Calculus does not have to be difficult. Dr. William Murray knows what it takes to excel in math and will show you everything you need to know about calculus. Dr. Murray demonstrates his extensive teaching experience by clarifying complicated topics with a wide array of examples, helpful tips, and time-saving tricks. Topics range from Advanced Integration Techniques and Applications of Integrals to Sequences/Series. Dr. Murray received his Ph.D from UC Berkeley, B.S. from Georgetown University, and has been teaching in the university setting for 10+ years.

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Integration by Parts 24:52
Intro 0:00
Important Equation 0:07
Where It Comes From (Product Rule) 0:20
Why Use It? 0:35
Lecture Example 1 1:24
Lecture Example 2 3:30
Shortcut: Tabular Integration 7:34
Example 7:52
Lecture Example 3 10:00
Mnemonic: LIATE 14:44
Ln, Inverse, Algebra, Trigonometry, e 15:38
Integration of Trigonometric Functions 25:30
Intro 0:00
Important Equation 0:07
Powers (Odd and Even) 0:19
What To Do 1:03
Lecture Example 1 1:37
Lecture Example 2 3:12
Half-Angle Formulas 6:16
Both Powers Even 6:31
Lecture Example 3 7:06
Lecture Example 4 10:59
Trigonometric Substitutions 30:09
Intro 0:00
Important Equations 0:06
How They Work 0:35
Example 1:45
Remember: du and dx 2:50
Lecture Example 1 3:43
Lecture Example 2 10:01
Lecture Example 3 12:04
Partial Fractions 41:22
Intro 0:00
Overview 0:07
Why Use It? 0:18
Lecture Example 1 1:21
Lecture Example 2 6:52
Lecture Example 3 13:28
Integration Tables 20:00
Intro 0:00
Using Tables 0:09
Match Exactly 0:32
Lecture Example 1 1:16
Lecture Example 2 5:28
Lecture Example 3 8:51
Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule 22:36
Intro 0:00
Trapezoidal Rule 0:13
Graphical Representation 0:20
How They Work 1:08
Formula 1:47
Why a Trapezoid? 2:53
Lecture Example 1 5:10
Midpoint Rule 8:23
Why Midpoints? 8:56
Formula 9:37
Lecture Example 2 11:22
Left/Right Endpoint Rule 13:54
Left Endpoint 14:08
Right Endpoint 14:39
Lecture Example 3 15:32
Simpson's Rule 21:08
Intro 0:00
Important Equation 0:03
Estimating Area 0:28
Difference from Previous Methods 0:50
General Principle 1:09
Lecture Example 1 3:49
Lecture Example 2 6:32
Lecture Example 3 9:07
Improper Integration 44:18
Intro 0:00
Horizontal and Vertical Asymptotes 0:04
Example: Horizontal 0:16
Formal Notation 0:37
Example: Vertical 1:58
Formal Notation 2:29
Lecture Example 1 5:01
Lecture Example 2 7:41
Lecture Example 3 11:32
Lecture Example 4 15:49
Formulas to Remember 18:26
Improper Integrals 18:36
Lecture Example 5 21:34
Lecture Example 6 (Hidden Discontinuities) 26:51
II. Applications of Integrals, part 2
Arclength 23:20
Intro 0:00
Important Equation 0:04
Why It Works 0:49
Common Mistake 1:21
Lecture Example 1 2:14
Lecture Example 2 6:26
Lecture Example 3 10:49
Surface Area of Revolution 28:53
Intro 0:00
Important Equation 0:05
Surface Area 0:38
Relation to Arclength 1:11
Lecture Example 1 1:46
Lecture Example 2 4:29
