you get e^{10π}.0755

*If you plug 0 into e*^{5θ} you get e^{0}, which is just 1.0763

*That is our answer for the arc length.*0770

*The key to that problem is recognizing that it is a length problem and then going to the arc length formula.*0776

*Which is this Pythagorean formula f'(θ)*^{2} + f(θ)^{2} square of that and then integrate it.0784

*We use the f(θ) that we are given, work it through, and then plug in the limits that we are given.*0792

*We are here to try some more examples of finding limits of sequences.*0000

*The first one is the sqrt(n*^{2} + 6n) - n.0006

*There is a trick here when you see square roots and you see them being added and subtracted with things.*0014

*The trick is to multiply the top and bottom by the conjugate.*0023

*What I mean by the conjugate is if you have an expression like a + b,*0040

*You multiply it by a - b/a - b.*0046

*So, if you have a + b, you multiply it by a - b,*0052

*If you have a - b, you multiply it by a + b.*0058

*The point of doing that is to take advantage of this difference of squares formula.*0062

*Remember a + b × a - b = a*^{2} - b^{2}.0067

*What that does is it squares out the expressions for you and that can often simply a square root formula.*0075

*Let us try that out with this example.*0083

*We have sqrt(n*^{2} + 6n) - n.0088

*I want to multiply that by its conjugate, the sqrt(n*^{2}+6n)+n.0094

*To pay for that multiplication, I have to do some division by the same expression,*0104

*sqrt(n*^{2}+6n)+n.0110

*The point of that is that in the numerator we get this a*^{2}-b^{2} expression.0115

*We get the sqrt(n*^{2}+6n)^{2}, so that is just n^{2}+6n.0124

*Minus n*^{2}.0134

*Then that same denominator.*0140

*sqrt(n*^{2}+6n) + n.0142

*This simplifies nicely down to 6n/sqrt(n*^{2}+6n) + n.0150

*We are not done yet with this because we still have something that is not so obvious.*0160

*We have 6n in the numerator, some square root stuff in the denominator.*0167

*The key thing to remember is to go through and look at the top and bottom and see what the biggest thing is in both the numerator and the denominator.*0171

*In the top I see the biggest thing here is n.*0180

*In the bottom, well I see sqrt(n*^{2}), that is approximately n.0186

*You know there is a 6n next to it.*0190

*The n*^{2} is bigger than the 6n, so the sqrt(n^{2}) is appromixately n.0194

*Then of course we have an n in the denominator.*0200

*The biggest thing top and bottom would be n.*0204

*What I want to do is divide top and bottom by n.*0206

*That will give me just 6 in the numerator and then in the denominator,*0213

*sqrt(n*^{2}+6n)/n + 1.0220

*Sqrt(n*^{2}+6n), I can write that as,0230

*Well I can make that a big square root.*0236

*n*^{2}+6n.0241

*Now if I want to make this n part of the square root, I have to change it to n*^{2}.0244

*That in turn becomes 1+6/n.*0248

*This whole thing becomes 6/sqrt(1+6/n)+1.*0258

*Now it is in a format that is very conducive to letting n go to infinity. *0269

*As n goes to infinity, the 6/n goes to 0.*0275

*We get 6/sqrt(1) which is just 1 + 1.*0278

*So, 6/2 and we get our answer, that the sequence converges to 3.*0283

*There are a couple clever idea in that one.*0291

*The first was that we had a square root of something minus something.*0296

*The trick to handling that is by multiplying that by its conjugate.*0301

*We multiplied it by the same expression with a plus there.*0306

*To pay for that multiplication, we had to do a division.*0311

*The point of multiplying by the conjugate,*0315

*Is we can exploit this difference of squares formula.*0319

*We get the sqrt(n*^{2}+6n)^{2}, that gets rid of the square root there.0324

*We get n*^{2}, and we get this difference of squares formula,0333

*Which simplifies down into 6n.*0339

*That was the first trick, multiplying by the conjugate.*0342

*The second trick was looking at the expression we got, and then dividing top and bottom by the biggest term that we saw.*0346

*Then recognizing that when we had the square root of n*^{2}, that was really approximately n,0350

*The biggest term in top and bottom was n.*0360

*We divide top and bottom by n.*0362

*Then do some algebra to simplify it, and we finally get our answer.*0366

*Our final example here is to calculate the limit of 1 + 1/n*^{2n}.0000

*This is an example that is quite common in Calculus homework sets.*0007

*It is also one that students often make mistakes on.*0013

*People think that as n goes to infinity, 1+1/n ought to go to 1 + 0.*0016

*So, you are 1+0*^{infinity}, so that is 1^{infinity},0026

*And people think that goes to 1.*0035

*That is a very common response that students give to problems of this form.*0038

*That is flat wrong.*0043

*Let me emphatically warn you away from making that mistake.*0046

*This is actually something a lot more subtle.*0053

*If you see an expression that looks like 1 to the infinity,*0056

*There is something deeper going on there.*0060

*We are actually going to need some more sophisticated tools for that.*0063

*The trick here is when you have an expression of the form a*^{b},0070

*Often, a very useful way to write that is,*0078

*Write it in the form, e*^{ln}(a^{b}).0087

*The point of that is that the ln(a*^{b}),0097

*Is the same as b*^{ln(a)}.0100

*Natural log is a way of separating an exponent out to the side of an expression,*0104

*So that it is no longer an exponent.*0113

*Then we have to put an e in there to cancel off the natural log.*0117

*The e is kind of paying for the natural log, *0120

*While the natural log is doing the actual work of moving the exponent out of the exponent,*0123

*So that we just have a multiplication.*0127

*Then, what you often do is you look at the exponent,*0131

*Work on the exponent,*0138

*Now, we have two things multiplied by each other.*0147

*Often, that turns out to be a 0 × infinity situation.*0152

*Which is something that can be manipulated into a l'Hopital's rule situation.*0158

*It is often not l'Hopital's rule immediately, but you can manipulate it into 0/0 or infinity/infinity.*0165

*Then you can use l'Hopital's rule.*0172

*Let us try that out with this example.*0175

*We are going to write it as e*^{ln}(1+1/n)^{2n}.0177

*That is e*^{2n}, we can pull the exponent out, × ln(1+1/n).0186

*Now, I am just going to look at the exponent, and we will come back and deal with the e part later.*0195

*2n ln(1+1/n), *0201

*Notice that 2n, 2n goes to infinity, as n goes to infinity.*0205

*ln(1+1/n), well 1/n goes to 0.*0215

*So we get ln(1), ln(1) = 0.*0221

*So this is an infinity × 0 situation.*0228

*Not quite ready for l'Hopital's rule yet, but we can write it as.*0230

*2 ln(1+1/n)/1/n.*0238

*What I did was I took this and, and I pulled it down into the denominator.*0246

*Now, remember ln(1+1/n) goes to 0, and 1/n itself goes to 0.*0253

*So what we have is a 0 over 0 situation,*0258

*And that is something where it is legitimate to use l'Hopital's rule.*0262

*So, we are going to do l'Hopital's rule,*0270

*Which means we are going to take the derivative of both the top and bottom separately, *0276

*Then we find the limit of what we get.*0278

*Let us take the derivative of 2 ln(1+1/n).*0281

*Derivative of ln(x) = 1/x.*0287

*This is 2/1+1/n × the derivative of 1 + 1/n by the chain rule.*0290

*That is the derivative of 1 + 1/n is -1/n*^{2}.0300

*Remember derivative of 1 is 0.*0306

*Derivative of 1/n, that is n*^{-n},0309

*So its derivative is -1n*^{-2}.0312

*In the denominator, the derivative of 1/n, is just -1/n*^{2} again.0315

*This now simplifies, the -1/n squares cancel.*0324

*We get 2/1+1/n.*0328

*Now that is something very easy to take the limit of.*0333

*As n goes to infinity, the 1/n goes to 0, so that is 2.*0337

*We are not finished that, remember this was all the exponent on the e.*0344

*The actual answer is, the limit is e*^{2}.0348

*That sequence converges to e*^{2}.0357

*Let us go back to the beginning.*0360

*If you had made this mistake of thinking, oh that is just 1*^{infinity},0362

*Which is just 1,*0367

*You would have gotten your answer to just be 1 which is very different from e*^{2}.0370

*That just shows that it really is a bad mistake.*0373

*The problem is a lot more subtle than that.*0378

*The real secret to this one is first to take e*^{ln} of the expression that you are given.0381

*Then you sort out the natural log by remembering that you can pull exponents outside of natural logs.*0390

*Then, the expression that you get turns into an infinity × 0 form.*0400

*Infinity × 0 is not ready for l'Hopital's yet,*0405

*But you can rewrite it as 0/0, by pulling n down into the denominator.*0409

*Then it is ready for l'Hopital's rule.*0416

*L'Hopital's rule says you take the derivative of the top and bottom separately.*0420

*After you do l'Hopital's rule, it simplifies a bit, it is an easy limit,*0425

*Then you just had to remember that you are actually working out the limit of the exponent.*0431

*So to find the limit of the whole thing, you had to do e to that power.*0436

*That is the end of our lecture on sequences.*0440

*This is educator.com.*0442

*Hi, this is educator.com and this is the chapter on sequences.*0000

*We are going to be exploring some different ways to find limits of sequences.*0005

*There are several definitions that lead us up to a big theorem that sometimes be a very powerful way to show that a sequence converges.*0011

*Also, to show its limit.*0021

*The definitions we have are monotonically increasing means that the terms of the sequence are getting steadily bigger.*0023

*The a*_{n} term is getting steadily bigger in other words the a_{n+1} term is bigger than or equal to the a_{n} term for all n.0030

*Monotonically decreasing, same idea except the terms are getting steadily smaller.*0040

*The word monotonic just means the sequence is either monotonically increasing, or monotonically decreasing.*0046

*If either one of those is true then you say the sequence is monotonic.*0055

*Bounded means that all of the terms of the sequence means that all of the terms in absolute value are less than some constant number m.*0060

*The big theorem that we are going to use here is that if you can show that a sequence is both bounded and monotonic,*0070

*Meaning that if you can show it is monotonically increasing or decreasing, and that is bounded,*0078

*Then the sequence converges.*0085

*Let us see how that works with an example.*0088

*The example here is a sequence that starts at the sqrt(2).*0091

*Then the way you get successive terms is you take a previous term is by adding 2 and taking its square root.*0095

*For example, a*_{1} would be the sqrt(2+a_{0}), so 2 + sqrt(2)0103

*a*_{1} would be the sqrt(2+a_{1}), and so on.0110

*We have to show that that sequence converges and then we are going to find its limit.*0120

*We will use our definitions and theorem.*0125

*Our first claim here is that the sequence is bounded above by 2.*0130

*In other words, I claim that every term in the sequence is < or = 2.*0155

*To prove this, note that the first term of the sequence, the a a*_{0} term,0165

*Is certainly less than 2 because the sqrt(2) is about 1.4.*0172

*That is certainly less than 2.*0177

*If a*_{n} is < 2, if one term < 2,0182

*Then I am going to a do a little arithmetic here.*0195

*2 + a*_{n} would then be less than 4.0196

*So the sqrt(2+a*_{n}) would be less than sqrt(4) which is 2.0202

*That is saying that a*_{n} + 1 < 2.0208

*If one term is less than term, then the next term is less than 2.*0213

*We already showed that the first term was less than 2.*0220

*So, every term is less than 2.*0225

*Each term forces the next term to be less than 2.*0230

*That proves the claim that the sequence is bounded above by 2.*0236

*That really is an argument by mathematical induction.*0240

*The second claim is that the sequence is monotonic.*0245

*I claim that this sequence is monotonically increasing.*0260

*So I claim that each term is bigger or equal to the previous term.*0270

*The check whether this claim is true, this is true if and only if *0280

*Well to be monotonically increasing, is to say that a*_{n} + 1 is bigger than or equal to a_{n}.0290

*Let us plug in what a*_{n} + 1 is.0300

*By definition that is 2 + sqrt(2 + a*_{n}), bigger than or equal to a_{n}.0305

*I am going to work with this a little bit.*0310

*I am going to square both sides, this is saying 2 + a*_{n} is bigger than or equal to a_{n}^{2}.0312

*If I move all of the terms over to the right, that is saying 0 is bigger than or equal to a*_{n}^{2} - a_{n} - 2.0322

*I can factor that into (a*_{n} - 2) × (a_{n} + 1).0330

*Let us remember that since a*_{n} < 2.0345

*Over on the left we showed that every term is < 2.*0355

*Since a*_{n} < 2, a_{n} - 2 < 0.0360

*Since all of these terms are positive, a*_{n} + 1 would be bigger than 0.0370

*a*_{n} - 2 × a_{n} + 1 is less than or equal to 0.0377

*Which means that this equation is true and this equation is true.*0388

*a*_{n} + 1 is > or = a_{n}.0395

*What we showed here is that the sequence is bounded and monotonically increasing.*0403

*Our theorem kicks in here and says a bounded monotonic series converges.*0410

*We can invoke the theorem.*0419

*By the theorem the sequence converges.*0428

*That shows the sequence converges, so we have done half of the problem there.*0438

*The sequence converges.*0444

*The second half is to find its limit. *0446

*We will do that on the next page.*0447

*To find its limit, we know now that it does have a limit because we showed that it converges on the previous page.*0450

*Let us call that limit L.*0460

*That is saying that a*_{n} converges to L.0465

*There is a little trick here which is that if a*_{n} converges to L, then the sequence of successive terms must also converge to L.0467

*But a*_{n} + 1 is the sqrt(2 + a_{n})0482

*So, a*_{n} + 1 = sqrt(2 + a_{n}).0491

*But, a*_{n} converges to L, so sqrt(2 + a_{n}) converges to 2 + L.0503

*Now we have a*_{n} converging to L, but it also converges to 2 + L.0512

*So sqrt(2 + L), L must be equal to sqrt(2 + L).*0518

*Now we want to solve that equation and figure out what L is.*0527

*L*^{2} = 2 + L, L^{2} - L - 2 = 0.0533

*(L-2)(L+1) = 0.*0537

*L is either 2, or -1.*0545

*The limit is 1 of those two possible numbers.*0550

*Remember that all of these terms are positive *0555

*So, you cannot have a bunch of positive numbers converging to -1.*0561

*We cannot have a*_{n} converging to -1, so a_{n} must converge to the other possible limit, 2.0572

*Let us recap that problem there. *0589

*We were given this fairly tricky sequence and we had to show that it converges first *0592

*The way we showed it converges was we tried to show that it was bounded and we showed that it was monotonic.*0600

*Monotonically increasing.*0612

*Then those two together allowed us to invoke the theorem which says a bounded monotonic sequence converges.*0614

*Those two conditions allowed us to say that the sequence converges. *0624

*Then once we know it converges we can assume that its limit is L and go through this little algebraic trick looking at what a*_{n} + 1 converges to.0632

*Set those two limits equal to each other because the sequence only converges to one limit.*0646

*Do a little algebra to find the possible limits.*0651

*Then see which limit makes sense.*0658

*The limit that makes sense was 2 and so the limit must be 2.*0662

*Let us try something a little more computational.*0668

*Here we are given the sequence 3n + 5n*^{2}/2n^{2} + 6.0671

*The trick here is to find the largest term in the numerator and the denominator and divide.*0682

*Here we look at the numerator 3n + 5n*^{2}, definitely the largest term there is 5n^{2}.0709

*In the denominator, 2n*^{2} + 6, even though the 6 is bigger than the 2, the n^{2} is going to be much bigger in the long run.0720

*So the 2n is going to be much bigger there.*0730

*In fact, what really matters here is not the coefficients, it is the powers on the n's.*0733

*The fact that we have an n*^{2} in the numerator and the denominator is the important thing.0740

*What we are going to do is divide top and bottom by n*^{2}.0745

*When you divide top and bottom by the same number, that is really multiplying all of it by 1.*0753

*That is legitimate and it is not going to change the limit.*0760

*We get (3n/n*^{2} + 5n^{2}/n^{2})/(2n^{2}/n^{2} + 6/n^{2}).0765

*(3n/n*^{2} is 3/n + 5)/(2 + 6/n^{2}).0783

*In the limit, as n goes to infinity, the 3/n is going to go to 0.*0793

*The 6/n*^{2} is also going to go to 0 so we end up with 5/2 as our limit.0799

*To recap when you have one of these fractional situations*0810

*Where you have an expression in the numerator and an expression in the denominator.*0814

*You want to look at both numerator and denominator and find the largest term in sight,*0819

*And divide both top and bottom by that largest term.*0822

*Here the largest term was n*^{2}, so we divide top and bottom by n^{2}.0828

*That really makes all the other terms go to 0, and it leaves you with the terms that you need to determine the limit.*0834

*Let us try another limit example.*0844

*We want to find the limit of the sequence of sqrt(n)/ln(n).*0845

*Here, both of these sqrt(n) and ln(n), when n goes to infinity, both of these go to infinity.*0853

*What we are really looking at is a situation of something going to infinity/something going to infinity.*0865

*The classic rule to use here that you might remember from Calculus 1 lectures is l'Hopital's Rule.*0873

*Remember, you can use l'Hopital's Rule in two situations.*0889

*You can use it when you have a limit going to infinity/infinity or 0/0.*0894

*If you have 0 × infinity, you cannot use l'Hopital's Rule directly,*0903

*But you try to rewrite it as an infinity over infinity or 0 over 0 situation.*0910

*Then you can use l'Hopital's Rule.*0916

*Let us remember what l'Hopital's Rule does.*0921

*It says you can take the derivative of the top and bottom.*0923

*The strange thing here is you do not take the derivative using the quotient rule you learned in Calculus 1.*0925

*Instead you take the derivative of the top and bottom separately.*0934

*The derivative of the sqrt(n), we think of that as being n*^{1/2}.0947

*Its derivative is 1/2n*^{-1/2}, that is square root of n in the denominator.0953

*The derivative of ln(n) is 1/n.*0962

*If we take the denominator and flip it up to the numerator, we get n/2sqrt(n).*0968

*n/sqrt(n) is the sqrt(n).*0977

*Now we want to take the limit as n goes to infinity of sqrt(n/2).*0981

*That clearly goes to infinity.*0997

*We say that this sequence diverges to positive infinity.*1001

*To recap what we did with that example, we were given sqrt(n)/ln(n).*1014

*We noticed both of those go to infinity, so we have a limit that goes to infinity over a limit that goes to infinity.*1021

*That is l'Hopital's Rule.*1026

*l'Hopital's Rule says you take the derivative of the top and bottom and you take those two derivatives separately.*1030

*You do not use the regular quotient rule that you use in Calculus 1.*1036

*We simplify it using a little bit of algebra, and then we talk the limit of the new expression that we get.*1044

*Then that tells us what happens to the original sequence, it diverges to infinity.*1052

*Let us try some more examples later.*1053

*This is educator.com*1054

*Hi, this is educator.com, and we are trying some examples of finding the limits of series.*0000

*Our example right now is sum of ln(n+1/n).*0008

*let us take a look at that.*0014

*The ln(n=1/n), we can invoke some rules of natural logs there.*0016

*We can write that as ln(n+1)-ln(n).*0023

*That will be very useful when we try to determine whether this series converges or diverges,*0029

*And what it converges or diverges to.*0034

*Remember, the way you determine whether a series converges or diverges is you look at the partial sums.*0038

*You write those out as a sequences,*0044

*Then you look at whether that sequence converges or diverges and where it goes.*0048

*Let us write out some partial sums here.*0052

*The first partial sum, s*_{1}, is just the first term here.0057

*So, ln(2) - ln(1).*0064

*We could simplify that, of course ln(1) = 0,*0068

*But I think it will be easier to spot a pattern, if we do not simplify that.*0072

*let me go ahead and find s*_{1} is just the s_{1} term, ln(2) - ln(1),0077

*+ the n = 2 term, ln(3) - ln(2).*0084

*s*_{3}, ln(2) - ln(1) + ln(3) - ln(2) + ln(4) - ln(3).0092

*We start to see some cancellation here.*0109

*ln(2) cancels, ln(3) cancels, and then we are just left with the beginning and ending term.*0112

*Let us try one more, s*_{4}.0118

*Is ln(2) - ln(1) + ln(3) - ln(2) + ln(4) - ln(3) + ln(5) - ln(4).*0120

*Again, we have lots of cancellation.*0136

*ln(2), ln(3), ln(4), and we are left with beginning and ending terms.*0138

*A general formula for the nth partial sum would be -ln(1) +,*0147

*When we had n = 4, we had ln(5), so the general pattern would be ln(n+1).*0159

*Then that we could simplify down.*0164

*ln(1) is jsut 0, so that is just ln(n+1).*0166

*That is what we get when we look at the sequence of partial sums.*0173

*We want to ask what happens when n goes to infinity.*0179

*Take the limit of that as n goes to infinity.*0181

*As n goes to infinity, we are plugging in bigger and bigger values to natural log.*0184