Lecture Example 3 9:34
Hydrostatic Pressure 24:37
Intro 0:00
Important Equation 0:09
Main Idea 0:12
Different Forces 0:45
Weight Density Constant 1:10
Variables (Depth and Width) 2:21
Lecture Example 1 3:28
Center of Mass 25:39
Intro 0:00
Important Equation 0:07
Main Idea 0:25
Centroid 1:00
Area 1:28
Lecture Example 1 1:44
Lecture Example 2 6:13
Lecture Example 3 10:04
III. Parametric Functions
Parametric Curves 22:26
Intro 0:00
Important Equations 0:05
Slope of Tangent Line 0:30
Arc length 1:03
Lecture Example 1 1:40
Lecture Example 2 4:23
Lecture Example 3 8:38
Polar Coordinates 30:59
Intro 0:00
Important Equations 0:05
Polar Coordinates in Calculus 0:42
Area 0:58
Arc length 1:41
Lecture Example 1 2:14
Lecture Example 2 4:12
Lecture Example 3 10:06
IV. Sequences and Series
Sequences 31:13
Intro 0:00
Definition and Theorem 0:05
Monotonically Increasing 0:25
Monotonically Decreasing 0:40
Monotonic 0:48
Bounded 1:00
Theorem 1:11
Lecture Example 1 1:31
Lecture Example 2 11:06
Lecture Example 3 14:03
Series 31:46
Intro 0:00
Important Definitions 0:05
Sigma Notation 0:13
Sequence of Partial Sums 0:30
Converging to a Limit 1:49
Diverging to Infinite 2:20
Geometric Series 2:40
Common Ratio 2:47
Sum of a Geometric Series 3:09
Test for Divergence 5:11
Not for Convergence 6:06
Lecture Example 1 8:32
Lecture Example 2 10:25
Lecture Example 3 16:26
Integral Test 23:26
Intro 0:00
Important Theorem and Definition 0:05
Three Conditions 0:25
Converging and Diverging 0:51
P-Series 1:11
Lecture Example 1 2:19
Lecture Example 2 5:08
Lecture Example 3 6:38
Comparison Test 22:44
Intro 0:00
Important Tests 0:01
Comparison Test 0:22
Limit Comparison Test 1:05
Lecture Example 1 1:44
Lecture Example 2 3:52
Lecture Example 3 6:01
Lecture Example 4 10:04
Alternating Series 25:26
Intro 0:00
Main Theorems 0:05
Alternation Series Test (Leibniz) 0:11
How It Works 0:26
Two Conditions 0:46
Never Use for Divergence 1:12
Estimates of Sums 1:50
Lecture Example 1 3:19
Lecture Example 2 4:46
Lecture Example 3 6:28
Ratio Test and Root Test 33:27
Intro 0:00
Theorems and Definitions 0:06
Two Common Questions 0:17
Absolutely Convergent 0:45
Conditionally Convergent 1:18
Divergent 1:51
Missing Case 2:02
Ratio Test 3:07
Root Test 4:45
Lecture Example 1 5:46
Lecture Example 2 9:23
Lecture Example 3 13:13
Power Series 38:36
Intro 0:00
Main Definitions and Pattern 0:07
What Is The Point 0:22
Interval of Convergence 2:42
Lecture Example 1 3:24
Lecture Example 2 10:55
Lecture Example 3 14:44
V. Taylor and Maclaurin Series
Taylor Series and Maclaurin Series 30:18
Intro 0:00
Taylor and Maclaurin Series 0:08
Taylor Series 0:12
Maclaurin Series 0:59
Taylor Polynomial 1:20
Lecture Example 1 2:35
Lecture Example 2 6:51
Lecture Example 3 11:38
Lecture Example 4 17:29
Taylor Polynomial Applications 50:50
Intro 0:00
Main Formulas 0:06
Alternating Series Error Bound 0:28
Taylor's Remainder Theorem 1:18
Lecture Example 1 3:09
Lecture Example 2 9:08
Lecture Example 3 17:35

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/x2+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/x2+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 = x2+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 x2+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 x1/2,0035

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

X2 e3x dX. 0211

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

We will let U equal X2 and dV be e3x dX.0218

Again, we have got to figure out dU and V, so dU, if U is X2 is 2x dX. 0228

v is the integral of e3x dX. 0238

That is one third e3x. 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 x2 e3x 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 x3x 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 e3x dX, and again fill in dU equals dX and V is 1/3 e3x 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 e3x minus the integral of V dU, minus the integral of 1/3 e3x dX. 0347

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

This is minus 2/9 X e3x and then the 2 minuses give you a plus 2/9. 0370

Now the integral of e3x is 1/3 e3x.0381

If we put all of those parts together we get 1/3 x2 e3x. 0391

Minus 2/9 e3x plus 2/27 e3x.0398

2/27 X e3x 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 x2 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 x2 e3x dx. 0471

Here is the secret shortcut. 0477

What you do is write x2 e3x, and you make a little table here. 0482

On the left hand side you write down derivatives. 0488

The derivative of x2 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 e3x is 1/3 e3x. 0503

The integral of that is 1/9 e3x. 0507

The integral of that is 1/27 e3x. 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

So, you get x2 times 1/3 e3x. 0543

Now, the next diagonal line is minus 2x/9 e3x. 0553

The next diagonal line is plus 2/27 e3x. 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 x2, times something like e3x 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 ex 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, ex or cosin(x), we would never get down to 0.0612

The shortcut is not available here. 0620

Let U equal e3x, or sorry, ex, and dV equal cosin(x) dX, and fill in dU is ex 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, ex sin(x) minus the integral of V dU. 0667

That is ex sin(x) dX. 0677

That gives us a new integral ex sin(x) and we are going to do integration by parts again. 0681

We are agian going to let U equal ex. 0688

This time dv is going to be sin(x) dx, and dU is ex 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 ex 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

ex cosin(x) dx.0733

I am just going to get rid of the brackets and bring the negative sign through so we get ex times sin(x) plus ex 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 ex sin(x) plus ex cosin(x). 0810

That means we can solve for i. 0820

I is just one half the quantity ex sin(x) plus ex 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 ex and cosin(x).0857

You have eax, 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 x2, 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 ex.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 tan3(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 tan3(x), so we will write that as tan2(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 tan2(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, tan2(x) + 1 = sec2(x).0081

The tan2(x) converts into sec2(x) - 1.0090

tan(x) sec(x) dx.0099

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

Sec2(x) - 1 is u2 - 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 u3/3 - u.0121

Then we substitute back into sec3(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 sec3(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 = sec2(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 sec2 is the derivative of tangent. 0046

So, by making dv be sec2 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) × tan2(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

tan2(x) + 1 is sec2(x).0116

I am going to write that tan(x) as sec2(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 sec3(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, sec2(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 sec2(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 sin4X, 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, sin4X just becomes U4. 0136

All of a sudden you have a really easy integral, the integral of U4.0144

Well, that is just U5/5 + C.0149

Then you substitute back, and you get sin5X + 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

Cos4X sinU3X 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 sinU3 and we can split that up into a sin2X × 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 sin2, we can convert that into 1 - cosin2x.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, cos4 becomes u4.0276

The sin2 becomes 1 - cos2, and then becomes 1 - u4. 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 u4 × u2 - 1 dU.0304

Combine those into u6 - u4 dU. 0310

Then, that is an easy integral because then we can just use the power rule, u7/7 - u5/5.0321

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

So, this is cos7X/7 - cos5X/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

Cos2(x) = 1 + cos(2x)/2 0403

Sin2X = 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 sin2x cos2x 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, sin2x is 1 - cos(2x) over 2.0438

Cos2x 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 - cos2(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 cos2(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 cos2(2x), but if you think of that as being U, you can convert that into 1/4 the integral of 1 - cos2(u).0500

cos2(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 sec2(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 sec2(x) tan2(x) dx.0709

And, I remember that the derivative of tangent is sec2(x).0724

If I use the substitution u = tan(x) than dU is sec2(x) dX.0730

The point of this is that I can convert this integral into, I have got sec2(x) dX, so that is dU, tan2(x) is u2.0744

Tan2(x) + 1 is sec2(x).0764

Sec2(x) can be written as tan2(x) + 1 which I am going to convert directly into u2 + 1.0773

This integral converts into u3, sorry, u4 + u2 dU.0786

Again, that is a very easy integral.0800

That just integrates to u5/5 + u3/3.0803

Then we can substitute back, u is tan(x), so that is tan(x5/5) + tan(x3/3) + 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 = sec2(x).0841

That even power of sec2(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 x2 dx.0006

We have a quadratic under the sqrt.0012

The problem is we do not have a constant - x2.0016

We have 6x - x2.0020

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 - x2 - 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(-x2-6x+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-u2) 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 u2, 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-u2), well, u2 = 92θ.0171

If we pull a 9 out of the radical we get 3 × (1-sin2θ).0190

1-sin2θ is cos2θ, 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 9cos2θ 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 - u2 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 - sin2(θ)), which is sqrt(1) -, sin(θ) is u/3, so we know sin(θ) is u2/90430