*Natural log goes to infinity itself.*0190

*So the sequence of partial sums diverges to infinity.*0195

*So the series by definition,*0200

*We say that diverges, it does the same thing that the sequence of partial sums does.*0212

*It diverges to infinity as well.*0220

*Let us try one more example of a series problem.*0000

*We are given the summation of n=2 to infinity of 3*^{n}/4^{3n+2}.0004

*Again, I want to try to investigate this by writing out a few terms and seeing what happens.*0013

*If we plug in n=2, we get 3*^{2}/4^{3×2 + 2}.0019

*So that is 4*^{8}.0028

*Now n=3 gives us 3*^{3}, over 4^{3×3+2}, so that is 4^{11},0031

*+ 3*^{4}/4^{3n+2}, so 4^{14}.0041

*What you might notice here is that each one of these terms is a common ratio multiplied by the previous term.*0056

*This second term = the first term × 3/4*^{3}.0064

*To get from the second to the third term, we multiply by 3/4*^{3}.0073

*So, each one of these terms, you get it by multiplying the previous term by 3/4*^{3}.0082

*So, what we have here is a geometric series.*0090

*Our common ratio is r = 3/4*^{3}.0097

*Which is certainly less than 1.*0107

*The reason I bring that up is because we want to check whether the geometric series converges,*0110

*You have to check whether the absolute value of 3 < 1.*0118

*And, certainly 3/4*^{3} < 1.0122

*So, it converges and we have a formula for what a geometric series converges to. *0125

*Remember that I said the easy way to remember that formula,*0135

*Is the first term/1-the common ratio.*0142

*That is the sum of a geometric series.*0161

*i think that is easier and more reliable than any numerical formula you can get.*0164

*Here, our first term is 3*^{2}/4^{8}.0170

*The common ratio is 3/4*^{3}.0176

*So, that is a little bit messy.*0186

*We can clean it up a little bit by multiplying top and bottom by 4*^{3}.0189

*That will give us 3*^{2}/4^{5}.0195

*On 4*^{3} - 3 in the denominator.0199

*That is 9/4*^{5}/64-3.0205

*So, that in turn becomes 9/61×4*^{5}.0214

*Again, the principle of dealing with this series is to write out the first few terms.*0226

*Recognize that it is a geometric series.*0235

*Recognize that each term is the previous term multiplied by a common ratio.*0238

*Identify the common ratio, see if it less than 1, and if it is, you can say right away that the series converges.*0244

*Then you can invoke this formula, first term divided by 1-common ratio.*0252

*Do a little bit of simplification, and find the sum of the series.*0259

*This has been Will Murray for educator.com.*0265

*Hi this is educator.com*0000

*We are going to talk today about series*0003

*There are several bits of notations and definitions before we look at some examples*0005

*The first is this big sigma notation.*0014

*The notation this symbol sigma stands for the series where you plug in different values of n.*0017

*This is just short hand a*_{0}, a_{0} + 1, a_{0} + 2.0025

*The way you want to think about these series is by thinking about the sequence of partial sums.*0030

*What you do is you add up these series 1 term at a time.*0039

*First you start with the 0 term if there is one. *0041

*Then the second partial sum is a*_{0} + a_{1},the third is a_{0} + a_{1} + a_{2} and so on.0047

*The s*_{n} is a_{0} + a_{1} up to a_{n}0054

*You call this the sequence of partial sums.*0060

*s*_{0}, s_{1}, s_{2} up to s_{n}.0064

*You think of that as a sequence, not a series.*0069

*This is a sequence s*_{n}.0075

*We want to say what it means for a series to converge.*0081

*You have to be quite careful when you make this definition.*0091

*It is not quite as obvious as you think.*0094

*What you say is you look at the sequence of partial sums.*0097

*We view that as a sequence*0102

*Then we go back to our definition for a sequence converging.*0105

*if the sequence of partial sums converges, to a particular limit.*0111

*Then we say the series converges to that limit.*0117

*By definition, a series converging means the sequence of partial sums converges.*0122

*In several of the examples that we will see later on,*0127

*We will be given a series and the way we will handle it is we will write out the partial sums and then we will think of them as the sequence.*0130

*We will see what happens to that sequence.*0136

*The same thing holds for all the other possibilities.*0141

*Diverging, diverging to infinity, or diverging to negative infinity.*0145

*You look at what the sequence of partial sums does.*0151

*Whatever behavior that does, you say the series does the same thing.*0155

*There is a very common type of thing that you will see.*0162

*We call it a geometric series.*0165

*It is one where each term is equal to the previous term multiplied by the same common ratio.*0167

*In practice that looks like a + some a × r + a × r*^{2} and so on.0175

*The important thing is that each term is getting multiplied by the same number every time.*0182

*We have a formula for the sum of a geometric series.*0189

*What you have to do is determine first of all if the ratio and absolute value < 1 or > 1 or = 1.*0193

*If the ratio and absolute value < 1, the series adds up to a/1-r.*0204

*This is the formula you will see in a lot of books.*0212

*I think this formula can be a bit misleading.*0214

*Because it depends on whether you start the series at n = 0, or at n = 1.*0217

*Some books have the a/1-r formula, some books have a slightly different formula.*0226

*It depends on whether they are using n = 1 or n = 0 as the first term in the series.*0234

*I do not like that formula so, I will give you a fail safe formula that will work in all situations right here.*0240

*First term, divided by 1 - the common ratio of the series.*0246

*That formula always works in all geometric series.*0252

*When your ratio is < 1.*0255

*I think that is the one to remember, even though it is words instead of an equation.*0260

*That is the one that will get you through any geometric series.*0265

*If the common ratio is bigger than 1 in absolute value*0270

*Or if it is equal to -1, than the series just diverges.*0274

*If the common ratio, if r = 1, then that means that what you are doing is you are adding a + a + a.*0280

*That clearly diverges either to infinity, if a > 0 , or -infinity, if a is negative.*0292

*You will probably not get geometric series with r - 0 because they are too simple to be given in a calculus exercise. *0303

*One more theorem that we are going to be using is called the test for divergence.*0311

*It is usually the first thing that you want to test with every series.*0318

*If you are given a series, what you do is you look at the individual terms.*0320

*You see whether the individual terms converge.*0326

*If they converge you ask what do they converge to.*0332

*If they converge to anything other than 0, then you can immediately say the series diverges.*0336

*Also, if the sequence of terms diverges, you can say the series diverges.*0347

*To recap, you look at the individual terms, if they converge to something other than 0, or if they diverge.*0354

*Immediately you can say the series diverges.*0360

*The flip side that often mixes up students.*0366

*If the sequence of terms does converge to 0, people think that you can use that to conclude that the series converges.*0369

*That is not true and we will see examples of that later on.*0391

*Then just from that information you cannot conclude anything about the series.*0398

*You have to go and find one of the other tests that we will discuss later on.*0416

*If the sequence converges to 0, you are really stuck.*0421

*You cannot use the test for divergence, however, if the sequence converges to something other than 0,*0425

*You can use the test for divergence immediately to say that the series diverges.*0430

*To emphasize here, the test for divergence can tell you that a series diverges, but it can never tell you that a series converges.*0439

*This is a common mistake made by students.*0482

*People will say oh a series converges by the test for divergence.*0488

*That is a very bad mis-use and your teachers will have no patience with that.*0491

*You can use the test for divergence to say that a series diverges,*0498

*But if you get that the series converges, if the sequence converges to 0, the test for divergence tells you nothing and you have to find something else.*0500

*Let us try some examples.*0510

*First example here is the series n-1/n*0513

*Right away we will look at the test for divergence and see if it works. *0516

*a*_{n} = n-1/n.0520

*We rewrite that as 1-1/n.*0525

*The limit of the sequence a*_{n} is well the 1/n will go to 0 is 1.0531

*This is not 0.*0541

*So the sequence of terms converges to something other than 0.*0545

*The sequence a*_{n} converges to something other than 0.0560

*The series diverges by the test for divergence.*0580

*The test for divergence says you look at the sequence of terms, see if they converge to something other than 0.*0599

*If so, then the whole series diverges.*0608

*If this had come out to be 0, if the limit had been 0, then we would not know and we would have to go on and find some other test there.*0614

*Let us try another example here.*0626

*The series of 1/n so again you will look at the test for divergence.*0627

*a*_{n} = 1/n and the limit of that as n goes to infinity is 0.0637

*The sequence converges to 0. *0645

*The test for divergence, if the sequence converges to 0, tells us nothing.*0649

*So, t(d) fails to give us an answer here.*0656

*We cannot say anything yet.*0663

*Instead we will look at the partial sums.*0664

*s*_{1} is just the first partial sum, that is just 1. 0676

*s*_{2} is 1 + 1/2.0679

*s*_{3} is 1 + 1/2 + 1/3.0684

*s*_{4} is 1 + 1/2 + 1/3 + 1/4.0691

*I am going to start adding these numbers up and see what kind of sums we get.*0703

*Just look at 1, it sounds kind of silly to say it, but 1 > or = 1.*0708

*We will see why I am making a big deal out of that later on. *0715

*1 + 1/2 > or = 3/2, in fact it is equal to 3/2.*0721

*I will skip s*_{3}, I will not look at that.0726

*s*_{4}, if you look at 1/3 + 1/4 now 1/3 is bigger than 1/4.0730

*So this is > or = to 1/4 + 1/4.*0737

*What we have here is 1 + 1/2.*0743

*Plus something bigger than 1/2.*0747

*This is bigger than 2.*0750

*Now I am going to start skipping, I am going to look at s*_{n}.0756

*s*_{n} is 1 + 1/2 + 1/3 + 1/4 + 1/5 all the way up to 1/8.0760

*Again, we have 1, that is bigger than 1.*0774

*1/2 is at least 1/2.*0777

*1/3 + 1/4 is bigger than or equal to 1/2.*0783

*Here we have 4 terms, each one of those is bigger than or equal to 1/8.*0788

*Collectively they are bigger than or equal to 1/2.*0794

*What we have here is 1 + 1/2 + 1/2 + 1/2.*0802

*This whole thing is < or = to 5/2.*0807

*Now I will skip to s*_{16}.0812

*Without showing all the individual terms, it is bigger than 1/2 + 1/2 + 1/2 + 1/2 and then we will have 8 more terms, all bigger than 1/16.*0816

*So 1/2, this is bigger than or equal to 3.*0829

*If you look at the sequence of partial sums as a sequence, we have got 1, 3/2, 2, 5/2, 3.*0834

*Clearly, the sequence of partial sums s*_{n} diverges to infinity. 0845

*By definition, the series 1/n diverges to positive infinity also.*0868

*There are a couple of points we want to make to recap that.*0884

*We tried to use the test for divergence.*0886

*That says you use the sequence for individual terms, but those went to 0, so the test for divergence tells us nothing.*0891

*Instead, we look at the sequence of partial sums.*0897

*That means we start adding up these terms and we kind of group them together in clever ways.*0900

*When we group them together in clever ways, we notice that the sequence of partial sums is going to infinity.*0912

*So, we say the entire series diverges to infinity.*0921

*A couple more things that I want to say about this series.*0923

*One is that it is very well known, it is called the harmonic series.*0927

*It is called the harmonic series because it arises in looking at musical notes.*0934

*This is well known and is called the harmonic series. *0941

*People will tell you the harmonic series diverges and the reason is by this proof here.*0945

*Another important thing to note about this series is we could write this as the sum of 1-n*^{p} where p=1.0952

*It is really 1/n*^{1}.0960

*This is a special case of something we are going to see later called the p series.*0963

*That is kind of a preview of something we will see later when we talk about the integral test.*0972

*We will be talking about p series.*0976

*The harmonic series is a special case of p series, with p=1.*0978

*Let us try another example here.*0986

*We are trying to determine if the series 1/n+1 converges or diverges.*0990

*Again, the trick here is to look at the partial sums and before we right out the sequence of partial sums.*0999

*We are going to do a little algebra here.*1012

*We are going to try to use partial fractions on 1/n × n+2.*1016

*Partial sums is an algebraic trick that we learned back in the partial sums section.*1023

*It says we can separate 1/n × n+2.*1029

*It says we can separate a/n + b/(n+2).*1034

*Then we can try and solve for constants a and b.*1041

*We learned this in detail in the lecture on partial fractions.*1046

*If you are a little fuzzy on partial fractions, we might want to go back and review that lecture,*1050

*In the meantime, I am not going to work it out but I am going to tell you the answer is a=1/2 and b=-1/2.*1057

*That comes from applying partial fractions here.*1069

*There is a little bit of algebra and solving 2 equations and 2 unknowns that I am suppressing here.*1073

*But, you can work it out and get these values for a and b.*1080

*We can write this as 1/2/n - 1/2/n+2.*1085

*If I factor out the 1/2 there, I get 1/2 × 1/n - 1/n+2.*1092

*That is going to be very useful in trying to figure out what the partial sums are.*1102

*Let me write down what a few of those partial sums are now.*1106

*s*_{1}, plug n=1 in here and we get 1/2 × 1 - n=1, gives us 1/3.1108

*I could simplify that but I am not going to because later on it will be easier to spot a pattern if we do not simplify that.*1119

*s*_{2} is 1/2 × 1 - 1/3, + the n = 2 term, is 1/2 - 1/4.1127

*s*_{3} is 1/2 × 1 - 1/2 + 1/2 × 1/2 - 1/4 + 1/2 × the third term is 1/3 - 1/5.1140

*Now you start to notice some cancellation here.*1160

*This 1/3 will cancel with that 1/3, which is why I did not want to simplify earlier.*1164

*Let me write out 1 more term, s*_{4}.1167

*1/2 × 1/3 + 1/2 × 1/2 - 1/4 + 1/2 × 1/3 - 1/5 + 1/2 × 1/4 - 1/6.*1172

*Now you start to see a lot of cancellation.*1191

*This 1/3 cancels with this 1/3, this 1/4 cancels with this 1/4. *1196

*If there were another term, the next term would have a 1/5, and that would cancel with that 1/5.*1200

*So, it looks like there is going to be a lot of cancellation as we write out more partial sums. *1205

*Let me try to write a general term for s*_{n}.1216

*It is going to be 1/2, well after we cancel everything, the only terms left are this 1*1218

*That never cancels, the 1/2 never cancels.*1226

*Then the very last terms do not cancel, but everything in the middle is all cancelled.*1230

*We get 1/2 × 1 + 1/2 -, well the last two terms.*1240

*When n=4, the last 2 terms were 1/5 and 1/6.*1249

*Those last two terms are n+1 and 1/n+2.*1255

*So, this simplifies down to 1/2 × 3/2 - 1/n+1 -1/n+2.*1262

*The limit as n goes to infinity of the partial sums.*1279

*Remember we think of the partial sums as a sequence now.*1284

*Those terms just go to 0.*1289

*We are left with 1/2 × 3/2 = 3/4.*1292

*We have the sequence of partial sums going to 3/4.*1298

*We say the series converges to 3/4.*1304

*To recap there, when you are given a series,*1325

*You want to determine whether it converges or diverges.*1331

*That depends on what the sequence of partial sums does.*1336

*That is s*_{1}, s_{2}, s_{3}, s_{4}.1338

*What we did with this particular one is we did kind of some algebraic cleverness.*1343

*In breaking n/n+2 up using partial fractions.*1348

*That gave us an expression that when we wrote out the partial sums,*1355

*They all cancelled in the middle and just left us with these beginning terms, and these ending terms.*1360

*This happens often and it is called a telescoping series.*1366

*It is called a telescoping series when you write out the sequence of partial sums and the middle terms all cancel,*1381

*Leaving you with just the terms at the beginning and the end.*1391

*Once you simplify it down to something in terms of the beginning and the end,*1392

*You can take the limit, whatever limit you get is, the limit of the sequence of partial sums.*1399

*By definition that is also the limit of the series.*1405

*We will try some more examples later.*1410

*This is educator.com.*1413

*Hi, this is educator.com.*0000

*We are trying a couple more examples of the integral test.*0002

*The first one here is the summation from n=2 to infinity of 1/n × ln(n).*0007

*You are probably used to seeing series that start at n=1.*0013

*Here I had to start it at n=2.*0018

*If you plug it n=1, to the natural log, ln(1) = 0.*0021

*If we started at n=1, we would get 1/0 and that would not make sense.*0026

*That is why I had to start at n=2.*0030

*So the integral test says you look at what you are given for a series and you convert it into a function.*0032

*f(x) = 1/x ln(x).*0037

*There are three things you have to check.*0047

*One, whether it is continuous.*0050

*This is a continuous function, because we are only looking at values of x bigger than 2.*0056

*There is no question about dividing by 0 in there.*0062

*Whether it is positive and this is positive as long as, well x is always positive.*0066

*This is positive as long as ln(x) is greater than 0, which it is,*0079

*Since we are only looking at values of x bigger than 2.*0083

*The third question is whether the function is decreasing.*0093

*To check that, we have to look a the derivative f'(x).*0098

*The derivative of that, if we think of f(x) as x × ln(x)*^{-1}.0105

*The derivative is -x log(x)*^{-2} × the derivative of x log(x).0112

*That is a product so we have to use the product rule.*0125

*That is x × the derivative of log(x) + log(x) times derivative of x,*0126

*The derivative of x is just 1.*0136

*This simplifies down into -1 + log(x)/x*^{2} × log(x)^{2}.0137

*If we look at that, 1 + log(x) > 0.*0151

*x*^{2} × log(x) ^{2} > 0.0157

*We have got numerator, denominator positive, but then we have got a big negative sign on the outside.*0162

*So, this whole thing is going to be negative.*0169

*Remember that was going to be the derivative, that means it has a negative derivative.*0173

*That means the function is decreasing.*0179

*Having checked those 3 conditions, it is ok to use the integral test.*0183

*We are going to look at the integral from 2 to infinity of 1/x log(x) dx.*0193

*That is not such a bad integral to do.*0205

*We can use u = ln(x).*0208

*Then, du = just 1/x dx.*0211

*We have got the integral from x=2 to x=infinity of 1/u du.*0217

*So, that is the integral of ln(u) evaluated from x = 2, well a value of t going to infinity.*0230

*That is the ln(ln(x)), because we have to change u back into the ln(x).*0244

*Again, evaluated from x=2 to x=a value that goes to infinity.*0253

*The second part of this is very easy.*0261

*- ln(ln(2)) is no problem there.*0264

*- ln(ln(2)) is just some number.*0268

*However, when we take ln(of some number going to infinity), that will go to infinity.*0272

*We actually want ln(that), but then we have ln(something going to infinity),*0277

*This whole thing diverges to infinity.*0291

*The integral diverges to infinity.*0298

*So, we can say the series 1/n log(n) diverges to infinity by the integral test.*0302

*Let us recap what happened there.*0321

*We were given the series of 1/n log(n).*0330

*We converted that into a function and checked these 3 conditions, continuous, positive, and decreasing.*0335

*The trickiest one there was decreasing.*0340

*We had to look at the derivative and work it through and check that it was negative.*0343

*Once we verified those 3 conditions, it is OK to use the integral test.*0347

*We look at the improper integral of 1/x log(x) from 2 to infinity.*0353

*That, we work out the calculus and that turns out to be ln(ln(x)).*0358

*Then when we plug in large values of x to ln(ln(x)), we get something that diverges to infinity,*0365

*So the integral test tells us that the original series diverges to infinity as well.*0374

*Let us try one more example of a problem that suggests the integral test.*0000

*Here we have the sum from n=1 to infinity of 1/n*^{2} + 1.0005

*We would set up f(x) = 1/n*^{2} + 1.0012

*Now, there are three conditions we need to check, continuous, positive, decreasing.*0020

*The continuous is pretty clear.*0032

*The positive is also pretty clear, no matter what x you put there that is going to be positive.*0038

*The decreasing, you should work out f'(x) and check that it is negative.*0044

*I am not going to work it out here, but it does turn out to be negative so it is ok to use the integral test.*0049

*Let us look at the integral from x=1 to x going to infinity of 1/x*^{2}+1 dx.0062

*There are several ways to do that integral.*0074

*You can do it with a trig substitution, you would use x = tan(θ).*0077

*You could look it up in the table of integrals in the back of your calculus book.*0081

*Probably for this integral, it is one that you are going to see often enough that it is good to have the answer memorized.*0087

*The integral of 1/x*^{2}+1 = arctan(x).0096

*That is one that if you have not memorized it yet,*0103

*You will probably see often enough as you are doing calculus, that it is worthwhile to memorize it.*0104

*Then we want to evaluate that from x=1 to a value that is going to infinity.*0109