Cos(θ) is sqrt(1 - u2/9).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 - u2/9 is, 1 - u2 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 x2 - 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 - x2 + 6x.0538

If you pull that 9 out, we get a 1/3 6x-x2, 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 - x2) + 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 θ0618

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(x2 + 1).0004

Remember, when you see x2 + 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 = sec2(θ) dθ.0027

x2 + 1 = tan2(θ + 1), which by the trigonometric identity is sec2(θ). 0043

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

That we would get tan2 + 1 and could convert it into sec2(θ).0064

The sqrt(sec2(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 sec2(θ) dθ.0087

In the denominator we have the sqrt(x2 + 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 x2 + 1.0156

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

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

Again, the key step there was recognizing that we had x2 + 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-bu2).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 u2 is a/(bsin2θ), 0053

And, so a-bu2 will be a - while bu2 will be asin2(θ)0063

That is a × (1-sin2(θ))0079

And 1-sin2(θ) of course is cos2(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 au2-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 tan2(θ)+1=sec2(θ).0117

Sec2(θ)-1 = tan2(θ) so once you make this substitution you are going to end up with sec2(θ)-1 under the square root.0131

The last one you will have a+bu2=sqrt(a/(btan(θ))) and so again you are taking advantage of this trigonometric identity tan2(θ)+1.0144

You will end up with that under the square root, and that will convert into sec2(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 - 9x2 dx.0223

That example matches the first pattern that we saw before, which was the square root of a minus b u2.0229

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 9x2 is 9, x2 is 4/9 sin2(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 sin2(θ) and we can pull the 4 out of the square root.0334

We get 4 times the square root of 1 minus sin2θ.0343

That is 2 times 1 minus sin2(θ) is cos2(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 cos2(θ) because we have one cos here, and one cos here, dθ.0363

We learned how to integrate cos2(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 sin2(θ), so 1 minus 3/2 sin(θ) was x 9/4 x2.0532

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

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 + x2 0605

The key thing to remember here is that when you see the square root of 1 + x2, or even 1 + x2 without a root, you want to use the tangent substitution.0611

X equals the tan(θ)0629

The reason you do that is that x2 + 1 is tangent2(θ) + 1 but that is sec2(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 sec2(θ) dθ0653

If we plug that into the integral, we get dX is sec2(θ) dθ.0663

1 + x2 is sec2(θ) 0672

We kind of lucked out on this one because the sec2'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/x2 + 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 x2 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/u2 + 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 u2 + 9 will be 9tan2(θ)+9.0844

And, that is 9+tan2(θ)+1.0856

Which is 9 sec2(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 sec2(θ) dθ.0871

We will plug all of that in, the dU was 3 sec2θ dθ.0881

The u2 + 9 converted into 9 sec2θ0890

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 x2 + 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 x2 + 4x + 5 down into linear terms.0052

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 = x2 + 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/x2 + 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 x2 + 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 x2 + 4x + 5.0240

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

So, 4/2 = 2, 22 = 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 w2 + 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 x2 + 4x + 5. 0370

Then we will add on a constant as before.0382

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 - u2.0453

You use secant when you have something like u2 - 1.0460

Now, either one of those can be factored.0465

u2 - 1 factors as (u-1)(u+1).0469

1- u2 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

x2 + 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, 8x2 + 30x + 30/x3 + 6x + 13x + 10 = a/x + 2,0276

Then it would be nice if we could factor x2 + 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

x2 + 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 × (x2 + 4x + 5).0339

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

The x3 term that we started with. 0348

What you would get is a × x2 + 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 8x2 + 30x + 30. 0367

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

So, my x2, I am going to have an a here,0384

Then from the b and the x here, we get a b x2.0389

+, 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 8x2 +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 x2 term, an x term, and a constant term.0455

On both sides.0465

The x2 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/x2 + 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/x2 + 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 denominator0621

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(x2 + 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 x2 + 5x + 6.0084

We are going to forget about calculus for a while, and just write this as 1/(x2 + 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/(x2 + 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=10258

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) dX0307

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

(2x2 - 7x)/(x2+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 (x2-6x+8) into (2x2-7x) and I am going to stick on 0 as a constant term there.0480

We look at x2 and 2x2 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 2x2-12x+16, subtract those the 2x2 cancel, subtracting is the same as changing signs and adding.0505

So, minus 7x + 12x gives us 5x - 16.0518

We can write (2x2 - 7x)/(x2-6x+8) as our answer 2 plus a remainder term (5x-16)/(x2-6x+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

(x2-6x+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)20914

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 u2, 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 - x2 dx.0006

The thing that makes that different from the basic forms,0009

The basic forms look like a2 - u2, or a2 + u2, or u2 - a2. 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 - u2) du = u - a/2 × sqrt(2au - u2) + a2/2, 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 - u2) = 2x - x2 + now a2/2, 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 e3x × 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 eau × cos(bu) du = eau/a2 + b2 × 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

We have eau, OK, that is e3x.0105

Now a2 + b2, that is 32 + 52,0111

32 is 9 and 52 is 25.0116

9 + 25 = 34.0121

a = 3 so, 3cosine, b=5, so 5x, + b is 5, 5sin(5x) + C.0127

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(x2+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 a2+u2 dU. 0094

Then that converts, or that solves to, (u/2) × sqrt(a2+u2)+a2/2 × ln(u+sqrt(a2+u2)+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 x2+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 12=1, so I am going to write 1.0164

But actually, what we had here was x2+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)x2+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 ux2+4 dU.0200

Now, that matches this formula from the integration table.0213

The a2=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(a2+u2).0232

Well a2+u2 was u2+4 which came from x2+2x+5.0244

+a2/2, that is 4/2 = 2, times ln(u), u is x + 1, + that square root again, a2+u2, came from u2/4 came from x2+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-u2)+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=1/3dU0410

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-u2) is 1-9x2 + 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(1-9x2)+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, tan2x × 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 sec3u dU = 1/2 secu × tanu + 1/2 ln(secu + tanu) + C.0552

So, I do not have exactly the integral of sec3u, but I know that sec2, I know that I can write that sec2x-1. 0585

I can write this integral as the integral of sec2(x-1) × secx dX, which in turn is the integral of sec3x dX - the integral of sec(x) dx).0603

Now, I do have sec3x dX, which is the same as sec3u 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 tan2, 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 curve0014

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 x0, x1, x2 all the way up to xn is b.0115

That is where the next part of this formula comes from.0124

x0 is a, x1 is a + δX because it is a and then you go over δx, x2 is a + 2 δX all the way up to xn 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 b0156

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(base1 × base2 × 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 δX0225

Then you have base1 + base2 is, I am going to show this in red, base1 + base2, that is for the first trapezoid.0243

For the next trapezoid, base1 + base2, and so on, up to the last trapezoid, base1 + base2.0260

What you are doing is you are plugging each of these x0, x1, x2 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(x0) + 2f(x1) + 2f(x2) + 2f(x3) + only 1 of f(x4). 0352

These x0's are the division points in between 1 and 2, so this x0 is 1, so that is sin(1) + 2, now that is 1, x1 × δx is 1/4.0385

x2 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 = 0.9514620486

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 x0 = a, xn = 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 x1* and x2* and xn*.0596

Those represent these midpoints, so that is x1*, there is x2*, and so on.0616

Those are the midpoints so x1* is just x0 + x1/2, x2* is just x2 + x2/2, 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 x1× and the x2× 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 0.9589440805

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, x0, x1, up to xn-1.0865

You do not go up to xn.0876

For the right endpoint rule it is the same formula except you use the right endpoints.0879

The right endpoint would be x1, x2, up to xn.0885

You do not have x0 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 x0, x1, up to xn-1, so that is the left endpoint rule.0902

The right endpoint, I will do that one in blue, is x1, x2, up to xn-1, and xn.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, 0.67233487080150

To reiterate there, we are estimating an integral as the area under a curve using the Simpson's Rule formula.0160

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 -x2/4 + x2/2 × 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

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 x0, x1, x2, up to xn, 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 1/40265

Our points that we are going to check are x0 = 1, x1 = 5/4 because we are going over by 1/4 each time, x2 = 3/2, x3 = 7/4, and x4 = 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 Rule0383

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 x0 = 1, x1 = 9/8, x2 = 5/4, all the way up to x8 = 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 ex.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 - 8e0.0073