*So, this is arctan(a number going to infinity) - arctan(1).*0121

*Now, remember what the arctangent function does.*0132

*It is the inverse of the tangent function.*0136

*There is the tangent function with asymptotes at -π/2 and π/2.*0139

*That is tan(x).*0147

*Arctan(x) says you flip that and so you get these horizontal asymptotes at -π/2 and π/2.*0149

*That is arctan(x).*0163

*When x goes to infinity, arctan(x) approaches π/2.*0168

*The limit there is π/2.*0173

*The arctan(1), well you think what angle as arctan(1) and if you remember your common values that is π/4.*0177

*So this improper integral converges to π/4.*0187

*What that tells us is that the series, 1/n*^{2}+1 converges by the integral test.0190

*What it does not tell us is what it converges to.*0211

*Remember, the integral test does not tell you that.*0221

*I will remind you that we do not know from anything we have learned so far, what that series converges to.*0225

*Even though the integral came out to be π/4, that does not tell you anything about what the series converges to.*0235

*All we can say from the integral test is that the series converges.*0244

*To recap there, we were given a function, a series,*0249

*We convert it into a function,*0255

*We check ti see if it is continuous, positive and decreasing,*0256

*Decreasing is the challenging part, we have to check that its derivative is negative.*0259

*Once we verify those, we do the integral.*0265

*We find that the answer turns out to be a finite limit.*0269

*That does not turn out to be infinity.*0274

*That allows us to conclude that the series converges by the integral test.*0278

*But we do not know exactly what the series converges to.*0284

*This has been educator.com*0286

*Hi, this is educator.com and we are here to talk about the integral test.*0000

*The integral test is a way of determining when a series converges or diverges.*0006

*The way it often works is you will be given a series in this form.*0013

*f(n) is just some expression in terms of n.*0017

*What you do is convert that into a function of x.*0020

*Then there are 3 things you have to check before you can apply the integral test.*0025

*You have to check whether it is a continuous function,*0029

*You have to check whether it is always positive.*0033

*The integral test only works for series with positive terms.*0036

*You have to check whether it is decreasing.*0040

*In other words, is the derivative negative?*0043

*The derivative has to be negative to pass the integral test.*0046

*If you checked all three of those conditions, the integral test tells you that you can look at the integral of f(x) from 1 to infinity,*0050

*And if it converges, then your series converges.*0060

*If it diverges, then your series diverges.*0064

*We will try some examples of that in a moment.*0069

*In the meantime, let me give you 1 important definition that comes out of the integral test.*0073

*We are going to talk about p series.*0077

*That is a series of the form 1/n*^{p} where p is a constant value.0080

*P series lend themselves very nicely to integral tests.*0088

*Because, we figured out back in the integration section on improper integrals.*0096

*We figured out when the integral of dx/x*^{p} converges.0106

*What we figured out was that the integral from 1 to infinity of dx/x*^{p} converges exactly when p > 1 and diverges to infinity if p < or = 1.0109

*We will use that result to look at some series today.*0127

*You will use it also in your calculus homework.*0131

*Let us try out some examples of the integral test.*0135

*First one is to determine whether the series 1/sqrt(n) converges or diverges.*0140

*That one we can write it as 1/n*^{1/2}.0148

*n = one to infinity.*0157

*That is now a p series.*0162

*With p = 1/2, 1/2 < or = to 1, so the series diverges to positive infinity.*0169

*Let me mention something about p series, because we will be seeing several examples of them.*0198

*We have also seen another kind of series called a geometric series.*0204

*These are frequently mixed up by students.*0210

*Let me mention what the difference is.*0211

*This n*^{1/2} is a p series because the variable n is in the base, and the constant 1/2 is in the exponent, so that is a p series.0213

*A geometric series, the general form of a p series is 1/n*^{p}.0231

*A geometric series, if you had something like 1/2*^{n}, where the constant is the base and the variable is in the exponent,0236

*That is a geometric series.*0250

*We saw some examples of those earlier.*0253

*The general form of a geometric series would be the sum of r*^{n},0258

*Where r is the constant.*0262

*Remember in the p series, the exponent is the constant, and in the geometric series, the base is the constant.*0269

*Those two are very frequently mixed up by students.*0277

*They are different kinds of series and the rules for determining when they converge or diverge look similar on the surface.*0280

*You really have to remember separate rules for geometric rules and p series.*0288

*You have to be careful not to label a geometric series as a p series, or a p series as a geometric series.*0297

*This one is the series 1/n*^{3/2}.0309

*Again, this is a p series, because the constant is in the exponent.*0315

*Here p is 3/2, which is bigger than 1.*0324

*So, the series converges.*0332

*Let me mention another feature of the integral test in a p series.*0345

*Which is that the integral test and the rules for when a p series converge,*0351

*None of those tell you what the series converges to.*0357

*We do not know from anything we have learned so far what the series converges to.*0367

*We do not know the limit of the series.*0378

*In that sense, the integral test and the p series rules are limited.*0384

*They will tell you whether the series converges or diverges, but if a series does converge, it will not tell you what it converges to.*0389

*Let us try another example here.*0398

*The sum of n*^{2}/e^{n}.0402

*This one is a little more complicated and is not just a simple p series.*0405

*Let us look at f(x) = x*^{2}/x^{n}.0409

*We want to use the integral test to tell whether or not the series converges or diverges.*0420

*It will be easier to integrate this thing if we look at it as x*^{2}e^{-x}.0424

*I want to look at the improper integral from 1 to infinity of x*^{2}e^{-x} dx.0431

*That is a classic integral to solve by integration by parts.*0444

*I am going to use the trick of tabular integration that we learned in the integration by parts lecture to integrate this.*0447

*If you do not remember the tabular integration trick, you might want to go back and look at th integration by parts lecture.*0460

*I will also go through it slowly here.*0468

*If we take the derivatives of x*^{2}, we get 2x, 2 and then 0.0472

*If we take the derivatives of e*^{-x}, the first integral is -e^{-x}, the integral of that is just e^{-x} again.0479

*The integral of that is -e*^{-x}.0488

*Then we write these diagonal lines and the signs +, -, +.*0493

*Then we multiply along the diagonal lines to get the answer x*^{2}e^{-x} - 2xe^{-x} -2e^{-x}.0500

*We want to evaluate that from x=1 to x=some value t which will go to infinity.*0516

*This tabular integration is a shorthand way of doing integration by parts.*0527

*You can also do integration by parts by using u and dv.*0535

*You should get the same answer here.*0542

*If you plug that in, e*^{-x}, this is the same as saying x^{2}/e^{-x}.0543

*As we plug in infinite values there, larger and larger values of x,*0551

*The e*^{-x} dominates the x^{2}.0560

*When we take the limit as t goes to -infinity,*0568

*Since the e*^{-x} dominates the x^{2}, we actually get 0 on each of these terms.0569

*0-0-0, and now if we plug in x=1.*0575

*We get - a -, so that is a + 1/e + 2/e +2/e.*0581

*If you add those up to get 5/e, which is a finite number.*0604

*The conclusion we can draw from that is the series that we were given n*^{2}/e^{n} converges by the integral test.0613

*That is our answer.*0644

*If this integral had diverged to infinity, then we would have said the series diverges to infinity.*0645

*The integral test is nice in that it can give you either answer.*0654

*Unlike some of the other tests.*0657

*If this integral had diverged to infinity, we would have said the series diverged to infinity.*0660

*Where the integral test falls, and does not give us an answer, is it does not tell us exactly what the series converges to.*0667

*Even this 5/e is attempting to say, well that is the value of the interval, that must be the value of the series.*0674

*That is not safe.*0682

*This 5/e, we do not actually get that much from the numerical value 5/e.*0684

*The important thing is just that it is finite, and so it tells use the series converges.*0689

*We do not know using what we have learned so far what it converges to.*0698

*So, that is how the integral test works.*0713

*You take the series that you are given, convert it into a function, work out the integral from 1 to infinity of that function.*0717

*Then, if you get a finite answer, you can say the series converges.*0727

*If you get an infinite answer, you can say the series diverges.*0733

*We will try a couple more examples of that later on.*0736

*Hi, we are checking out some more examples of the comparison test and the limit comparison test.*0000

*What we have right here is the sum of e*^{n}/n!.0005

*That one is a little tricky.*0011

*I want to write that one as e*^{n}/n×1×2×3...×n.0013

*What I notice here is this 3, 4, up to n, all of those numbers are bigger than or equal to 3.*0031

*You will see why the 3 is significant as a cut off in a second.*0039

*This is < or = e*^{n}/1×2×3, and there were n-2 factors there.0044

*So 3*^{n-2}.0062

*Just to make the algebra a little bit cleaner, I am going to multiply the top and bottom by 3*^{2}.0064

*3*^{2} × e^{n}/1 × 2, now I can say that is 3^{n}.0071

*I just multiplied in 3*^{2} × the top and bottom.0078

*So that I can have an n in the exponent instead of an n-2.*0082

*This in turn is, 3*^{2} is 9/2 × e^{n}/3^{n}.0085

*This is (e/3)*^{n}.0094

*That is the series that I am going to use as my b*_{n}.0100

*Remember a*_{n} is always the series that you are given.0102

*The point here is that a*_{n} is < or = to b_{n}.0109

*The sum of b*_{n} converges.0114

*Why does it converge?*0118

*Because, b*_{n} is just a constant 9/2 × e/3^{n}.0121

*When you have a constant raised to the n power, it is a geometric series.*0128

*So you look at the common ratio, well here r = e/3.*0136

*Now e is about 2.7.*0146

*I do not know the exact value of e, but I know it is about 2.7.*0149

*The key thing there is that 2.7/3 is less than 1.*0154

*Remember 1 is the cutoff to check when a geometric series converges.*0159

*We have that e/3 < 1, so our common ratio < 1.*0165

*That means the sum of b*_{n} converges.0171

*So, our smaller series a*_{n}, must also converge by the comparison test0175

*Not the limit comparison, the original comparison test.*0185

*So, now it should be evident why I cut things off between 2 and 3 here.*0193

*I was kind of looking ahead to this common ratio.*0202

*I knew that e is about 2.7.*0206

*So, I saw that I had a bunch of numbers bigger than that in the denominator,*0211

*If I cut it off between 2 and 3.*0218

*That is why I cut off all of these number bigger than 3.*0221

*I used those to build the geometric series,*0225

*And then I just had to save the 1 and 2 separately.*0229

*1 × 2 just gave me a two in the denominator and did not really affect things later on.*0232

*The important thing was to save the e/3, and compare it with a geometric series.*0238

*Since we know the geometric series converges, and we have something smaller,*0242

*We can say our own series converges by the comparison test.*0250

*The last example that I want to do here is a little more complicated.*0000

*We have the series of n*^{4}+3n-8/3n^{7}+6n^{5}.0006

*You see something complicated like this and it is important to focus on which terms are really going to affect it,*0014

*And which terms are not really going to play much role. *0020

*The key thing here is the two biggest terms, n*^{4}, in the numerator, and n^{7} in the denominator.0024

*It looks like if you strip away the extraneous stuff there,*0030

*It looks like the sum of n*^{4}/n^{7}.0038

*That in turn would simplify down to 1/n*^{3}.0041

*OK, I am going to call my 1/n*^{3}, that is going to be the b_{n} that I create.0051

*This entire given series, that is a*_{n}.0057

*We will look at a*_{n} vs. b_{n},0062

*We will divide them together and try to go for something using the limit comparison test.*0065

*a*_{n}/b_{n} is n^{4}+3n-8/3n^{7}+6n^{5}.0069

*b*_{n} is 1/n^{3},0081

*But we can flip that up into the numerator so we can write that as n*^{3}/1.0084

*That turns into n*^{7}+3n^{4}-8n^{3}/3n^{7}+6n^{5}.0094

*Now you look at something like this and identify the biggest term anywhere,*0107

*Which is an n*^{7} in the top and the bottom and you divide by it.0113

*So, you get 1+3/n*^{3}-8/n^{4}/3+6/n^{2}.0116

*All of these other terms, go to 0.*0134

*This whole thing goes to 1/3.*0140

*The key thing about the 1/3 there, is only that it is a finite number, it is not infinity and it is not 0.*0144

*If you get a finite number there, the limit comparison test says that your given series does whatever the other series does.*0152

*It does the same thing that the other series does.*0160

*In this case, since the series we made ourselves, the b*_{n},0164

*Well that is 1/n*^{3}, that is a p series and p is 3.0171

*That converges.*0176

*The important thing about 3 there is that it is > 1.*0187

*Anything > 1 makes that converge.*0190

*Since that converges, we can say that the sum/a*_{n} also converges.0192

*Our justification there is the limit comparison test.*0203

*Either both series converge, or both series diverge.*0208

*Since the one we introduced converged, the given one converges as well.*0211

*To recap there, we are given some complicated series here.*0217

*We identify the important elements and use those important elements to build our own series that we introduce.*0222

*We then divide them together, take the limit,*0230

*If we get a finite non-0 limit,*0234

*Then we can say that both series do the same thing.*0237

*Hopefully the series that we introduced is a simple enough one where we can look at it and quickly say if it diverges or converges.*0240

*In this case it is a p series, and p is bigger than 1, so it converges.*0248

*So, we can say that the given series converges as well.*0254

*Thanks for watching, this is educator.com*0259

*Hi, this is Will Murray and we are here today to talk about the comparison test.*0000

*The way this works is that you will be given a series that we are going to call a*_{n},0006

*And you create your own series that we will call b*_{n}.0015

*Then you will try to compare these 2 series to each other. *0020

*The way this works is, the one you create, if that one converges,*0024

*And the one that you are given is smaller than it,*0029

*Then you can say that the given series converges as well.*0032

*On the other hand, if the one you create diverges,*0037

*And the given series is bigger than the one you created,*0040

*Then you can say the given series diverges.*0045

*Now it is very important to get these inequalities going the right way.*0048

*If these inequalities are going the wrong way, then the comparison test does not tell you anything.*0053

*We will see some examples that illustrate the difference between those inequalities going the right way and going the wrong way.*0057

*We will be using a second test called the limit comparison test.*0067

*It starts out the same way,*0069

*You will be given a series, then you create your own series,*0071

*But then instead of checking the inequalities,*0074

*What you do is you divide those 2 series together and then take the limit as n goes to infinity.*0077

*If you get a finite and positive number, it cannot be 0, it has to be a positive number,*0084

*Then you can say whatever the series you created does, converges or diverges,*0091

*You can say that the given series does the same thing.*0097

*let us check that out with some examples.*0101

*The first one here is the series of 1/n+1.*0105

*That is the given series, the a*_{n}.0109

*When you look at this series, the simplest series that this seems to resemble,*0113

*Is the series 1/n.*0123

*Because 1/n+1, is about the same as 1/n.*0126

*So that is the series we create ourselves and we call the b*_{n}.0131

*Now let us try comparing those to each other.*0134

*1/n+1 works as 1/n, well n+1 has a bigger denominator.*0138

*Which means that 1/n is bigger.*0146

*So, 1/n is bigger, the b*_{n}.0149

*So, the a*_{n} is less than the b_{n}.0155

*The sum of b*_{n}, that is 1/n, diverges.0162

*We know that we saw that one before and that was the harmonic series.*0170

*Or, you can think of it as the p series, with p = 1.*0178

*We know that series diverges.*0182

*What we have is a series that is less than it.*0185

*If a series is less than a divergent series, that is not going the right way to use the comparison test.*0189

*So, comparison fails to give us an answer here.*0197

*Tell us nothing.*0203

*So, the comparison test does not tell us what this series does.*0207

*Instead, you have to try another test,*0212

*For example, you could try the integral test and that actually will give you a good answer.*0215

*You can check that out with the integral test.*0223

*It turns out that this series diverges,*0226

*But the important thing is that you cannot use the comparison test on this one.*0228

*Let us try it out with another one. *0230

*3*^{n}/2^{n} - 1, 0233

*That looks like the sum of 3*^{n}/2^{n}.0240

*So a*_{n} is the given series, for our b_{n}, we will use the sum of 3^{n}/2^{n}.0249

*Then we will compare those to each other.*0255

*So, 3*^{n}/2^{n}-1 vs. 3^{n}/2^{n}.0260

*Well 2*^{n}-1 is a smaller denominator which means that 3^{n}/2^{n}-1 is actually a bigger number.0268

*What that is saying is that a*_{n} is bigger than b_{n}.0278

*We know that the sum of the b*_{n}'s diverges.0285

*The reason we know that is because it is a geometric series.*0295

*The common ratio is 3/2, because 3/n/2*^{n} is just 3/2^{n}, and 3/2 is bigger than 1.0300

*So that is a geometric series that diverges,*0309

*And here we have a bigger series, so this bigger series, we can say it diverges by the comparison test.*0313

*In this case, the inequality did go the right way.*0329

*So, we get a conclusion by the comparison test.*0333

*Just a recap there, we looked at the series we were given, we tried to find the series that was similar to it,*0338

*And simple enough that we could answer pretty quickly whether it converged or diverged.*0346

*In this case it was a geometric series.*0350

*Then we compared them to each other.*0354

*The inequality does go the right way, so we can make this conclusion.*0356

*Next example I wanted to try is the sum of sqrt(2n+17)/n.*0361

*Now, this one most closely looks like,*0369

*Well, the sqrt(2n+17) that is really more or less the sqrt(n).*0376

*sqrt(n)/n, and so that in turn is sqrt(n)/n 1/sqrt(n).*0385

*That is the series we are going to use as our b*_{n}.0398

*This series is our a*_{n}.0401

*What we are going to do is divide those together because we are going to try to use the limit comparison test this time.*0404

*So we look at a*_{N}/b_{n},0411

*And that is the sqrt(2n+17)/n.*0414

*All of that divided by b*_{n}, which is 1/sqrt(n).0420

*We can flip that fraction in the denominator up the numerator,*0424

*So we get sqrt(2n+17).*0429

*The sqrt(2n+17) × sqrt(n)/n,*0435

*Which is sqrt(2n*^{2}) + 17n/n.0444

*If we look at the top and bottom there, the biggest terms we have in the top, we have an n*^{2} with a square root over it.0451

*That is essentially n and in the bottom we have n.*0457

*So we would divide top and bottom by n, and we get,*0463

*If we bring that top end under the square root, we get 2+17/n.*0467

*Because when the n comes under the square root it turns into an n*^{2}.0472

*Then just 1 in the denominator.*0477

*17/n goes to 0.*0480

*So, the whole thing goes to sqrt(2).*0483

*The important thing about the sqrt(2) is that it is a finite number, it is not infinity.*0488

*It is not 0, so the limit comparison test applies.*0493

*Since the sum of b*_{n} diverges, well how do we know that,0498

*That is because we can think of 1/sqrt(n) as 1/n*^{1/2}.0509

*That is a p series, with p equal to 1/2.*0515

*The key thing there is that 1/2 < or = 1.*0522

*That means that b*_{n} is a divergent series.0527

*The limit comparison test says that the given series a*_{n} does the same thing that your series b_{n} does.0530

*So, we can conclude that the sum of a*_{n} also diverges by the limit comparison test.0539

*Again, the key point there is that we look at the series that we are given, and we try to find,*0550

*So that is the given series,*0562

*We try to find the given series that behaves like the series that we are given but is simpler to deal with.*0565

*What I did here was essentially strip away the 2 and the 17,*0569

*Because those were not so important, and then I called the new series b*_{n}.0574

*Then we try to compare those series to each other.*0578

*This time by dividing.*0582

*If we get this finite non-zero number at the end of it,*0584

*That says that both series behave the same way.*0588

*Since we know the b*_{n} diverges, because it is a p series,0593

*Then we can say the a*_{n} diverges as well and justify that conclusion using the limit comparison test.0597

*So, another example I would like to look at is the sum of sin(1/n).*0605

*The way we might think about that is,*0610

*let us think about the graph of sin(x).*0614

*As n goes to infinity, 1/n goes to 0.*0618

*So let us look at the graph of sin(x) when x is very near 0.*0625

*What we see is that the graph of sin(x),*0629

*This is sin(x), is very close to the graph of x as x approaches 0.*0637

*Sin(x) is almost the same as x.*0647

*Sin(1/n) the a*_{n} might behave like the series b_{n} = 1/n.0650

*Because sin(x) behaves like x.*0660

*Let us try that out.*0662

*We will take the series b*_{n} = 1/n, and then we will look at a_{n}/b_{n}.0665

*Is sin(1/n)/1/n.*0673

*Remember as n goes to infinity, 1/n goes to 0.*0679

*So this is sin(0) which goes to 0.*0684

*1/n also goes to 0, so we have a 0/0 situation.*0686

*That is a situation where you can use l'Hopital's rule.*0690

*So l'Hopital's rule says you can take the derivatives of the top and bottom,*0695