Sorry, subtracting a negative sign, so + 8e0.0088

Then we take the limit as t goes to infinity of this.0093

So this is minus 8e-infinity + 8e0 = 1.0101

Well e-infinity, this is 1/einfinity, 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

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

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

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 integration0444

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 cancel0852

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 x2 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 = x2, 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 make1000

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/(xp).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/(xp), 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/xp 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

We will convert that into the integral of 1/xp 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)x1/3.1297

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/u1/3 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 = u1/3 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/u1/3 du.1428

Plus the integral from 1 to 5 of 1/u1/3 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/u1/3.1476

That is exactly the integral that we were looking at on the previous page.1482

I will just write it as 01 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 xp,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 × 53/2.1584

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/x2 - 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/x2 - 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/x2 - 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 denominator2090

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 = x4/8 + 1/4x2.0005

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 x4 is 4x3/8 is just x3/2.0026

Now, if you think of x2 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 x3 in the denominator.0060

So, that is what we get for the derivative.0066

We need to square that, f'(x)2 is x6/4 + 1/4x6, squaring out both terms.0069

Then, minus 2ab, so minus 2 × 1/2 × 1/2 is 1/2.0085

Then the x3 cancel, and this term is what we get by doing the -2ab of the a - b2 formula.0094

So, that is where that - 1/2 comes in.0107

1 + f'(x)2 is x6/4, still + 1/4x6.0109

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-b2, this is now going to turn into a+b2.0137

So this is x3/2 + 1/2x3 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 x3/2 + 1/2x3.0173

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 x3/2 + 1/2x3 dx.0199

If we integrate x3 that is x4/4, so this is x4/8.0210

Now, the integral of x-3 is x-2/-2, using the power rule.0217

So this is -, because the -2 in the denominator, 1/4x2, 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 24/8, that is 16/8, so that is just 2.0251

Minus 1/4 × 22, 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 length0294

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-x2/4.0000

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 x2 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 a2, so 1/4x2,0035

+ b2 + x2/4.0041

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 1/4x2,0066

+ x2/4, 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 x2/4 and we evaluate that whole thing from x=1 to x=e.0136

So we get ln(e)/2 + e2/4 - 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 + e2/4 - 1/4.0168

So that is 1/4, because that is 1/2 - 1/4, + e2/4.0176

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 know0005

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 formula0035

The integral from x=a to x=b of the sqrt(1+x'(x)) that is the derivate of x2 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 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 x2/8 - 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 x2/8, derivative of x2 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)20160

It is x2/16 + 1/x2 - remember that (a+b)2 or (a-b)2 is a2 + b2 - 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 x2/16 + 1/x2 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, x2/16 + 1/x2 + 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 x2/8 and 1/x integrates back to ln(x).0288

We evaluate that whole thing from x=e down to x=1.0307

We get e2/8 + ln(e) - 1/8 - ln(1).0316

Of course ln(e) is just e, ln(1) is just 0.0327

So, we get e2/8 + 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'(x2))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 = 4x3/2 + 1.0390

Again, we have to look at f'(x), so we have to take the derivative of that.0397

The derivative of x3/2 is 3/2 x1/20401

So this is 4 × 3/2x1/2, derivative of 1 is 0.0406

This just simplifies into 6x1/2.0416

f'(x)2 would be 36, (x1/2)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 u1/20509

So the integral of u1/2 is u3/2, 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 u3/2 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 373/2 - x=0, just gives us 13/2.0582

This reduces down to 1/54 × 373/2 is the same 37 times the square root of 37 - 13/2 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 = 4x3/2 + 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 tan2(x)0685

So, (1 + f'(x))2 = 1 + tan2(x).0691

But, if you remember the pythagorean identity, that is sec2(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/30789

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

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 2x3 from x=0 to x=1 around the x axis.0004

So here, our f(x) is 2x3, we want to find f'(x), which is 6x2,0012

f'(x)2 is 36 x4.0025

1 + that is 36x4 + 1.0034

The square root of 1 + f'(x)2 is just the square root of 36x4 + 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 2x3.0060

× the radical expression, 36x4 + 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 36x4 looks pretty nasty, but its derivative is exactly x3.0101

Well not exactly x3, but x3 times a constant.0110

That suggests the substitution, u = 36x4 + 1.0115

Then du is 4 × 36 x3 dx.0123

We kind of have the du here when we have x3 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 x3 dx gives you du.0145