*I will take the derivative of the bottom first because it is easier.*0703

*The derivative of 1/n is just 1/-n*^{2}.0707

*The derivative of sin(1/n) is cos(1/n).*0710

*× derivative of 1/n by the chain rule which is -1/n*^{2}.0718

*Those -1/n*^{2}'s cancel.0724

*We get cos(1/n) and if we take the limit of that,*0727

*As n goes to infinity, that is cos(0) which is 1.*0735

*The key thing about 1 here is only that it is a finite non-zero number.*0742

*If it is a finite non-zero number, that says whatever one series does, the other series does.*0752

*So, let us ask ourselves,*0760

*The sum of b*_{n} does what?0763

*Well b*_{n} is just 1/n, and we know that is the harmonic series.0765

*And it is also a p series.*0771

*So, either one of those is just vacationed since we already showed that it diverges.*0775

*It is a p series with p = 1.*0780

*So that is a divergent series.*0788

*Since that series diverges,*0792

*We can say that the given series a*_{n} also diverges by the limit comparison test.0795

*So that one was a little bit trickier, probably was not so obvious what series we should compare it too.*0810

*The key thing there was realizing that sin(1/n) is a lot like 1/n when n goes to infinity.*0816

*Because 1/n was going to 0.*0825

*sin(x) was very similar to x.*0828

*Once we figure out which series we want to compare it to, we divide them together, take the limit,*0832

*Which uses l'Hopital's rule, we get a finite non-zero number,*0838

*And that says the two series do the same thing.*0843

*Since one of them diverges, we can say that the given series diverges as well.*0847

*I would like to try another example here of the alternating series test.*0000

*We are given -1*^{n+1}/n.0005

*We want to show that it converges and determine how many times would be necessary to estimate the sum within 0.01.*0010

*So, here let us take our b*_{n} to be 1/n.0018

*We do have that that is decreasing, b*_{n}+1 is less than b_{n}.0025

*We do have that the limit of b*_{n}=0.0033

*Our two conditions are satisfied.*0038

*That means the alternating series converges.*0040

*So, the sum of -1*^{n+1} b_{n},0045

*We can say that that certainly converges by the alternating series test.*0055

*That is the first part of what we had to do there.*0065

*We had to show that that series converged.*0070

*Then we have to determine how many terms are necessary to estimate the sum within 0.01.*0072

*Remember the error when you use the alternating series test, is bounded by a*_{n}+1.0079

*By the first term that you cut off.*0086

*We want that to be less than 0.01.*0090

*We want a*_{n} to be less than 0.01.0103

*Well a*_{n}+1 in absolute value is 1/n+1.0107

*That should be less than or equal to, 0.01 is the same as 1/100.*0113

*So if we take the reciprocal of that, remember that reverses the inequality.*0119

*So this n+1 > or = 100.*0124

*So n would have to be bigger than 99.*0130

*We would need 99 terms.*0138

*We would need to add up 99 terms of this series to get an estimate that is within 0.01 of the true answer.*0147

*In fact, on the computer earlier, I did add up 99 terms of these series.*0155

*It turns out that s*_{99} is approximately equal to 0.698172.0162

*If you actually add up all those 99 terms, that is the approximate answer that you get.*0173

*This series is another one that we can figure out using some techniques that we are going to learn later on.*0180

*Later, we are going to learn about Taylor Series.*0186

*I do not expect you to understand these yet because we have not developed the machinery for it yet. *0191

*But, later we will learn that the Taylor Series for the function ln(1-x) is the sum from n=1 to infinity,*0196

*Of -x*^{n}/n.0208

*If you plug in x=-1, then we get ln(2)=the sum of -1*^{n}/n,0216

*Which is exactly the series we have been trying to estimate.*0237

*The true sum of the series is in fact ln(2).*0241

*That turns out to be approximately equal to 0.693147.*0247

*The true sum using techniques that we have not really developed the machinery for yet is 0.693147.*0257

*The sum we got was 0.698172, so in fact we were within 0.01 of the true sum.*0266

*Just to recap there,*0274

*What we were trying to show is how many terms do we need to estimate the true sum within 0.01.*0277

*Well, we have this error formula that says the error is bounded by the first term that we cut off, the n+1 term.*0284

*We set that less than 0.01, we would like the error to be less than 0.01,*0290

*Then we solve that out, get a value of n, and then that tells us how many terms we need to take to estimate our sum.*0298

*The last example I would like to look at is to show that the series -1*^{n}/e^{n} converges.0000

*To determine what partial sum we would need to estimate the true sum within 0.001, so very very accurate estimate there.*0008

*Here, for our b*_{n}, we just take everything except the -1^{n}.0018

*That is 1/e*^{n}.0024

*Then we check our two conditions, one is that it is decreasing.*0026

*bN+1 should be less than b*_{n}.0030

*That should not be an e*_{n} that should be an e^{n}.0037

*Certainly that is decreasing.*0040

*e*^{n+1} is bigger than e^{n}, so 1/e^{n+1} is less than 1/e^{n}.0042

*Certainly, the limit of that as n goes to infinity is equal to 0.*0049

*So, that certainly justified.*0058

*The series, we can say the alternating series, *0061

*-1*^{n}/e^{n} converges by the alternating series test.0068

*That is the first half of the work that we needed to do there.*0084

*Now we have to figure out what partial sum we would need to estimate the true sum to within 0.001.*0088

*Again, we are going to use our error estimation formula.*0097

*It says the error is less than whatever term we cut off.*0102

*The absolute value of whatever term we cut off.*0108

*We would like that to be less than 0.001.*0110

*So, a*_{n}+1, if we take the absolute value, that knocks away any possible negative signs.0117

*So this is 1/e*^{n} < or = 0.001 is 1/1000.0124

*If we take reciprocals, remember that reverses the inequality.*0135

*So we get e*^{n} > or = to 1000.0140

*Now there are 2 possible ways you can finish this.*0144

*If you do not have a calculator handy, if you are on a test or you are not allowed to use calculators,*0147

*Or for some reason you are trying to make a quick estimate without any mechanical tools,*0150

*We want to figure out what n is, is it safe to say that e*^{n} is bigger than 1000.0157

*Here is what I know.*0165

*I know that e is about 2.7.*0166

*That is certainly bigger than 2, right?*0171

*Sorry, a small correction here, this is a*_{n}+1,0176

*So my exponents on the e's should be n+1's.*0180

*Now I am trying to make that bigger than 1000, *0187

*And if I do not have a calculator, I estimate that e is bigger than 2.*0192

*I know that 2*^{10} = 1024.0198

*One way I know that is that when people talk about a kilobyte, or 1K of memory, is that they really mean 1024 bytes.*0204

*So that is bigger than 1000.*0217

*So certainly, e*^{10} is bigger than 1000 because e is bigger than 2.0219

*So, if n+1 = 10, then n = 9 is enough.*0227

*So that is just a rough estimation based on no calculator.*0235

*If we do not have a calculator handy.*0240

*We would get n = 9 is enough. *0246

*If you do have access to a calculator and you can make this more precise, we could actually solve for n.*0250

*So, if you do have a calculator,*0255

*We can take e*^{n+1} > 1000, we can solve for n.0260

*We can take ln(both sides) and get n+1 > or = ln(1000).*0264

*So n should be > or = ln(1000)-1.*0271

*If you check on a calculator, ln(1000) = 6.9077.*0280

*So - 1, we get n > 5.9077.*0288

*But, since n is the number of terms we are taking, it has to be a whole number, so n=6 is enough.*0297

*So if you have access to a calculator, you would use n=6 there.*0310

*If you do not have access to a calculator, you probably would get a slightly rougher answer.*0314

*The conclusion there is that you would use s*_{6}, the partial sum for n=6.0322

*That is really the answer to the question that we were asked, the partial sum would have to be s6.*0330

*Let us go ahead and see if we can check out how close that actually gets us.*0335

*If you actually add up s*_{6} for this series,0340

*Add up terms and find the partial sum, you get s*_{6} is approximately equal to 0.731725.0347

*That is the partial sum if you just run this series from n=0 up to n=6.*0361

*You get this partial sum s*_{6} approximately 0.731725.0367

*That would be your estimate for the true sum of this series.*0375

*Now this series is another one where you can actually find the true sum using the different tools.*0380

*Because, notice that this series is actually a geometric series.*0387

*It is just the sum of -1/e raised to the n power.*0395

*We have a formula for the sum of a geometric series.*0404

*The sum, remember, is, I like to think of the formula in terms of words.*0407

*If you write the formula in terms of variables, you get slightly different formulas depending on whether you start at n=1 or n=1,*0416

*You will see in different books different version of this formula.*0426

*The version that always works is to write it in words.*0432

*First term over one minus the common ratio.*0434

*Here, the first term is n=0, you plug in n=0 and you get 1.*0439

*The common ratio there is 1/e, so 1-1/e,*0446

*Sorry, the common ratio is -1/e, so 1- that is 1+1/e.*0455

*If you multiply top and bottom by e, you get e/e+1.*0465

*That in turn you can plug into a calculator and get 0.731059.*0473

*So, remember we were trying to estimate this series to within 0.001.*0482

*Our estimate was 0.731725.*0487

*The true sum, which we found out using geometric series techniques was 0.731059.*0491

*So, in fact, we did get within 0.001 of the true sum as we wanted to.*0500

*Just to recap there, the first thing we had to do was show that it converged.*0507

*We did that by checking these two conditions to invoke the alternating series test and they both worked.*0511

*The second thing was to find how many sums we need to estimate this series within, estimate the true sum within a certain error tolerance.*0519

*We used our alternating series error formula and set it less than the error we are willing to allow,*0528

*Because we want to have less error than 0.001,*0535

*We solve that out for n.*0540

*Now, that could go different ways depending on whether you have to make a rough estimate without a calculator,*0541

*Or whether you can find an exact value with a calculator.*0548

*We got a value of n, n=6,*0551

*That was really all we were asked for.*0555

*Just to check it, we did add up the partial sum s*_{6} to get this 0.731.0558

*Because this happened to be a geometric series, we could also use the geometric series formula to get the true sum.*0566

*In fact, both answers are 0.731 so they are within 0.001 of each other.*0575

*That is the end of our section on alternating series, this is educator.com.*0582

*This is educator.com and we are here to talk about alternating series.*0000

*The main test that we are going to be using is called the alternating series test.*0005

*In some Calculus classes, this is called the Leibnitz alternating series test.*0011

*I will be referring to it as AST for short.*0015

*Sometimes you might see it as LAST for short.*0019

*The alternating series test works like this.*0023

*You start out with some positive terms, and then you make that into an alternating positive and negative series,*0025

*By either multiplying it by -1*^{n}, or -1^{n+1}.0033

*If you started positive, that turns it into a series that alternates positive and negative.*0038

*There are two conditions that you have to check.*0044

*You first check is b*_{n} decreasing?0047

*Meaning, are they getting smaller, in other words, is each subsequent term smaller than that previous term?*0052

*Then you also have to check if the limit of b*_{n} is 0.0056

*If both of those terms work, if both of those conditions are satisfied, *0060

*Then you can say that the series converges by the alternating series test.*0065

*By the way, you can never use AST to say a series diverges.*0073

*That is a very common mistake that Calculus students make.*0084

*But, you will not make it because you should know that this is a one-way test.*0089

*It can only tell you that something converges.*0094

*It can NEVER tell you that a series diverges.*0097

*So, if these conditions work then you say that it converges.*0100

*If they do not work, you are out of luck and you have got to find another test.*0104

*The second thing we are going to do with the alternating series test is estimates of sums.*0110

*If you have a series that satisfies the alternating series test,*0116

*And you want to estimate the sum of that series,*0120

*What you might do is add up the first few terms and say well that is my guess as the sum of the series.*0124

*In other words, you would use the partial sum as an estimate of the total sum that we will call S.*0131

*S is the true answer but we do not know what that is.*0137

*SN is the partial sum, is your estimate.*0140

*Then we ask, what is the possible error that you could make by making that estimate.*0144

*The area, we do not know, it could be positive, it could be negative.*0150

*But, the area is bounded by the abs(a*_{n}+1).0153

*That means that this is the first term that you did not include in your estimate, in your partial sum.*0158

*That you cut off.*0174

*So, you will be adding up terms of a series.*0179

*At some point you stop, you cut it off, and you say I am not going to look at any more terms. *0180

*What is my possible error by just taking the terms I have looked at so far?*0185

*The answer is, your error is the first term that you cut off,*0191

*So it is sort of the next term.*0194

*We will see some examples of that so that you get some practice.*0199

*First up is a series here, -1*^{n}/n!.0200

*Here the b*_{n} is just 1/n!.0205

*We have to check our two conditions, whether the b*_{n} is decreasing, and whether the limit is 0.0210

*If we look at b*_{n}+1 vs. b_{n}, b_{n} should be less than b_{n}, well that is 1/n+1!.0218

*Vs. 1/n!. *0228

*Certainly n+1! is much bigger than n!,*0230

*So n+1/n+1! is less than 1/n!,*0238

*So that condition is certainly satisfied.*0243

*The limit of the b*_{n} is 1/n!.0247

*That certainly equals 0.*0253

*So, condition 1 and condition 2 are both satisfied.*0255

*The series, when you make that series alternating by attaching a -1 to the n, *0260

*The series converges by the alternating series test.*0270

*So, let us try another example there.*0284

*Here we have -1*^{n} × sin(n)/n^{2}.0287

*So, I sort of strip off -1*^{n}.0294

*We look at this part,*0296

*b*_{n} is going to be sin(n)/n^{2}.0298

*Now, right away we have a problem because remember I said that we want to have b*_{n} > 0.0306

*Here, the n*^{2}, no problem there, that is always positive.0315

*But, the sin(n) that kind of oscillates all over the place.*0320

*Remember, sin oscillates between -1 and 1,*0325

*And sin(n) is going to be bouncing all around between -1 and 1.*0329

*So this is not always positive.*0332

*One of the requirements, before you even think about the alternating series test, *0340

*Was that the b*_{n} be positive.0348

*That fails right here so the alternating series test does not apply, which does not necessarily mean that the series diverges.*0351

*It just means that we do not know based on any of the tools that we have so far.*0367

*We have no conclusion.*0375

*We do not know based on the tools that we have learned so far,*0377

*Whether this series converges or diverges.*0379

*You cannot use the alternating series test.*0384

*So, let us try another example. *0387

*The sum of -1*^{n}/n!.0391

*Here, we want to find the partial sum s4,*0394

*Then determine how far away that might be from the true sum.*0396

*Let us write out some terms here.*0403

*-1*^{n}/n!, the n = 0 term, gives us -1^{0}, that is 1.0407

*0! by definition is 1.*0413

*The n=1 term is -1*^{1}, so -1, and 1 factorial is also 1.0419

*n=2 gives us positive 1 again, and then 2! = 2.*0427

*n=3 gives -1/3! is 1/6.*0437

*n=4 gives us +1 over 4! is 24.*0443

*Then we will write one more term, n=5.*0449

*n=5 gives us -1/5!, which is 1/120.*0454

*Now s4 means that you go up to the n=4 term, so we are going to add that up. *0462

*s4=1 - 1 + 1/2 - 1/6 + 1/24.*0467

*The 1 - 1 cancel, and if we put everything in terms of 24ths here, that is 12-4+1/24.*0480

*That is 9/24.*0490

*Which is 3/8, which is 0.375.*0496

*That is s4, our first answer.*0502

*Then they say, what is the maximum possible error in using that as an estimate of the true sum.*0506

*The error, remember is bounded by a*_{n} + 1, the abs(a_{n}+1).0515

*That is the first term that you cut off.*0525

*The first term that we cut off is, well it was -1/120, but since we take the absolute value, that is 1/120.*0528

*The error is at most 1/120, if you are looking for a decimal approximation, I can tell you that is less than 1/100 which is 0.01.*0542

*We are accurate to 2 decimal places.*0554

*We are accurate there to at least 2 decimal places.*0567

*In fact, later we are going to be studying Taylor Series, so this is not something that you are supposed to understand yet.*0574

*We will find that the Taylor Series for e*^{x} is the sum of x^{n}/n!.0582

*That is Taylor Series stuff, we have not learned it yet.*0595

*When we do learn it, we will realize that e*^{-1} is the sum of -1^{n}/n!.0598

*Which is exactly the series we have been looking at.*0607

*And e*^{-1}, that is 1/e, if you plug that in, then you get approximately 0.3679.0610

*That is really the true sum of the series, 0.3679.*0624

*What we got was 0.375, so we did get pretty close.*0629

*We did in fact get closer than 0.01, so our estimate is accurate.*0636

*Hi we are here trying more examples of the ratio test and the root test.*0000

*We have a kind of complicated one, -100*^{n} n^{4}/n! × ln(n).0004

*That one looks a little nasty but you will see that the ratio test handles it pretty well.*0014

*So, ratio test says you look at a*_{n}+1/a_{n}.0022

*I am going to work out this whole expression, but substituting an n+1 for n.*0028

*We get -100*^{n+1} (n+1)^{4}/n+1! × ln(n+1).0036

*All of that divided by the same thing just with n, so -100*^{n} n^{4}/n! × ln(n).0060

*Again, I am going to flip the denominator and bring it up and multiply it into the numerator,*0070

*I am going to go ahead and pair the factors off with factors that look like them.*0078

*We get 100*^{n+1}/100^{n}.0084

*By the way, I am getting rid of the negative signs because we are taking absolute values there.*0089

*n+1/4/n*^{4}.0096

*Let us see, this n! is going to flip up into the numerator,*0103

*So we have n!/n+1!.*0109

*Then the ln(n) is going to flip up into the numerator.*0112

*ln(n) × ln(n+1).*0117

*Then let us look at each one of those factors separately.*0122

*100*^{n+1}/100^{n} is 100.0126

*Now (n+1)*^{4} is going to be some big complicated polynomial.0130

*The key thing here is that n*^{4} plus some smaller terms.0136

*We are dividing it by n*^{4},0145

*So both in top and bottom there, the biggest term is n*^{4}.0148

*We can divide top and bottom by n*^{4},0154

*So, that will make the n*^{4} terms go to 1 when you take the limit.0161

*All of the smaller terms go to 0.*0168

*This just goes to 1 as n goes to infinity.*0171

*Remember, that was the pattern we saw before with polynomials.*0177

*When you take the ratio of n+1 and n in a polynomial, you always get 1.*0180

*That is very useful for the ratio test.*0186

*n+1!, remember we can write that as n!.*0189

*× n+1, so the n! cancels with the n! in the numerator.*0195

*We get 1/n+1.*0201

*Then ln(n)/ln(n+1).*0204

*For that we could say that that goes to infinity/infinity.*0209

*We could use l'hopital's rule.*0215

*L'hopital's rule says you take the derivative of the top and bottom,*0219

*That is 1/n divided 1/n+1.*0220

*So that is n+1/n.*0226

*The limit of that as n goes to infinity is just 1.*0231

*You put all of these things together and you get, 100/n+1.*0240

*Now, you can take the limit as n goes to infinity.*0249

*Now 100 is constant and n is getting bigger and bigger, so this limit is 0.*0255

*We can say l=0, so the series converges absolutely.*0259

*By the ratio test.*0280

*The key thing about the 0 there is that it is less than 1.*0288

*That was the cut off to check for the ratio test.*0295

*There is a lot of things we can learn from this example.*0300

*If you look at the different terms here, which ones work well for the ratio test?*0305

*Well, the -100, if we follow the -100,*0312

*If we kind of trace those terms through the ratio test,*0317

*That gave us a 100, which was very very good for the ratio test.*0321

*I like using that for the ratio test.*0328

*Good in the sense that it did contribute to the answer.*0330

*n*^{4}, if you look at n^{4} and follow that through,0335

*That just gave us 1 at the end.*0340

*Remember, in the ratio test, if you get 1 as an answer, it does not tell you whether or not the series converges.*0343

*So n*^{4} was not useful for the ratio test.0350

*n!, we have got n+1! and n!, that cancelled away nicely, we got 1/n+1,*0355

*Which definitely had an influence on the answer,*0366

*So the n! was good for the ratio test.*0370

*ln(n), if you trace those through, ln(n), just ended up giving us 1.*0374

*Remember, if you get 1 as the answer to the ratio test, it gives you no information.*0383

*If you have just things like polynomials like this n*^{4}, or natural logs,0388

*Your answer for the ratio test is going to be 1 which gives you no information and it means you did a lot of work for nothing.*0396

*On the other hand, if you have geometric terms, like -100*^{n},0405

*Or factorial terms, like n!,*0410

*Then the ratio test is very useful in giving an answer.*0415

*Let me summarize that there.*0418

*If you have geometric terms, like 100*^{n}, or factorial terms,0421

*You definitely want to use the ratio test.*0434

*Because it will give you a useful answer.*0439

*On the other hand, if you have polynomial terms,*0444

*Or natural log terms, remember polynomials like n*^{4},0450

*When you run those through the ratio test, what you get for your answer is 1.*0456

*The ratio test gives you no information if your answer comes out to be 1.*0463

*If you see those turns, do not use the ratio test.*0470

*You will be wasting your time and you will not get an answer.*0473

*Do not use the ratio test on those terms.*0480

*Unless you also have some of the good terms, the geometric and factorial stuff.*0483

*Good factors.*0502

*To summarize that, the ratio test works very well on geometric and factorial stuff.*0506

*It works badly on polynomial and natural log stuff.*0512

*However, if you have a mix of both,*0516

*You do want to use the ratio test.*0520

*Because the ratio test will sort of sweep away the polynomial and the natural log stuff.*0523

*It will leave you with the good stuff, the geometric stuff and the factorial stuff that will actually give you a meaningful answer.*0530

*That is kind of when to use the ratio test and when not to use the ratio test.*0541

*How you can tell ahead of time whether the ratio test is going to be a useful one,*0546

*Or whether it is going to be a big waste of time.*0550

*We are going to try one more example here.*0000

*The sum of 1-2/n*^{n2}.0004

*What I notice right away about this one is that I have some stuff involving n, the 1-2/n,*0010

*Raised to some stuff involving n in the exponent.*0016

*I have n both in the base and the exponent.*0022

*That is not something I want to use the ratio test.*0027

*Do not use the ratio test for something like that.*0028

*Instead, you want to use the root test for something like that.*0032

*When you have something involving n raised to something involving n in the exponent,*0039

*You want to use the root test on that.*0044

*Let us try this out using the root test.*0047

*Remember that says you look at the absolute value of a*_{n}^{1/n},0050

*That is the nth root of the terms.*0055

*This is 1-2/n raised to the n*^{2}, all raised to the 1/n power.0059

*I am going to get rid of the absolute values right away because 1-2/n,*0074

*After n gets big enough, that is going to be a positive number.*0080

*This is 1-2/n.*0085

*Now n*^{2} raised to the 1/n,0090

*Remember you multiply those exponents.*0092

*That is n*^{2}/n, that is just n.0097

*Now this is a tricky expression, 1-2/n*^{n}.0103

*We have seen an example like this before in another lecture. *0108

*The trick here is when you have something complicated raised to a complicated exponent,*0112

*You try to separate them out.*0117

*You write this as e*^{ln(ab)}, the point of doing that is you can pull the exponent b out of the natural log.0121

*You can write it as e*^{b}, ln(a).0132

*Here the a is the 1-2/n and the b is n.*0138

*We can write this as e*^{n} × ln(1-2/n).0142

*Now we are just going to focus on the exponent.*0155

*Because that will take a little work to sort out.*0158

*We will come back later and incorporate the e term.*0162

*Let us just look at that exponent.*0165

*n × ln(1-2/n).*0166

*n × ln(1-2/n), well n goes to infinity.*0175

*ln(1-2/n), well 2/n goes to 0.*0180

*That is ln(1) which is 0.*0186

*We have an infinity × 0 situation.*0189

*You cannot do anything with infinity × 0, it could come out to be anything.*0193

*What I mean is you cannot do anything directly.*0197

*What you can do is write it as ln(1-2/n)/1/n.*0200

*That is going to, the numerator is still 0, 1/n is now 0, that is going to 0 over 0.*0214

*That is a situation where you can use l'Hopital's rule.*0223

*We are going to invoke l'Hopital's rule here.*0228

*Remember l'Hopital's rule, you take the derivative of the top and bottom separately.*0233

*It is not like the quotient rule from Calculus 1.*0241

*It is kind of different.*0243

*You take the derivative of the top and bottom separately.*0244

*I will take the derivative of the bottom first because it is easier.*0248

*Derivative of 1/n is -1/n*^{2},0250

*Now the derivative of the top, natural log of something,*0256

*Is 1/that something, 1-2/n.*0261

*× derivative of the inside part.*0263

*That is the derivative of 1-2/n is - 2/n*^{2}, and the negative of that turns that into a positive.0267

*The derivative of 2/n is -2/n*^{2}, derivative of -2/n is positive 2/n^{2}.0283

*Let us clean this up a little.*0294

*The 1/n*^{2} cancels with that 1/n^{2} and we are left with negative from the denominator 2/1-2/n.0296

*Now, as n goes to infinity, the 2/n goes to 0.*0307

*This whole thing goes to -2.*0317

*Remember, this was the exponent.*0320

*If we incorporate back in this e that we had before,*0324

*The true limit here, a*_{n} to the 1/n goes to e^{-2}.0329

*So, our limit l that we checked for the root test is e*^{-2}.0340

*That is the same as 1/e*^{2}.0347

*We have to figure out approximately how big that is.*0350

*e is approximately 2.7.*0354

*1/e*^{2}, I do not know exactly what it is, but I know that it is definitely less than 1.0357

*That is the cut off for the root test, and so,*0365

*If l < 1, that tells us that the series aN is absolutely convergent by the root test.*0370

*Let us recap what we did here.*0402

*The first thing we did when we looked at this series was to say,*0407

*I see an n in the base and an n in the exponent.*0410

*That is something complicated, not something that the ratio test is going to work well on.*0414

*That tells me right away, try for the root test.*0421

*I set up aN to the 1/n.*0425

*That simplifies down a little bit and we get to this step right here.*0428

*When we look at that step, again looks like something a little complicated.*0433

*I can sort it out using this e*^{ln} rule.0439

*You sort it out using this e*^{ln} rule,0442

*And if you look at the exponent, you get infinity × 0.*0447

*You can rewrite that as 0/0.*0451

*Which allows you to invoke l'Hopital's rule.*0454

*You work l'Hopital's rule through and you get -2.*0459

*But that -2 is actually just the exponent because you still have this e from earlier.*0462

*So you get e*^{-2} as our final limit.0470

*The important thing about that was that it was less than 1,*0472

*And, the root test says that if your limit is less than 1, then the whole series is absolutely convergent.*0477

*This is educator.com, thanks for watching.*0486

*Hi, this is educator.com and we are here to learn about the ratio test and the root test.*0000

*Now, there are some definitions we need before we get started.*0007

*From now on, before you are given a series,*0010

*You are going to be asked two different questions packed into one.*0014

*The first is whether the original series converges.*0017

*The second question is if you take the absolute value of each term of the series, does it still converge?*0021

*The point there is that you are making all of the terms positive,*0031

*That potentially makes the series bigger and might make it diverge.*0033

*There are three classifications that we are going to be using for series from now on.*0039

*We are going to say it is absolutely convergent if the original series converges and the absolute series converges.*0047

*The words absolutely convergent are supposed to make you think of something that is very strongly convergent.*0054

*In the sense that, I say super-convergent, even if you make all of the terms positive,*0064

*Even if you make all of the terms as big as you can, that series still converges.*0070

*That is absolute convergence.*0075

*Conditional convergence means that the series converges,*0078

*But the absolute series diverges.*0084

*The way you want to think about that is the series just barely converges.*0090

*It does converge, but if you make all of the terms positive,*0092

*Then it gets so big that it diverges.*0096

*It only just barely converged because some of the original terms were negative.*0100

*It had to have some of the negative terms to make it converge.*0105

*If you make all of the terms positive, it diverged.*0108

*Then finally, we will talk about divergent series,*0110

*Means that the series diverges and the absolute series also diverges.*0115

*Now if you think about this, there is one case missing from those 3 categories.*0122

*The case that is missing is what if a*_{n} diverges,0127

*But the absolute series converges.*0137

*Now, what about that possibility?*0145

*The answer is that that possibility cannot happen.*0148

*Essentially, you want to think about it as the absolute series is always at least as big as the original series.*0150

*If the absolute series converges, then the original series must converge as well.*0158

*That is what this theorem is saying.*0164

*If the absolute series converges, then the original series must converge as well.*0166

*You cannot have the situation where the original series diverges but the absolute series converges.*0172

*That never happens.*0178

*There are really only three possible answers for all the series that we are going to study today.*0180

*We have a couple tests to help you determine those.*0185

*You will be given a series,*0190

*And, one nice thing here is that you do not have to check anything about whether the series has positive terms or anything like that.*0191

*This is a very strong test.*0200

*It works for any kind of series.*0203

*For any kind of series where you can calculate this limit.*0206

*What you do is you look at the n+1 term, and you look at the n term,*0208

*You divide them together, you take the absolute value, and then you take the limit of that.*0213

*Whatever you get, you call it l.*0218

*Now if l is less than 1, then the series converges absolutely.*0221

*If l is bigger than 1, the series diverges.*0226

*The natural inclination at this point is to say that if l=1 then the series converges conditionally, because that is the remaining case.*0230

*That is not true.*0238

*If l=1, then it kind of breaks down.*0240

*The ratio test gives you no answer, and there are examples where the series converges absolutely.*0244

*There are other examples where the series converges conditionally.*0251

*There are others where it diverges.*0254

*Essentially if l=1, then you have wasted your time using the ratio test,*0256

*Because it does not give you an answer, you have to find something else.*0262

*That is the downside of the ratio test, if you get l=1.*0266

*However, if you get l = bigger than 1 or less than 1, that includes infinity and 0, *0268

*Those are OK to use here, then you do get a good answer.*0272

*You can say that the series converges absolutely, or diverges. That is the ratio test.*0278

*We will do some examples in a second but I want to introduce you also to the root test.*0284

*Which looks a lot like the ratio test but instead of calculating a ratio, you calculate a root.*0287

*You look at absolute value of the terms, and then you take the nth root.*0297

*Remember, that is the same as taking a*_{n} to the 1/n power.0302

*You take the limit again and you call that number l.*0306

*Then you have the exact same conclusions as the ratio test.*0310

*If l < 1, it converges absolutely, if l > 1, it diverges, and if l = 1, then you might hope to say it converges conditionally,*0314

*But that would be very unjustified.*0326

*You would have to look at some other tests,*0330

*Because just based on the root test, we do not know yet whether it converges conditionally, converges absolutely, or diverges.*0332

*Let us try that out with some examples.*0343

*You will see how it works.*0345

*First example I have is -e*^{n}/n!.0347

*So we are going to go for the ratio test.*0351

*We are going to look at, remember a*_{n} +1/a_{n} , and the absolute value of that.0353

*aN+1 means you look at each term of the series and you plug in n+1 wherever you see n.*0360

*So that is -e*^{n+1}/n+1!.0369

*a*_{n} just means the original term, -e^{n}/n!.0376

*Take absolute value of that.*0382

*What I am going to do is flip these two fractions.*0385

*Flip the denominator so we can compare it with a numerator.*0388

*I will write that as e*^{n+1}/e^{n} × n!/n+1!.0393

*There are a couple things to explain here.*0403

*Notice with the n! and the e*^{n},0406

*I got that just by flipping the denominator up and rearranging them to match their respective factors in the numerator.*0410

*The other thing that I noticed was you took out the negative signs and that is because of the absolute values.*0419

*The absolute values make everything positive so I can just throw away the negative signs there.*0425

*Now, let us work with this a bit.*0428

*e*^{n+1}/e^{n} is just e.0431

*Now n!/n+1!, remember n+1 factorial.*0435

*That means you multiply all the numbers together, up to n × n+1.*0443

*You can write that as n! × n+1.*0452

*We will write that as n! × n+1, and now the n! cancel.*0458

*So you get, e/n+1.*0464

*Now, remember the ratio test says you take the limit of this as n goes to infinity.*0468

*We will take the limit as n goes to infinity.*0475

*e is just a number, it is 2.7, well 2.7 and on and on and on.*0480

*n+1 is getting bigger and bigger and bigger,*0487

*So we are dividing a constant by a bigger and bigger number.*0491

*That limit is 0. Our l=0.*0493

*Now the important thing in the ratio test is you check whether that limit is < or > 1.*0499

*0 is certainly less than 1, so the series is absolutely convergent by the ratio test.*0506

*That is the conclusion that we can make here.*0520

*Just to recap here, we look at the series that we are given,*0530

*We call that a*_{n} .0535

*Ratio test says you divide a*_{n} +1/a_{n} .0540

*aN+1 means you plug in n+1 wherever you saw n.*0544

*You did a little bit of algebra, the absolute values got rid of the negative signs, the factorials cancel each other out,*0548

*The limit came out to be 0, that is less than 1, and the ratio test then says that it is absolutely convergent. *0556

*The next example, we have n! over n*^{2} + 2.0564

*Let us go for the ratio test again.*0568

*so a*_{n} +1/a_{n} .0573

*Is n+1!/n+1*^{2} + 2/n!/^{2} + 2.0578

*I am going to get rid of the absolute values because everything here is positive.*0596

*Again I am going to flip the denominator up and match it with the numerator.*0600

*We get n+1! divided by n! and then n*^{2} + 2/n+1^{2} + 2.0604

*These are multiplied together.*0617

*n+1!/n!, remember you can write n+1! as n+1 × n!.*0621

*We learned that in a previous example so that cancels down to just n+1.*0634

*Now, n*^{2} + 2/n+1^{2} + 2.0640

*This is a very common pattern where you have a polynomial.*0645

*The n version of it, in one part and the n+1 part in the other part.*0649

*The same pattern always works in that you have n*^{2} + some other stuff divided by,0656

*Now if you work out n*^{2}, you can work it out, you are going to get n^{2} + some smaller terms, guaranteed.0664

*If you divide top and bottom by n*^{2}, in the limit that thing is going to go to 1.0673

*That always happens when you have a polynomial and you plug in n+1 and n.*0685

*It always goes to 1.*0692

*In that sense, the ratio test, what it is kind of doing, is sweeping a polynomial term away from our series because it is just sending that limit to 1.*0693

*What we have is n+1 × 1.*0705

*If we take the limit as n goes to infinity, of n+1 × 1, that is infinity which is certainly bigger than 1.*0710

*So l here, our l is infinity.*0722

*That is bigger than one.*0725

*The series diverges by the ratio test.*0729

*Remember in some of our other tests, like the limit comparison test,*0740

*You were not allowed to have infinite limits or 0 limits.*0747

*You were only allowed to have finite non-zero limits in the limit comparison test.*0752

*The ratio test is much more forgiving.*0754

*You can have any limit as long as the limit exists.*0758

*It can be 0, it can be any number, it can be infinity,*0762

*All you have to check is it less than 1 or is it bigger than 1.*0768

*If it turns out to be 1, that is the 1 case where you out of luck.*0772

*If it comes out to be one, then the ratio test does not give you an answer.*0776

*But 0 or infinity, they are ok for limits in the ratio test,*0780

*And they do give you an answer.*0783

*In this case, infinity is certainly bigger than 1 so we can say the series diverges by the ratio test.*0786

*Our third example is a*_{n} = n+1^{n}/9^{n}.0795

*That is one that is not very conducive to the ratio test.*0800

*The reason is that we have n's in the base and in the exponents.*0806

*I do not like using that for the ratio test.*0811

*Instead we are going to try the root test.*0814

*I am hoping that the root test will clear away some of those n's in the exponent.*0818

*Remember the root test says you look at a*_{n} , well the nth root of that,0829

*That is the same as a*_{n} ^{1/n},0832

*So that is n-1*^{n}/9^{n}, absolute value of that to the ^{1/n},0836

*The absolute values do not do anything here because the terms are positive.*0847

*n*^{1/n} is just 1, so we get n-1^{1}/9.0855

*Then we take the limit of that.*0863

*As n goes to infinity, 9 is just a constant, so that limit is going to infinity.*0866

*Here, the l is infinity, which is certainly bigger than 1.*0873

*The series, we can say it diverges, by the root test, because we got a limit bigger than 1.*0880

*So, again, what made that series something that we wanted to look at using the root test.*0900

*The answer is that we had something with n in it, raised to the n power.*0910

*That is almost always a case that you want to refer to the root test.*0917

*When you see something with n in it, raised to the n power, that is the kind of thing that you want to use the root test for.*0925

*Because the ratio test is going to be really ugly.*0935

*Here we had an n-1*^{n}, we know that we want to use the root test.0936

*The root test says you look at a*_{n} ^{1/n}, or the nth root of a_{n} ,0944

*You work out the limit and again if it comes out to be bigger than 1 it diverges,*0950

*Less than 1 converges absolutely.*0955

*If you are unfortunate enough that it is equal to one then you have no information and the root test fails you.*0958

*So we will try some more examples later on.*0966

*OK, here we are working with some more examples of power series.*0000

*In particular, finding the intervals of convergence.*0004

*Here I have one that looks very complicated.*0008

*x-2*^{n}/6^{n} × a big polynomial.0012

*So again, we will start out with a ratio test.*0015

*The ratio test says that you look at the n+1 term divided by the n term.*0019

*So I get x-2*^{n+1}/6^{n+1}, and then this big polynomial n+1^{2} + 3 × n+1 - 7.0027

*All of that divided by the a*_{n} term, so x-2_{n}.0048

*Divided by 6*^{n}, and n^{2} + 3n - 7.0054

*That is large and messy, but what I am going to do is I am going to flip the denominator,*0062

*And pair off the factors in the denominator with their similar factors in the numerator.*0070

*I will get x-2*^{n+1}/x-2^{n}.0078

*6*^{n} in the numerator, 6^{n+1} in the denominator.0089

*In the numerator I get n*^{2} + smaller stuff.0102

*In the denominator I get n+1*^{2} + smaller stuff.0109

*I still need absolute values at least on the x part.*0116

*In turn, that simplifies down the x-2, everything cancels there but x-2, in absolute value.*0123

*6*^{n}/6^{n+1} cancels down to 6 in the denominator.0134

*n*^{2}/n+1^{2}, that is a big polynomial, but both of the big polynomials,0142

*Their leading term is n*^{2}, so if you divide both terms by n^{2},0150

*What happens there is that the n*^{2} terms will go to 1 and everything else will have an n in the denominator,0159

*The whole thing will go to 1. Just 1/1.*0166

*Remember the whole point of doing the ratio test is that it responds very well to the geometric terms.*0171

*That is x-2*^{n}/6^{n}.0178

*It gives you a really nice answer with those terms.*0181

*When you do the ratio test on a polynomial, it just gives you 1 when you take the limit*0184

*This polynomial, this big polynomial, the ratio test kind of sweeps it away and kind of makes it into a 1.*0190

*Our answer, when we take the ratio test is just the absolute value of x-2/6.*0201

*The ratio test says that whatever you get for your l,*0207

*That should be less than 1 for the thing to converge.*0213

*So now I just have a little bit of algebra to figure out which values of x make that work.*0216

*I can solve that into x-2 < 6, so x-2 < 6 and less than -6.*0220

*So, x is between 8 and -4 there.*0238

*That is the region within which the series is absolutely convergent. Outside that region it is divergent.*0247

*And remember, the endpoints we never know, we always have to check those endpoints individually.*0256

*Let us check those endpoints.*0263

*We will check those endpoints individually.*0276

*The first is x=-4, so if you plug in -4, you get -4 - 2*^{n}.0278

*So that is -6*^{n}/6^{n}.0287

*Then, the n*^{2} term.0293

*The -6*^{n} and 6^{n} gives you -1^{n}/n^{2} term.0300

*That is a series that we know converges because it is an alternating series and it converges by the alternating series test.*0311

*That was checking the left endpoint, x=-4.*0327

*x=8, that gives us the series, ok 8-2*^{n} is 6^{n}/6^{n} × the n^{2} term.0332

*That simplifies down to 1/the n*^{2} term.0351

*That is a series that is very similar to 1/n*^{2}.0362

*We can use the limit comparison test with the sum of 1/n*^{2}.0369

*I will not work out the details of that.*0378

*You divide the two series together, you calculate the limit, the limit is going to turn out to be 1.*0382

*The important thing here is that 1/n*^{2} is a p series.0388

*With p=2. That is convergent.*0395

*The limit comparison test says whatever the new series does, the old series must have done too.*0405

*That one converges as well.*0415

*Both endpoints in this case turned out to give us convergent series.*0420

*The interval, our final answer here, is x goes between -4 and 8,*0426

*But since both endpoints make it converge, I am going to put equal signs on this.*0436

*We include both endpoints there.*0441

*If you write that in interval notation, it is -4 to 8, with both endpoints included there.*0443

*Let us recap there.*0455

*Again, power series we worked with the ratio test, so we worked out the big ratio.*0458

*It looks really nasty but remember the point of the ratio test was it kind of reduces you down to geometric terms.*0463

*The polynomial terms all get swept away to 1.*0472

*So, the polynomial terms get swept away, we are reduced down to x-2/6 < 1.*0476

*Do a little algebra to simplify that.*0483

*Then we just have to check whether these two endpoints are included or not.*0487

*The endpoints, remember, you cannot use the ratio test.*0491

*Because the ratio test, if you did use it, it would always give you 1.*0495

*Because those are exactly the cutoff points where the ratio test gives you 1.*0499

*You cannot use the ratio test, so for each endpoint we plug it in separately and use some other test,*0505

*It turns out that both of those do converge.*0512

*So, when giving the answers, we include both endpoints there.*0516

*One final example here is we want to find the function arctan(x) as a power series*0000

*Centered at 0, and find the interval of convergence of that.*0007

*Again, the trick here is to think about derivatives and integrals and remember that the derivative of arctan(x) is 1/1+x*^{2}.0012

*Then, that is something that we can write as 1/1-(-x*^{2}).0028

*The point of that is to make it look like the sum of a geometric series.*0038

*That is equal to the sum from n=0 to infinity of -x(*^{2})^{n}0042

*The easiest way to think about that is probably to work backwards.*0054

*Remember that its sum will be exactly given by that formula.*0058

*That is for -x*^{2} in absolute value less than 1.0063

*Motivated by that, we will write 1/1+x*^{2}.0071

*I am going to expand out this series in two individual terms.*0079

*That n=0 gives you 1, - x*^{2} + (x^{2})^{2}, so that is x^{4} - x^{6}, and so on.0085

*This is for x*^{2} in absolute value less than 1.0100

*That is x is between -1 and 1 there.*0108

*Arctan(x) is the integral of 1/1+x*^{2}.0118

*I am going to integrate both sides here and I get, the integral of this, taking the integral of the right hand side, is x - -x*^{3}/3 - x^{5}/5 - x^{7}/7,0128

*We still have the problem of adding on that constant, so I will add it on at the beginning this time.*0155

*C + x + x*^{3}, and so on.0162

*To find the value of C I am going to plug in x=0,*0166

*So I get arctan(0) = C + 0 + a bunch of zeroes.*0172

*arctan(0) = 0, so I get C = 0, so again I do not need to worry about my constant here.*0182

*I get arctan(x) = x - x*^{3}/3 + x^{5}/5 - x^{7}/7 and so on.0190

*If you want to write that in a nice concise form, you can write that as the sum from n=0 to infinity,*0208

*Now I just want to catch the odd powers here,*0216

*So, a way to catch the odd powers is to write x*^{2n+1}/2n+1.0220

*That will just give you the odd values.*0235

*To make it alternate in sign, I am going to put a negative 1 to the end here.*0238

*So, there is a power series for arctan(x).*0247

*Now, we still have to find the interval convergence.*0252

*What we know is that the original series converged for x between -1 and 1.*0257

*That means the new series, after we take its integral, is also going to converge for x between -1 and 1.*0264

*But we still have to check the endpoints,*0271

*Because when you take the derivative of the integral, that can change the convergence at the endpoints.*0272

*So we check the endpoints, always have to check them separately.*0279

*So x=-1, gives us, if we plug that into this series, -1.*0289

*Now, -x*^{3}, that is - -1^{3}/3 + -1^{5}/3 - -1^{7}/3,0301

*So that is -1*^{3} is -1, and then there is one more negative, so this is plus 1/3,0322

*Oops, I should not have put 3's in all the denominators, those should be 5 and 7.*0336

*So, minus 1/5.*0344

*So 7 -1's and one more -1, gives you + 1/7 and so on.*0345

*This is an alternating series.*0354

*The terms go to 0 and they get steadily smaller.*0357

*They are decreasing so this converges by the alternating series test.*0361

*If we check x=1, the other endpoint, plug that in and we get,*0370

*1 - 1/3, plugging that into the series, + 1/5, - 1/7, and so on.*0375

*That is also an alternating series and it also satisfies the condition of the alternating series test.*0385

*So this converges by the alternating series test.*0392

*So this time both endpoints converge so our interval is x between -1 and 1.*0398

*Both endpoints are included, or if you write it in interval notation, you use square brackets on both of those, form -1 to 1.*0410

*Again, the trick there for finding a power series for arctan(x),*0420

*Was to think about the derivative, 1/1+x*^{2},0424

*And that can be written in a form that lets you convert it into a geometric series.*0429

*Once it is a geometric series, we can undo the derivative by taking its integral and get back to a series for arctan(x).*0435

*To find the integral of convergence, well we just use the old interval of convergence.*0445

*Then, because we took the integral, we have to recheck the endpoints of the series,*0451

*So we check each one of those endpoints separately, they each converge,*0455

*And so we throw each of those endpoints into the intervals. *0460

*Thanks for watching, this is educator.com*0464

*Hi, this is Will Murray for educator.com and we are here to talk about power series.*0000

*A power series is a series of the form, it has a coefficient c*_{n},0007

*Then it has a power of either x*^{n} or x-a^{n}.0014

*The game with power series is you are trying to plug in different values of x and see what values of x.*0020

*When you plug in a value of x, you get a series just of constants.*0030

*Then the question is which values of x make that thing converge.*0035

*If you try out some examples, we will work on some examples soon.*0040

*You start to notice a pattern which is that you always have a center value at a.*0045

*Now you notice that if you plug in x=a here, then you get a-a*^{n}, so you just get a series of 0's, that always converges.0055

*It always converges at x=a.*0064

*It turns out that it always converges in some radius around that center.*0067

*So, there is always some radius r around that center.*0072

*And the series always converges absolutely between those values.*0080

*That is a+r, and this is a-r.*0085

*It always converges absolutely in that region.*0091

*That region is absolute convergence.*0098

*It is always divergent outside that region.*0100

*Then at the two end points, a-r and a+r, for those values of x, it is really unpredictable what the series does.*0105

*Sometimes it will diverge at both of them, it could converge at both of them.*0114

*It could converge at one and diverge at the other one.*0120

*There is no way to predict it ahead of time, just every particular example, you have to check each one of those endpoints separately.*0125

*To recap here, it always converges between a-r and a+r.*0132

*Another way of saying that is that x-a, in absolute value, is less than r.*0138

*Because it is saying x is within r units of a.*0146

*r is called the radius of convergence, it can happen that the radius is 0 or infinite.*0152

*We will see some examples of that.*0158

*Then the interval, what you can do is you can look at the interval from a-r to a+r.*0162

*That interval is called the interval of convergence.*0172

*That interval might include both endpoints or not include both endpoints, or it might include one of them and exclude the other one.*0178

*There are all of these different possibilities about whether the endpoints are included.*0188

*That interval, including whichever endpoint makes it converge is called the interval of convergence.*0194

*That is a lot of definitions to start with.*0203

*It will be easier once we look at some examples.*0206

*Here is an example, we want to look at the interval of convergence for the power series x*^{n}/2^{n} × n.0208

*The trick with power series is you almost always want to use the ratio test.*0216

*You almost always want to start out using the ratio test.*0223

*We will start out using the ratio test.*0229

*Remember the ratio test says you look at a*_{n}+1/a_{n} in absolute value.0231

*That in this case is x*^{n}+1/2^{n}+1 × n+1/x^{n}/2^{n} × n.0238

*All in absolute value.*0254

*I will flip the denominator up and multiply it by the numerator.*0255

*I am going to organize the terms with the terms that look like them,*0259

*We get x*^{n+1}/x^{n}.0264

*2*^{n}/2^{n+1}, and n/n+1.0269

*I cannot get rid of the absolute values completely because the x could be positive or negative.*0280

*This simplifies down into the absolute value of x.*0287

*n/n+1, when you take the limit as n goes to infinity, that just goes to 1.*0291

*Then 2*^{n}/2^{n+1} just gives you a 2.0298

*Remember, the ratio test says it converges absolutely whenever that limit is less than 1.*0304

*We look at abs(x)/2 < 1, and that is the same as saying the abs(x) < 2.*0313

*So, x is between -2 and 2.*0326

*Those are the values of l that give you a value less than 1.*0333

*Those values of x make this series absolutely convergent.*0340

*If x < -2, or x > 2, we know that the ratio will be greater than 1.*0344

*The series will be divergent.*0355

*Let me fill in what we have learned so far.*0358

*From -2 to 2, we know that it is absolutely convergent.*0361

*If it is less than -2, we know it is divergent,*0372

*Or if x is bigger than 2, we know it is divergent.*0381

*The question we have not answered yet is what happens at the endpoints.*0383

*A big rule here is you cannot use the ratio test at the endpoints.*0387

*The reason is the endpoints are exactly where the ratio test gives you 1.*0393

*The ratio test, when it gives you 1, tells you nothing.*0398

*You have got to use some other test at the endpoints.*0403

*Let us check the endpoints.*0408

*Without using the ratio test because I know if I use the ratio test, I am just going to get 1.*0413

*Let us check these separately.*0423

*If x=-2, then our series turns into -2*^{n}/2^{n} × n, which simplifies down into -1^{n}/n.0425

*That is a series that we have investigated before.*0442

*This converges by the alternating series test.*0448

*That is the alternating harmonic series, and the alternating series test applies to it.*0454

*x=2, gives us the series of 2*^{n}/2^{n} × n, which gives us the series the sum of 1/n.0460

*That is the harmonic series, or if you like, that is a p series with p=1 and we know that diverges.*0477

*We have seen that one before.*0481

*That is the harmonic series.*0486

*You can also think of it as a p series with p=1.*0491

*The important thing there is that 1 is less than or equal to 1.*0498

*Anything < or = 1 makes it diverge.*0501

*So x=-2 makes it converge, x=2 makes it diverge.*0508

*So, the interval of convergence is everything between -2 and 2.*0512

*And we include -2 because it makes it converge.*0533

*We exclude 2 because it makes it diverge.*0536

*Another way of writing that in interval notation is to say the interval from -2 to 2,*0540

*You put square brackets on the -2, that means it is included,*0546

*Round brackets on the 2, that means it is excluded.*0551

*That is your final answer there.*0558

*Let me recap here.*0560

*The important way to approach power series, most of the time you want to start out with the ratio test.*0563

*At least 95 times out of 100, you want to start out with the ratio test.*0571

*You work through the ratio test and you get your limit.*0577

*You set that less than 1, because remember less than 1 is what the ratio test looks for to tell you the series converges.*0581

*That tells you the basic interval except it does not tell you the endpoints.*0592

*It tells you that it is absolutely convergent between those points, divergent outside those points.*0597

*Then you check those two endpoints separately.*0605

*At first you do not know what the endpoints do, and remember, you cannot use the ratio test for the endpoints, because the ratio test will give you 1.*0610

*When the ratio test gives you 1, you get no information and you have to try something else -- you have to plug them in individually.*0619

*x=-2, it turns out that that one works,*0627

*x=2, it turns out that it makes it diverge.*0631

*When we are writing the interval we include x=-2, and exclude x=2, and if we write that in interval notation, we put square brackets on -2, and round brackets on 2.*0637

*Let us try another one.*0654

*We want to find the interval of convergence for the power series of -1*^{n} × x^{n}/n!.0656

*Again, we will start with the ratio test.*0663

*We will look at a*_{n}+1/a_{n}.0667

*I know that these -1's will get swept away by the absolute values anyway, so I am not even going to write the -1 terms.*0672

*I am just going to write x*^{n+1}/n+1! divided by the a_{n} term, x^{n}/n!.0680

*Still need my absolute values here.*0694

*If I flip my denominator and pull it up, we get x*^{n+1}/x^{n} × n!/n+1!.0697

*That simplifies down to x abs(x).*0711

*Now n+1!, remember you can write that as n! × n+1.*0718

*So, the n!'s cancel and we are just left with x/n+1.*0725

*Remember, we take the limit of this as n goes to infinity.*0733

*We are plugging in different values of x here, but whatever value of x we plug in,*0738

*It is a constant. n is the one going to infinity.*0747

*When n goes to infinity, no matter what value of x you start with, this thing goes to 0.*0752

*For all values of x, no matter what value of x you start with.*0764

*This thing goes to 0 because n is the one going to infinity.*0770

*So, l is 0 no matter what you start with.*0776

*The key point here is l < 1, so the ratio test says that no matter what value of x you put in there, it is absolutely convergent.*0778

*For all possible values of x, for all real numbers of x.*0810

*If you want to write that in interval notation, the answer would be (-infinity, infinity).*0816

*I put round brackets there because there is no question of endpoints here because infinity and -infinity are not actual numbers we can plug in there.*0824

*There is no question of plugging in endpoints or considering endpoints to be included or excluded.*0833

*There are no endpoints.*0840

*We say that the interval runs over all real numbers from -infinity to infinity.*0845

*Another way of saying that would be that x going from -infinity to infinity.*0853

*We started here by applying the ratio test.*0861

*We wrote down the ratio, we took the limit as n goes to infinity, because that 0, which < 1, no matter what value of x it is, we get that it converges for all possible values of x.*0866

*Our third example here, we are not given a power series, we have to write a power series ourselves.*0882

*For ln(1-x).*0891

*Then we have to find the interval convergence.*0892

*This one is not so obvious how to start unless you have seen something like this before.*0895

*The key point is to remember that the derivative of ln(1-x),*0901

*Is 1/1-x × derivative of 1/x, so that is -1/1-x.*0914

*1/1-x is exactly the sum of a geometric series.*0926

*That is the sum from n=0 to infinity of x*^{n}, put a negative there.0932

*That is a geometric series and that is true for abs(x) < 1.*0943

*That is probably easier to think about if you work backwards.*0952

*If you look at this geometric series, that sums up to the first term/1 - the common ratio,*0955

*So that is equal to 1/1-x.*0964

*Armed with that intuition, we can write ln(1-x) as the integral of 1/1-x dx.*0970

*Except there was that negative sign, so I will put that on the outside.*0981

*That is the integral, now 1/1-x we said was this geometric series.*0988

*That is 1+x+x*^{2}, and so on, dx.0995

*If you integrate that, you get negative, integral of 1 is x, integral of x is x*^{2}/2,1009

*Integral of x*^{2}, is x^{3}/3, and so on.1020

*But this is not an indefinite integral, so we always have to include a constant.*1024

*How do we figure out what the constant should be.*1032

*To find the constant, we are going to plug in x=0 to both sides.*1035

*We get ln(1-0) = -, well if we plug in 0 to a bunch of x terms, we get a bunch of 0's + C.*1043

*So, the constant is ln(1), which is 0.*1060

*So that is nice, the constant disappears.*1064

*We get ln(1-x) is equal to, I will distribute the - sign, -x - x*^{2}/2 - x^{3}/3, and so on.1067

*We can write that as the sum from n=1 to infinity, of -x*^{n}/n.1087

*Notice that that is what we would have gotten if we had integrated this geometric series directly.*1105

*If we took the integral of x*^{n}, we would have gotten,1111

*Well, x*^{n+1}/n+1, then shifting the indices from n=0 to n=1,1114

*Would have converted that into x*^{n}/n.1123

*That is the power series.*1129

*What we now have to do is find the integral of convergence.*1132

*What we knew before, is that x was between 1 and -1.*1137

*When you take the derivative of integral of a power series, it does not change its radius of convergence.*1144

*We know it still goes from -1 to 1.*1152

*However, it might change what happens at the endpoints.*1156

*So, we still have to check the endpoints of this series.*1162

*Sorry, check the endpoints of the integral, and see whether it converges or diverges at those two endpoints.*1171

*Wee have to check those separately.*1179

*Let us try x=-1 first.*1181

*that would give us the series of -1*^{n}/n,1183

*That is the alternating harmonic series.*1190

*We know that converges by the alternating series test.*1193

*x=1 gives us the sum of -, well 1*^{n}/n,1202

*That is the negative of the harmonic series,*1218

*And we have seen several times that that diverges, either by just saying that that is a harmonic series,*1220

*Or by saying it is a p series, with p=1,*1231

*Since 1 < or = to 1, that makes it diverge.*1235

*x=1 makes it diverge, x=-1 makes it converge.*1240

*So, the interval, you could write that as x between -1 and I am including -1 because it made it converge,*1245

*I am not including 1 because it made it diverge.*1256

*In interval notation that is [-1,1),*1260

*With a straight bracket on -1 because it made it converge and a round bracket on 1 because it made it diverge.*1265

*Our answers there, we found the series by taking the derivative of ln(1-x),*1275

*We get something we can convert into a geometric series,*1283

*Then to get back to ln(1-x), we took that geometric series and we integrated it back up to get ln(1-x).*1291

*Now we have a power series for ln(1-x).*1300

*To get the interval, we used the interval for the geometric series, but then because we are taking derivatives and integrating,*1303

*We know that the endpoints, whether or not it converges at the endpoints, that could possibly change.*1312

*We have to check those again, we check -1, it converges.*1318

*1 diverges, and so we reflect those in the answers for the interval.*1322

*We will try some more examples later on.*1330

*Here we were asked to find the example, to find the McLauren series for e*^{x}.0000

*This is one where we do not have anything memorized yet.*0009

*So, we are going to work out n and the nth derivative of x.*0014

*Since it is a McLauren series, we are going to plug in f*_{n}(0).0017

*Then finally our coefficient is the nth derivative of 0 divided by n!.*0021

*We will work this out for a few values of n.*0032

*f*_{n}(x), well e^{x}, that is very easy, that is e^{x} all the way down.0038

*So, f*_{n}(0), that is just plugging 0 into e^{x}, so that is 1 all the way down.0043

*cN is the coefficient, you divide that by n!.*0054

*We get 1/1, because 0 and 1! are both 1, 1/2!, 1/3!, and so on.*0060

*The McLauren series is, remember, we take the coefficient f*_{n}(a)/n!/x-a^{n}.0074

*Then for the McLauren series the a is just 0.*0092

*So that means each one of these coefficients gets multiplied by a power of x.*0094

*So, it is 1 + 1 × x + 1/2! × x*^{2} × 1/3! x^{3}, and so on.0099

*So, we get the McLauren series for e*^{x} is the sum from 0 to infinity of just x^{n}/n!.0115

*That, again, is such a common function and such a common Taylor Series.*0130

*It comes up in so many different places and in particular Calculus classes, that that is one you should really memorize.*0135

*The first time you ever work it out, you have to use this derivative formula.*0146

*You have to write down the derivatives, plug in x=0 and then you get these coefficients,*0152

*And you multiply each coefficient by a power of x.*0158

*That is kind of the first time that you work it out.*0162

*From the on it is probably worth memorizing that the McLauren series for e*^{x}, is just x^{x}/n!.0164

*OK, for this example we are asked to find the McLauren series for f(x) = (e*^{x})^{2}.0000

*You could try to write down the derivative, so let us try to write down a couple of derivatives.*0007

*We know f(x) = e*^{x}. We are going to write down n fn(x). Oops, f(x) = (e^{x})^{2}.0012

*So f0(x) is just (e*^{x})^{2}.0031

*First derivative is just (e*^{x})^{2} × 2x.0036

*The second derivative is 2x × derivative of (e*^{x})^{2}, so that is 2x × 2x × (e^{x})^{2}.0041

*Plus (e*^{x})^{2} × derivative of 2x that is 2 (e^{x})^{2}.0053

*That can simplify down to 4x*^{2} (e^{x})^{2},0062

*I guess 4x*^{2} + 2, we could combine those terms (e^{x})^{2}.0068

*What you can see is that this is already getting messy and it is going to be difficult to find a pattern.*0076

*Taking the third derivative, things are getting more and more complicated.*0082

*We are already getting into something messy and complicated and in particular, it is going to be difficult to find a pattern.*0088

*What I want to show you is a better way to find this McLauren's series.*0098

*A better way than trying to write down all of these derivatives, which are getting messy and complicated.*0104

*A better way is to remember that we just figured out a McLauren series for e*^{x}.0114

*e*^{x} was the sum from 0 to infinity of x^{n}/n!.0121

*A much easier way than writing down all of these derivatives is to substitute in x*^{2} for x.0136

*So (e*^{x})^{2} is the sum form n=0 to infinity of x^{2}^{n}/n!.0145

*You could write that as the sum from n=0 to infinity of x*^{2n}/n!.0158

*If you want, you could expand this out term by term if you plug in n=0, you get 1.*0168

*n=1 you get +x*^{2}/1.0176

*n=2 gives you x*^{4}/2!.0182

*n=3 gives you x*^{6}/3!, and so on.0187

*Notice that that is just what you would have gotten if you had looked at the series for e*^{x}.0193

*1+x+x*^{2}/2!+x^{3}/3!.0199

*If you had taken that series and just changed x into x*^{2}, that would have given you the series we had just gotten by expanding it out.0208

*That immediately gives us a very McLauren series for (e*^{x})^{2}.0220

*Much easier than writing down all of these derivatives which get messy and very hard to write a pattern.*0227

*The moral there that a lot of calculus students sometimes miss is that,*0236

*Using this derivative process of finding the McLauren series or the Taylor Series is often the slowest and most difficult way, to find the McLauren series.*0240

*A much better way is to have some prefabricated series at your disposal.*0252

*By that I mean, you have memorized some series, some key series to start with.*0258

*Then when you want to find other series, you just work them out based on the series you already have memorized.*0262

*In this case, you already had memorized a series for e*^{x}, and when you have to find (e^{x})^{2} you just do a quick substitution, and we get our series for (e^{x})^{2}0271

*Hi, this is educator.com and we are here to look at Taylor Series and Maclaurin Series.*0000

*So, a couple definitions to get us started here.*0008

*You start with a function f(x), and a value a, the center value.*0012

*Then you form this power series that we call the Taylor series of x.*0018

*You take the nth derivative, so that is the n*^{th} derivative there.0023

*Of the function, you plug in a, you divide it by n!, and then you multiply by x-a*^{n}.0030

*You kind of want to think about this entire term here as being a coefficient.*0041

*We will call that c*_{n}, that is the n^{th} coefficient of the Taylor Series.0047

*Then the x-a*^{n} part is the power part of the Taylor series.0052

*You will also hear about McLauren series, that is the exact same thing as the Taylor Series.*0060

*That just means that you take the special case where a=0.*0068

*So McLauren Series is just a special case of the Taylor Series.*0070

*You just plug in a=0, and so you get something slightly simpler there.*0073

*We are also going to talk about Taylor Polynomials.*0080

*The formula for the Taylor Polynomial looks exactly the same for the Taylor series,*0087

*Except what you do is instead of running it to infinity, you cut the thing off at the degree k term.*0092

*So, you call this t*_{k}(x), and what that means is, essentially that you get this thing that only runs up to x^{k} power.0101

*You have these coefficients that give you a polynomial of degree.*0117

*Well, usually it will be degree K, but if that last term, if the coefficient happens to be 0, it could be lower degrees.*0130

*So I will see degree less than or equal to k.*0138

*So the Taylor Polynomial is a polynomial of degree k, but what it represents is you just take the first few terms of the Taylor Series,*0140

*And you run it up until you see an x*^{k} term.0149

*Let us try some examples there.*0153

*The first example is to find the Maclaurin series for f(x) = cos(x).*0158

*That means, remember, the formula there is the n*^{th} derivative of f at a.0163

*Well a=0 here, because it is a Maclaurin Series.*0171

*fN(0)/n! × normally x-a*^{n}, but since a=0 it is just x^{n}.0174

*I want to figure out what these coefficients are, the c*_{n}'s.0184

*I will make a little chart here.*0188

*n and the n*^{th} derivative of x, then I will plug in the n^{th} derivative of 0.0191

*Then I will figure out what the coefficient n is.*0198

*That is the same as f*_{n}(0) divided by n factorial.0200

*When n=0, we have the 0 derivative of f, which just means the original function.*0208

*The 0 derivative is just cos(x).*0215

*cos(0)=1, and 1/0!, remember 0! is just 1. So 1/0!=1.*0218

*When n=1, we take the first derivative, that is -sin(x).*0231

*If you plug in 0 to that, it is 0, so the coefficient is just 0.*0236

*When n=2, derivative of -sin is -cos(x).*0241

*Plug in 0 you get -1. So -1/2! is -1/2.*0247

*When n=3, we get the derivative of -cos is just sin(x).*0255

*Plug in 0 to that and you get 0. 0/3!=0.*0264

*When n=4, we get cos(x) and so that derivative when you plug in 0 is 1.*0270

*1/4! is 1/4!, if you want you can write that as 1/24.*0279

*cos(x) is, we take these coefficients, and you attach on the relevant powers of x.*0290

*That is 1 × x*^{0} + 0 × x^{1}.0302

*So I will leave that out, - 1/2, 1/2! × x*^{2}, + 0 x^{3}, + 1/4! × x^{4}.0307

*Then you can start to see the pattern here, of course the derivatives of sin and cos are going to repeat themselves.*0325

*The next term will be 0 x*^{5} - 1/6! × x^{6},0333

*0 x*^{7} + and so on.0341

*So, as a series there, we can write this as the sum from 0 to infinity.*0345

*I just want to catch the even powers, because all of the odd ones are 0.*0352

*So I can write x*^{2n}/2n!.0358

*I am going to write -1*^{n} because that makes it alternative positive and negative there.0366

*That is our Maclaurin series for cos(x).*0374

*Now, there are several common Maclaurin series that it is probably worth memorizing for at least as long as you are a Calculus student.*0384

*This is one of them, cos(x) is one you should really memorize.*0395

*You should not have to work out this whole chart every time.*0397

*You should probably know the answer because it will almost certainly come up in your Calculus class.*0403

*Let us try another one. f(x) = sin(x).*0409

*Again we want to find the McLauren series.*0413

*One way to do that is to write a whole chart, like we did in the previous example, *0417

*Where you write the derivatives, you plug in 0 every time, then you divide by the factorials, you string them together and you make a series.*0421

*That is a lot of work; instead, we are going to find something easier,*0430

*Which is to notice that sin(x)=derivative of cos(x).*0435

*Except for one negative sign that I have to include there.*0446

*The point there is that I remember a series for cos(x) because we just did it in the last problem.*0452

*We memorized that series. Let me write down the series that I just memorized for cos(x).*0461

*1 - x*^{2}/2! + x^{4}/4! - x^{6}/6! and so on.0468

*So we get -, now I am just going to take the derivative of that.*0482

*Derivative of 1 is just 0.*0485

*Derivative of x*^{2}/2! is just x. Sorry 2x/2!0488

*+ now the derivative of x*^{4} + 4x^{3}/4!.0503

*- derivative of x*^{6} is - x^{5}/6! and so on.0511

*So you want to go ahead and distribute the - sign so we get +, now 2x/2!,*0520

*2! is just 2 so this is just x-, because of the - sign on the outside, 4x*^{3}/4!.0529

*Well, 4/4!, 4! is 4 × 3 × 2 × 1.*0540

*That 4 in the numerator just cancels away that first 4 and leaves us with 3!.*0548

*Now we have got x*^{5}, 6 in the numerator and 6! in the denominator.0555

*Well the 6 in the numerator just cancels the 6 out of the factorial, and we get 5!*0562

*You can see the pattern here.*0569

*The next one is going to be, sorry this x*^{5}/5! should have been positive,0570

*Because it was negative before but we had a negative on the outside so those two negatives cancel. *0577

*You can see that the next term following the same pattern is going to be,*0581

*x*^{7}/7!, and so on.0586

*If you want to write that in a nice closed summation form.*0591

*We have to find a way of capturing the odd numbers.*0598

*But, that is x*^{2n+1}/(2n+1)!.0601

*And, -1*^{n} will make it alternate positive and negative.0613

*That is our Maclaurin series for sin(x).*0621

*Again, this is such a common one.*0626

*Sin(x) is such a common function that this one is worth memorizing as well.*0628

*The point of asking you to memorize these things is that later on, we could get more difficult Taylor Series, and Maclaurin series.*0636

*It helps if you kind of have a stock of pre-fabricated examples and then you can use those examples to build up more complicated ones*0645

*By taking derivatives, making substitutions, doing other algebraic tricks,*0656

*And you will not have to work everything out from first principles,*0662

*Taking lots of derivatives and so on.*0667

*In particular, in this case, we figured out sin(x) by using the fact that we had already worked out the cos(x) before.*0669

*This is really exploiting the work we did before.*0680

*We remember the answer to that.*0683

*Then we can just take its derivative and figure out a Taylor Series for sin(x).*0688

*So, example 3 is to find the Taylor Polynomial t*_{4}(x) for sin(x).0698

*This time we are not centering it around 0, we are centering it around a = pi/3.*0705

*It is not a McLaurin Series, it is a full-blown Taylor Series.*0711

*This time I am going to start with my chart.*0716

*n f*_{n}(x), the n^{th} derivative, f_{n}(pi/3).0718

*Then the full coefficient c*_{n}, which is f_{n}(pi/3)/n!.0730

*The reason I am setting that up is I am remembering the master formula for Taylor Series, which remember is the n*^{th} derivative x, of a, you plug in x=a,0742

*Divided by n! multiplied by the power part x-a*^{n}.0755

*That is the master formula for Taylor Series and to get the Taylor Polynomial you just cut that off at the k term.*0760

*Let us work out what these values are for the coefficients, for the first few values of n.*0774

*I will work this out for n=0, 1, 2, 3, & 4.*0785

*That is because we only have to go up to the k=4 term, or n=4.*0789

*So sin(x), my derivatives of sin(x), well the 0 derivative is just sin(x) itself.*0796

*You have not taken any derivatives. *0803

*First derivative is cos(x). Second derivative is derivative of cosine, which is -sin(x). Derivative of that is -cos(x).*0805

*Then it starts to repeat at sin(x), if you plug in pi/3 to each of these, the sin(pi/3) is sqrt(3/2).*0817

*Cos(pi/3) = 1/2.*0830

*-sin = -sqrt(3/2),*0833

*-cos is -1/2.*0836

*Then sine is sqrt(3/2) again.*0840

*Now the coefficients c*_{n} says you take those derivative values and you divide them by n!. Well 0! is just 1.0844

*So, we get sqrt(3/2), we are not dividing by anything except 1, and 1! is just 1 so we get 1/2 here.*0855

*But now 2! is 2. We get -sqrt(3/2)/2!, so that is -sqrt(3)/4.*0864

*3!=6, so we get -1/2/6, that is -1/12.*0877

*4! is 24, so we get sqrt(3)/2/24, so sqrt(3)/48.*0884

*I am going to take those coefficients and plug them into my formula for the Taylor Polynomial.*0898

*I will have an n=0 term, an n=1 term, n=2, n=3, and n=4 term.*0906

*The n=0 says that you take sqrt(3)/2 and I am getting that from here.*0917

*Then you multiply by x-a*^{0} which is just 0.0924

*n=1 says we take the 1/2 × x-a is pi/3, so pi/3*^{1}.0931

*n=2 gives us -sqrt(3)/2 × x - pi/3. To the 2 power. That is an x-pi/3.*0942

*n=3 gives us -1/12.*0962

*x-pi/3*_{3} + sqrt(3)/48 × x-pi/3^{4}.0967

*That, all of that together, is t4(x). All of that together is our solution.*0985

*To recap, what we did was we invoked this formula for the Taylor Polynomials, which again is just the formula for Taylor Series.*0998

*We cut it off. Instead of going to infinity, we cut it off at the k term, because this is a=pi/3, it is not a Maclaurin series.*1010

*It is not something where we can use something that we have memorized.*1019

*We just kind of go through and work out these coefficients one by one using the derivative formula.*1022

*Then we plug everything back into the formula where we attach the powers of x-a*^{n},1028

*And we get our Taylor Polynomial.*1035

*We will see how we can use these later on in the next lecture on applications of Taylor Polynomials.*1037

*We will try some more examples later on.*1045

*OK, we are here working on examples of Taylor and Maclaurin series.*1050

*The example we are given here is the Maclaurin Series for f(x) = sec(x).*1056

*The way you want to think about that is you could remember the generic formula for Taylor Series,*1065

*Where you take the n*^{th} derivative of f at x over n! × x-a^{n}.1074

*That is the generic formula for Taylor Series. *1084

*The problem with that is if you start taking lots of derivative of sec(x) it is going to be really messy because those derivatives do not repeat.*1092

*They get uglier and uglier.*1095

*All we are asked to do here is find 3 non-zero terms, so what I am going to do is exploit the fact that sec(x) is 1/cos(x).*1097

*I remember a Maclauran series for cos(x) because we have worked it out earlier and I memorized the answer.*1114

*That is 1-x*^{2}/2+x^{4}/4! is 24.1125

*So, I am filling that in because we already worked out the series for cos(x) and we memorized it.*1140

*It is good if you memorize the series for some of these common ones like sin(x) and cos(x) and e*^{x} because they come up so often.1149

*This is an example of exploiting one we have used before. *1157

*Our sec(x) is 1/c0s(x).*1160

*We are going to do a little long division of polynomials just like you did in high school.*1162

*We do 1-x*^{2}/2 + x^{4}/24. That gets divided into 1.1168

*I am just going to fill in, think of 1 as a big polynomial,*1182

*So I am going to fill in, 1 + 0x*^{2} + 0x^{4}, I am only filling in even terms,1185

*Because I only have even terms in the cos(x), so we are not going to need to use any odd terms here.*1194

*Now we just do long division of polynomials which you might have learned in high school but you might be a little bit rusty. *1200

*We look at these terms, 1 and 1, so we get 1 there. *1210

*Now we multiply 1 by the whole thing, so 1 - x*^{2}/2 + x^{4}/24,1214

*And then we subtract just like we do long division of numbers,*1225

*So that is + x*^{2}/2, then we multiply across there, x^{2}/2.1228

*x*^{2}/2 × -x^{2}/2 is -x^{4}/4 + x^{6}/48.1237

*We subtract again and the x*^{2}/2's cancel.1258

*x*^{4}/4, I can write that as 6/24.1263

*So, we have 6/24 - 1/24, that gives us 5/24 × x*^{4} - x^{6}/481269

*I was only asked to find three non-zero terms,*1285

*So I just need to find one more.*1289

*That would be then, if we look at the leading term 5/24 x*^{4}.1292

*So, that is our answer.*1302

*sec(x) = 1+x*^{2}/2 + 5x^{4}/24.1305

*That is really just the first few terms of the Taylor Series.*1320

*To indicate that there is more to come there, I will put a ...*1323

*The question just asked us to find the first 3 non-zero terms. That is what they are.*1328

*This is an example of a problem where it would be quite difficult to find a general pattern, and to write it out in sigma form,*1336

*That is probably why the question only asked us to find the first few terms.*1345

*So, we just found the first few terms by manipulating these polynomials, using the fact that we already knew what the Taylor Series for cos(x) was.*1349

*Then we manipulated the polynomials, and we got the first few terms of a Taylor Series for sec(x).*1361

*OK, we want to do some more examples of Taylor Polynomial Applications.*0000

*The first problem here is to look at the Taylor Polynomial for cos(x) and we are cutting it off at degree 4, t*_{4}(x).0006

*We want to figure out if we want that to be accurate to 2 decimal places, what values of x can you plug in there.*0015

*Let us start out by recalling the Taylor Polynomial for cos(x).*0022

*That is one that we have memorized.*0028

*That is 1 - x*^{2}/2! + x^{4}/4! - x^{6}/6! and so on.0031

*We have been told to use the Taylor Polynomial t*_{4}(x).0046

*That means you take the Taylor Series and you cut it off at the degree 4 term.*0052

*That does not mean you take 4 terms, it means you look for degree 4, there it is, and you cut if off right there.*0057

*So, t*_{4}(x) is 1 - x^{2}/2! + x^{4}/4!.0066

*What we notice here is that no matter what x you plug into this series,*0078

*It is going to be an alternating series.*0084

*Even if x is positive or negative, you have even powers of x will always be positive.*0087

*So you have a minus, a plus, a minus, plus, and so on this series obeys the alternating series test.*0097

*We can use the alternating series error estimation.*0106

*That says that the error is less than the cut off term, the first term that you did not take.*0115

*So in this case, the cutoff term is x*^{6}/6!.0128

*Remember, in the alternating series error bound, that was a*_{n}+1.0143

*An easier way to think of that is just the cutoff term, so that is x*^{6}/6!.0148

*Now we have been told that we have to make this thing accurate to 2 decimal places.*0155

*Let us interpret that as meaning we want x*^{6}/6!, we want our maximum error to be less than 0.01.0158

*We do not want an error bigger than 0.01, we want to be accurate at the second decimal place.*0173

*We can solve this out.*0178

*x*^{6}, 6! is 720, so 0.01 × 720, we got that from doing 6!.0181

*So, we get abs(x*^{6}) < 7.2,0195

*If we solve that for x we get x < 7.2*^{1/6}.0205

*That is just a value that you can throw in your calculator.*0211

*That turns out to be 1.3896.*0216

*So, the conclusion there is that the Taylor Polynomial t*_{4}(x),0222

*Is an accurate estimate of cos(x) to 2 decimal places.*0234

*In other words, the error will be at most 0.01.*0251

*As long as you stick to values of x that are inside this range.*0260

*For values of x whose absolute value is less than 1.3896, we can plug in those values of x into the Taylor Polynomial, the 4*^{th} degree Taylor Polynomial.0268

*What we will get are answers that are close to cos(x), to an error tolerance of 0.01.*0295

*Just to recap there, we are given a degree and we are asked to make the Taylor Polynomial accurate to 2 decimal places.*0304

*The first thing there is to write out the Taylor Series, cut it off so that you get the Taylor Polynomial.*0315

*We notice that it is an alternating series so we can use the alternating series error bound. *0326

*The error bound says that the error is smaller than the first term that we cut off.*0330

*So here that is x*^{6}/6! and we want that to be less than the allowable error, 0.01.0336

*We solve that out and we get a range of x, and so those are the values of x that we can plug in and get an estimate of cos(x) accurate to 2 decimal places.*0346

*Our last example here is to use the Taylor polynomial for sin(x) around a=pi/3.*0000

*We are going to use that to estimate the sin(1 radian).*0010

*Then we are going to use Taylor's Remainder Theorem to estimate how accurate our approximation was.*0014

*The first thing to do here is to find the Taylor Polynomial for sin(x).*0022

*Actually that was one of the examples we did before, so I am not going to work that out again.*0027

*I am just going to remind you of the answer here.*0030

*t4(x), we did this before, earlier in the lecture.*0034

*was sqrt(3)/2 + 1/2 × x - pi/3.*0039

*This is not a Maclaurin Series, it is not centered at 0.*0050

*It is centered at pi/3, + sqrt(3)/4 × (x-pi/3)*^{2} - 1/12 × (x-pi/3)^{3} + sqrt(3)/48 × (x-pi/3)^{4}/0055

*That was the Taylor Polynomial that we worked out on the previous example.*0084

*I had not done all the work again to work that out.*0088

*To estimate sin(1) just means you plug x=1 in there.*0092

*So, t*_{4}(1)=sqrt(3)/2 + 1/2 × (1-pi/3), and so on.0099

*You plug in x=1 into every term here.*0113

*(1-pi/3)*^{4} × sqrt(3)/48 would be your last term.0120

*You plug that in and I am not going to show you the details of that,*0126

*That is just something that you can work out using a calculator.*0130

*It turns out to be, I am going to write down a lot of decimal places because it turns out that this answer is going to be pretty accurate.*0133

*0.841470985.*0142

*OK, so that is our estimate for sin(1) using the Taylor Polynomial.*0154

*That is the first part of the problem, that was the easy part of the problem.*0161

*Now we have to estimate how accurate that is.*0166

*Remember, we are not allowed to cheat and look up sin(1) on the calculator,*0168

*Because the whole point of this is that we are using the Taylor Polynomial to guess what sin(1) is.*0170

*Instead we are going to use Taylor's Remainder Theorem, and let me remind you how that goes.*0180

*That says that the error is at most m/k+1! × (x-a)*^{k+1}.0188

*We have to address each one of those individual terms.*0208

*Here the k, that is the degree of the Taylor Polynomial, so that is the 4 here. So k=4.*0211

*a is the place where you are centering the Taylor Series, so that is a, a=pi/3.*0221

*The x value that you are plugging in is 1.*0231

*Those are kind of the easy parts.*0235

*The tricky part is m, what do we do about m?*0238

*Well, remember, m is bound on the k+1 derivative.*0242

*M, we want to bound, ask how big can the k+1 derivative,*0250

*Well k=4, so this is the 5th derivative of f(x).*0261

*How big can that be on the interval between x and a.*0266

*So between pi/3 and 1.*0280

*Now, that can be a tricky problem to find out exactly what the maximum value of that derivative could be.*0288

*However, there is an easy way to give us a simple upper bound.*0295

*Note that all derivatives, including the 5th one, of f(x)=sin(x), that is the function we are talking about.*0301

*All of the derivatives are just, well either plus or minus sin(x), depending on where you are in the rotation.*0314

*Or + or - cos(x).*0324

*All of these derivatives are + or - sin(x) or cos(x).*0327

*+ or - sin(x) and + or - cos(x), they are all bounded by 1.*0332

*They never get bigger in absolute value than 1.*0337

*So, any of these derivatives, no matter what value we plug in and no matter what derivative it is, these derivatives are always less than or equal to 1.*0343

*This is a very common estimation when you are talking about sin(x) or cos(x), you have to make an estimate on how big the derivatives are.*0360

*You know that the derivative of sin(x) or cos(x) will always be either sin or cos, and sin or cos never gets bigger than 1 in absolute value.*0369

*So we can take m=1.*0380

*Let me remind you again. *0382

*We are trying to work out this formula, m/k+1! × x-a*^{k+1}.0385

*We figured out now that we can use m=1, and k was 4, x is 1, and a is pi/3.*0400

*Let us work that out.*0415

*x-a is something we are going to have to work out.*0417

*x-a is 1 - pi/3.*0420

*That is 1 - well pi is about 3.14/3.*0428

*3.14/3 is definitely less than 1.05, it is safely less than that.*0440

*So 1 - that is safely less than 0.05.*0450

*That is 1/20.*0461

*So, I can say that this x-a term, this term of the formula is safely less than 1/20.*0465

*Let me plug all of these values in.*0473

*Our error in absolute value is < or = to, m=1, k+1! is 5!, *0476

*Now the x-a term, we just said that is less than 1/20. Then k+1=5.*0490

*So this in turn is equal to, 5! is 1/120.*0500

*1/20*^{5}, is, 20^{5}, well 2^{5}=32 and 10^{5} gives me 5 0's there, so we get 3,200,000 there.0512

*Just to make a rough estimate, this is < or = 1 over, well 120 is less than 100.*0535

*Sorry, 120 is bigger than 100 so 1/120 is less than 1/100,*0548

*And 3,200,000 is bigger than 3,000,000 so we can safely say that 1 over that is less than 1/3,000,000.*0555

*So this entire thing is < or = 3,000,000 × 100.*0569

*So that is 1/300,000,000.*0580

*What that is saying is that, by Taylor's Remainder Theorem,*0595

*Our error is less than 1/300,000,000.*0605

*In other words, the value we got from plugging 1 to t*_{4}(x), t_{4}(1),0622

*Is close to the actual value of sin(1).*0632

*It is so close that the difference is less than 1/300,000,000.*0640

*What I invite you to do is to look at t*_{4}(1).0648

*I will remind you that that was 0.841470985.*0652

*Then, actually take a calculator and work out sin(1) on your calculator.*0664

*I will let you plug that in on your own, and look at how close those values are to each other.*0670

*They really are extremely close and so we can say for sure, even without checking on our calculator, that our answer is less than 1/300,000,000.*0674

*Let us recap that problem.*0685

*What we did here was we first found the Taylor Polynomial t*_{4}(x) for f(x)=sin(x) around a=pi/3.0688

*That was actually something that we did in a previous example problem.*0698

*You can go back and check the previous example problem to see the mechanics of that.*0701

*We took that Taylor Polynomial and we plugged in x=1.*0705

*Into that polynomial, so we did not actually work out sin(1) on a calculator.*0715

*We plugged in x=1 into the Taylor Polynomial, and that is where this value 0.8414 and so on came from.*0720

*Then to estimate the accuracy we had to look at this Taylor Remainder Formula.*0729

*This Taylor Remainder Formula, m/k+1! (x-a)*^{k+1} and we filled in all of our values.0737

*x is 1 because that is the value we are estimating, a=pi/3 because that is the center of the series,*0745

*k=4, that comes from right there, the degree of the polynomial,*0752

*Then m was this bound on the derivative,*0760

*And technically it is a bound on the 5*^{th} derivative of f.0764

*But what I know is when you take derivatives of sin and cosin, they are all bound by 1.*0767

*So just for convenience I used m=1 here.*0772

*Then I worked out x-a, I said that is safely less than 1/20, so I plugged that in.*0778

*k+1! is 1/120, and then just to give myself round numbers, I rounded all the denominators down,*0785

*Which means that I am rounding the entire numbers up, which is still safe to say that the error is less than our answer of 1/300,000,000.*0795

*Our final conclusion there is that our estimate using the Taylor Polynomial must be close to the true value of sin(1), to within 1/300,000,000.*0806

*That concludes our section on applications of Taylor Polynomials, and in fact that concludes all of our Calculus 2.*0820

*So, this has been Will Murray, I hope you enjoyed it,*0827

*This is educator.com.*0829

*This is educator.com and we are here to talk about applications of Taylor Polynomials.*0000

*The idea here is that we are going to write down some Taylor Polynomials, and plug in some values of x and we will see how close we get to the original function values.*0007

*That will probably make some more sense after we study some examples, but I want to give you a couple of tests that we are going to be using to check how accurate we are.*0020

*The first one is one that we have seen before, if you look back at the lecture on alternating series.*0028

*We learned the alternating series error bound.*0033

*What that said is that if you have a series that satisfies the conditions for the alternating series test, then if we just take a partial sum, as an estimate of the total sum, then our error can be positive or negative, but it is bounded by aN+1.*0037

*In other words, the first term that we do not take, the first term that we cut off.*0059

*So, if you are a little rusty on that, you might want to go back and look at the lecture on alternating series.*0068

*That is where we did some practice with error bounds for alternating series.*0074

*The second series we are going to use is a new one.*0078

*It is called Taylor's Remainder Theorem.*0082

*It says suppose you use the Taylor Polynomial t*_{k}(x) to estimate a function at a value of x near a.0083

*Then, it gives you this formula for the error but we have to break that down piece by piece because it is a little bit complicated.*0093

*The first thing to look at is what is the m in this formula.*0100

*What you do is you find the k+1 derivative of f.*0105

*That is the k+1 derivative, and look at the interval between a and x, and you ask yourself, how big can that derivative get on that interval. *0110

*In other words, how big can that derivative get.*0121

*You find the maximum value, or at least an upper bound.*0126

*You call that upper bound m.*0135

*That is what the m is in this formula.*0137

*Then the rest of the formula looks a lot like the original Taylor Formula.*0140

*You have a k+1 factorial in the denominator.*0148

*Remember, that k comes from the degree of the Taylor Polynomial that you are looking at.*0151

*Then x-a*^{k+1}.0158

*Again, the k is the degree of the Taylor Polynomial.*0162

*The a is the coming from the center of the series, so that is that a right there.*0167

*The x is the value that you are plugging in to use the Taylor Polynomial to guess the value of the function.*0174

*So that comes from that x right there.*0180

*That is all a little complicated, but it will probably be easier after we work out some examples. *0183

*Let us move straight to some examples.*0188

*The first one, we are going to use the Taylor Polynomial t*_{2}(x) to estimate cos(1/2).0190

*The idea here is we are trying to work out cos(1/2) without a calculator or in fact...*0198

*If you are a programmer, if you are writing the programs for calculators, you would be using Taylor Polynomials to work these values out.*0204

*We are going to try to estimate the value of cos(1/2).*0212

*Then we are going to give an error bound to say how accurate we are.*0215

*We start out by remembering the Taylor Series for cos(x), which is one of those common ones we memorized.*0222

*cos(x) = 1 - x*^{2}/2! + x^{4}/4! - x^{6}/6! ...0230

*So, t*_{2}(x), that means we cut off the Taylor series at the degree 2 term.0246

*Because that is the degree 2 term, t*_{2}(x) says we just look at 1 - x^{2}/2!.0257

*So we cut off the series at the degree 2 term and that is the Taylor Polynomial.*0270

*We are going to estimate cos(1/2) using the Taylor Polynomial, that is t*_{2}(1/2).0277

*That is 1 - 1/2*^{2}/2, which is 1 - 1/4-2, which is 1 - 1/8. Which is 7/8. Which is 0.875.0284

*That is our estimate of cos(1/2).*0305

*We think that cos(1/2) is approximately equal to 0.875.*0309

*How do we know if that is an accurate estimate.*0320

*Of course we could cheat and look at cos(1/2) on the calculator, but the whole point of this is to know how accurate we are without cheating.*0324

*Because if we know what cos(1/2) was, we would not need the Taylor Polynomial in the first place.*0331

*We want to give an error bound for the estimate.*0337

*What we are going to do is look at this series and notice that it is an alternating series, because these terms alternate from positive to negative.*0340

*So, we will use the alternating series error bound.*0355

*Which says that the error is < or = the first term that we cut off.*0365

*In this case that is x*_{4}/4!, and we plug in 1/2 there.0375

*1/2*_{4}/4!, so that is the cutoff term there, that is aN+1.0384

*That in turn, that is 4!, so this is 1/2*_{4} is 1/16, 1/16 × 24.0395

*Then we can just multiply that together.*0412

*If you want a quick off the cuff estimate of that, I know that is certainly, 16 × 24 is way bigger than 100.*0415

*So this is less than 1/100, considerably. 1/100 is 0.01, so our error is less than 0.01.*0427

*Our estimate is accurate at the very least, to 2 decimal places.*0446

*So, without checking on our calculator, I know that the cos(1/2) is approximately 0.875 and I know that my error there is definitely less than 0.01.*0462

*In fact I can say that it is even less, but I know for sure that I am certainly accurate to 2 decimal places.*0480

*In fact, if you check on a calculator, cos(1/2) = 0.87758.*0488

*So, in fact we did get the first 2 decimal places right there in our estimation there.*0500

*And that was without using a calculator at all, we knew that we were going to be right.*0507

*So, to recap there, what we did there was we took the Taylor Series for cos(x),*0511

*We cut it off at the degree 2 term to get the Taylor Polynomial for cos(x).*0518

*We plugged in 1/2, we got a value for cos(1/2) or at least an approximate value.*0525

*Then we noticed that the Taylor Series was an alternating series, so we could use the alternating series error bound.*0530

*That says you look at the cutoff term, work that out to be less than 0.01.*0537

*That tells us our error < 0.01.*0543

*Let us try another application of Taylor Polynomials.*0550

*Which is to do some integrals that would be very very difficult or even impossible without Taylor Polynomials.*0552

*Here is an integral of essentially 1/ I will write it as 1+x*^{5}.0561

*That would be a really nasty integral to try to do without the use of Taylor Polynomials.*0570

*If you look at some of the integration techniques we learned, they would not do you very well for 1/1+x*^{5}. 0575

*Instead, we are going to notice that this is the same as 1/1-(-x*^{5}).0585

*That is the sum of the geometric series.*0593

*We can write that as the sum from n=0 to infinity of (-x*^{5})^{n}.0596

*Of course that is only for values of x within the radius of convergence.*0610

*The radius of convergence of that geometric series is 1.*0618

*So, we can write that as 1, now + -x*^{5}, so -x^{5}.0625

*+ (-x*^{5})^{2}, so + x^{10} - x^{15} and so on.0635

*That was just the function 1/1+x*^{5}.0645

*What we really want is the integral of all that.*0650

*The integral of 1/1+x*^{5} dx is the integral of 1-x^{5} + x^{10} - x^{15} and so on dx. 0655

*I can integrate that just term by term so I get x - x*^{6}/6 + x^{11}/11 - x^{16}/16.0672

*By the way, I do not need to put the arbitrary constant on here because this is a definite integral.*0690

*I am going to be plugging in the values of x here, so if I put an arbitrary constant on there,*0697

*The values of the constant would just cancel each other off.*0704

*What I am going to do, is evaluate this integral from 0 to 1/2 of 1/1+x*^{5}.0707

*That is 1/2 - (1/2)*^{6}/6 + 1/2^{11}/11 and so on.0719

*Now, I want to estimate what that series adds up to.*0733

*I have already been told that I want to make sure it is accurate to three decimal places.*0740

*I have got to make sure this thing is accurate to 3 decimal places.*0748

*What I notice here is that this is an alternating series, so we are going to use the alternating series error estimation.*0750

*We want, remember the alternating series says that the error is at most, whatever term we cut off, and we want that to be less than 0.001 because that was given to us in the problem. *0765

*Let us just look at these terms and see if we can figure out which one will be less than 0.001.*0785

*Obviously 1/2 is not less than 0.001.*0795

*Next possible term is 1/2*^{11}/6, which is the same as 1/ 2^{6} is 64 × 6.0799

*Now remember 0.001 is 1/1000, if we ask ourselves whether that is less than 1/1000 and 64 × 6 is not bigger than 1000, so this fails.*0814

*Let us try the next term, 1/2*^{11}/11.0834

*Now, that is 1 over 2*^{11} × 11.0842

*I know that 2*^{10} is about 1000, if you multiply out 2^{10} it actually turns out to be 1024.0850

*So 2 *^{11} × 11 is certainly bigger than 1000, so this is certainly less than 1/1000.0856

*So we have found the term that is less than 1/1000.*0867

*That means that you can use this as the cutoff term.*0874

*And just keep the rest of the series.*0881

*Let us look at the rest of the series.*0887

*1/2 - 1/2*^{6}/6 is, if we simplify that it is 1/2 - 1/64 × 6, as we figured out before.0890

*1/2 - 1/64 ×6 is 384.*0908

*We can write 1/2 as 192/384 - 1/384.*0916

*That simplifies down to 191/384.*0928

*So, that is our estimate for the value of this integral.*0937

*I know without even checking anywhere else that that answer is accurate to 3 decimal places.*0942

*In fact if you put 191/384 into a calculator, it turns out to be 0.497396.*0950

*And, if you put this integral into sophisticated integration software, you get the true answer approximates to 0.497439.*0964

*So, if you look at those, in fact they do match up to the first three decimal places.*0982

*They are not so far apart even in the next decimal place, so our answer really was accurate to within 0.001.*0988

*Just to recap that example, we are trying to solve this integral but it is a really nasty integral.*0996

*Instead what we did was we looked at the function, we managed to write it as a geometric series, then expand that into the terms of a Taylor Series.*1003

*Then you can integrate those terms term by term.*1015

*We get an answer and we plug in our values of x=1/2 and x=0.*1018

*Then we get what turns out to be an alternating series, and so we try to find out what term could we cut off that would be less than our desired error of 1/1000.*1024

*Turns out that this term is small enough, the previous term was not small enough, so this term is small enough.*1038

*Then we can work out the rest of the series, the stuff before that, and get our answer.*1045

*We know that is accurate to within 0.001.*1050

*Let us try another example here.*1057

*We want to find the Taylor Series for sqrt(x) around a=4.*1060

*We are going to use that to estimate the sqrt(3.8) again to within 0.001.*1063

*We want 3 decimal places of accuracy here.*1070

*I am going to find my Taylor Series, I am just going to use the generic formula for Taylor Series.*1075

*Make a little chart here, n*^{th} derivative of x,1079

*The n*^{th} derivative, now I am going to plug in the value a=4,1084

*The whole coefficient is obtained by taking that n*^{th} derivative and dividing by n!.1090

*So I am going to do this for several values of n.*1101

*I am going to take it down to n=3 and we will see if that is enough.*1105

*The 0*^{th} derivative just means you take the original function, sqrt(x).1110

*First derivative, if you think of that as x*^{1/2} we will write the derivative as 1/2 x^{-1/2}.1116

*Second derivative is 1/2 × -1/2 so that is -1/4 x*^{-3/2}.1123

*The third derivative is -1/4 × -3/2, that is 3/8 x*^{-5/2}.1130

*Now we plug in 4 to each one of those.*1144

*Well sqrt(4) is just 2.*1147

*Here we have 4*^{-1/2}, so that is 1/sqrt(4) that is 1/2 × 1/2, so this is 1/4.1153

*x*^{-3/2}, that is 4^{3/2}, that is 1/8 × -1/4 that is -1/32.1165

*x*^{-5/2}, that is 4^{5/2}, sqrt(4) is 2, 2^{5} is 32, so we get 3/8 × 32, 1177

*Which is 2*^{5} × 2^{3}, 2^{8} is 256.1181

*Then to get the full coefficients, we divide those numbers by the factorials.*1199

*So, 2/0! is just 2.*1204

*1/4/1! is just 1/4.*1208

*-1/32/2!, 2! is just 2, so that is -1/64.*1211

*3/256/3!, so that is divided by 6, the 3 and the 6 cancel, so you get 1/512.*1220

*Those are our Taylor coefficients.*1245

*We want to use the Taylor Remainder Theorem to estimate the error of this series.*1251

*Let me remind you how the Taylor Remainder Theorem works.*1261

*That said that the error is bounded by m/k+1! × x-a*^{k+1}.1265

*There are a lot of things that we need to talk about here.*1282

*Let us first of all, first of all we need to figure out what k we might want to use.*1286

*This is just a guess at this point, so let us try, we are going to try k = 2.*1292

*We will see if that makes the error small enough, so we will plug k=2 into this error formula and see what we get.*1302

*Now, m, m comes from, remember, m comes from the maximum of that derivative.*1311

*We look at the k+1 derivative, so that is f the third derivative because k=2 we are going to look at the third derivative of f.*1322

*So the third derivative of f is, I am reading that here,*1332

*That is 3/8/x*^{5/2}.1338

*Now to get m, you have to look at that derivative and you have to look at it on the interval between x and a.*1350

*Here, our x that we are interested in is 3.8. And a=4.*1359

*We are looking at the interval between 3.8 and 4.*1367

*We want to ask what is the maximum possible value of that derivative between 3.8 and 4.*1375

*Well, we have got an x in the denominator here.*1382

*So, that is going to be maximized when x is as small as possible.*1388

*Let me write that down.*1395

*This is maximized when x is as small as possible.*1397

*In the interval 3.8 to 4, the value you want to plug in there is the smallest possible value, so that is 3.8.*1405

*That is 3/8/3.8*^{5/2}.1421

*OK, so I can write that as 3/8, now I do not want to work out 3.8*^{5/2}.1430

*What I can tell you for sure though, is I am just going to give a very rough estimation.*1439

*I know that 3.8 is way bigger than 1.*1445

*I know that 3.8 in the denominator will be much smaller than 1*^{5/2}, that is not 1 and 5/2, that is 1^{5/2}.1453

*This is because 3.8 is way bigger than 1, so in the denominator, 3.8*^{5/2} will be less than 1^{5/2}.1467

*So this is equal to, just simplifies down to 3/8.*1477

*We are going to take m=3/8.*1486

*Our error using this formula now, is less than or equal to 3/8, that is from m.*1493

*k+1!, remember we said we were going to try k=2.*1508

*So 2+1! is 3!, that is 3!. x-1, these are the values of x and a.*1514

*So 3.8-4*^{k+1}, so that is 3, because k=2.1525

*This simplifies down a bit to 3/8 × 6, because 3! is 6.*1537

*Now 3.8-4 is -0.2, and the absolute value of that is just 0.2*^{3}.1547

*0.2*^{3} is 0.008, over, the 3 and the 6 cancel, so we get 16. 1560

*This gives us 0.008/16, which is 0.0005.*1572

*The important thing here is 0.0005 is less than 0.001.*1582

*OK, so let us keep going with this on the next page.*1592

*What we did was we tried k=2.*1598

*Then we looked at the error and we did a bunch of simplifications,*1604

*But what we got was that it was less than 0.001.*1611

*That worked OK, that was acceptable for the error tolerance that we were given.*1615

*If that had not worked, I kind of pulled out k=2 seemingly randomly and that was actually because I worked out this problem ahead of time, of course.*1620

*If that had not worked, if it does not work, You just try a bigger value of k.*1632

*You just try k=3 or 4, just keep trying values of k until you get a value that works.*1645

*In this case k=2 works.*1653

*That means we have to find the Taylor Polynomial of degree 2.*1655

*What I am going to do is remind you of those coefficients from the table on the previous page.*1661

*That was 2, this was the c*_{n}, 2 and then 1/4, and then -1/64.1667

*1/5/12, that was the last part of the table from the previous page.*1676

*t*_{2}(x) means we read off these coefficients and we attach each one to a power of x. 1681

*So, 2 × x*^{0} is just 1.1690

*+ 1/4 ×, sorry not a power of x, but x-a, so that is x-4*^{1}.1694

*We go up to the 2 term, so that is -1/64 × x-4*^{2}, and then we cut if off there, because we are using k=2.1704

*Finally we want to estimate the sqrt(3.8), so we are going to plug 3.8 into t2(x).*1719

*3.8 gives us 2+1/4, 3.8 - 4 = -0.2.*1729

*-1/64 × -0.2*^{2}.1743

*We can simplify that a little bit, this is 2 - 0.2/4, 0.2*^{2} is, sorry 0.04/64,1752

*At this point it is probably worthwhile to just put these numbers into a calculator. We get the estimate of 1.94937.*1772

*That is our estimate for f(3/8), in other words sqrt(3.8).*1785

*That was a lot of work for one problem, let me recap what we did there.*1795

*First of all we wrote down the terms for the Taylor Series.*1799

*Then we had to use Taylor's Remainder Formula, the error formula, to try to find where should we cut this series off, so that our error is less than 0.001.*1804

*We kind of arbitrarily tried k=2, it turned out to work, but if it did not work we would try a different value of k,*1820

*We worked out Taylor's Error formula.*1828

*That involved finding the m term, the maximum value of the derivative, plugging in all the other terms to the error formula, and we get that it was less than 0.001.*1830

*That tells us that we can use the Taylor Polynomial of degree 2, so we fill that in using these coefficients that we worked out.*1842

*Then we used that Taylor Polynomial and we plug in our value of 3.8, and we work it down and find our answer for our estimate of 3.8.*1852

*We will try some more examples of that later.*1863