We now have the square root of 36x3 + 1, that is u.0149

So this is pi/36.0156

Integral of sqrt(u) is not too bad, you think of that as u1/2.0161

So the integral is u3/2 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 u3/2, (36x4 + 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 36x4 + 1, we will get 37 3/2.0218

- x=0 in there just gives you 13/2.0235

This can be slightly simplified to pi/54 ×, 373/2 is the same as saying 37×sqrt(37) 0243

- 1/23/2 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 x4 + 1,0286

We basically have our du set up for us with 2x3, 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 2x1/2, 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 u1/2, and the integral of u1/2 is u3/2/3/2 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+13/2, 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 33/2,0223

- x=0 in there, into 2x+1, just gives you 13/2.0240

Again it simplifies down to 2pi/3.0247

33/2 is the same as 3sqrt(3) - 13/2 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 cone0107

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 + 32 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 x2/2,0203

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 = x3 around the x axis.0272

Our f'(x) is 3x2.0279

Our f'2 is 9x40288

1 + f'2 is 1 + 9x4.0297

And, the square root of that is sqrt(9x3 + 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 x3, so 2pix3 × integral of 9x4 + 1 dx.0326

Now, this looks like a rough integral but it is actually not so bad.0342

If you notice, you have 9x4 + 1 under the radical.0348

The derivative of that would 36x3, and we have an x3 outside.0352

What we can do is let u = 9x4 + 1.0358

Then du is 36x3 dx.0368

So, what this converts into is, I guess I can write dx or x3 dx, is 1/36 du.0376

That takes care of the dx and the x30390

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, 9x4 + 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, u1/2 is the same as sqrt(u).0420

If we integrate that, that integrates to u3/2/3/2 or, 2/3u3/2.0429

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 9x4 + 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 9x4 + 1.0495

If we plug in 1 to 9x4 + 1, that is 103/20500

X=0, if you plug it in just gives you 1 - 13/2.0507

This gives us pi/27 × 103/2, 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 formula0548

Then, we had a pretty tricky integral there.0554

The key thing there was observing that the derivative of 9x4 + 1 was more or less x3,0556

So we could make this substitution that gave us a very nice integral to solve.0568

Let us try another example0575

This time we are rotating the graph of y = 2x2 + 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 formula0605

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 = 2x2 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 something0722

So its derivative is 1/20728

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 -1/20736

I will write it down here in the denominator, y-1/20739

× the derivative of that inside stuff by the chain rule0745

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 numerator0760

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'(y2)),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 u1/2 du.0974

That is pi/2 × 1/8 u1/20986

The integral of u1/2 is u3/2/3/2, 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/241019

Now, u was 8y-73/21022

Finally, we get pi/24 × 1034

Now, if we plug in y=3 to 8y-7.1044

That is 24-7, which is 173/21047

If we plug in y=1 to 8y-7 which is 8-7, so -13/21055

We can clean that up a little bit into pi/24 × 17 sqrt(17).1063

13/2 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 axis1093

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-y2).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 - y2, 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-y2) 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-y2) dy.0233

Now, this integral unfortunately is not so pleasant.0245

We have to integrate sqrt(1-y2).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 a2 - u2 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-y2),0360

Is y/2 × sqrt(1-y2) + 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-y2) 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

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

Density is 300 kg/m3.0000

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) y2/3 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 y2 dy.0209

That is 1960 × sqrt(3)/3y3.0214

The integral of y2 is y3/3.0222

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 problem0270

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/m3 × 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/s2.0100

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/m2×s2, 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/m3, 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-y2).0285

The total width is l(y), which is twice that, which is 2×sqrt(1-y2).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-y2) × 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 - y2, 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 u1/2.0440

The integral of that is u3/2, 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-y2)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-y2, 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/m2 per s2 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-y2)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 sin2(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 sin2(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 x2 - 2x + 1 dx.0270

That is a very tractable integral.0279

That is 6 × x3/3 - integral of 2x is x2 + 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 equation0077

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 area0085

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 x2 + y2 = 1.0199

If we solve that for y in terms of x.0207

We get y = sqrt(1-x2)0210

In this case, if we plug that into our formula, we get 1/2 × area0215

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 - x2 dx0241

This in turn becomes 1/pi × 0250

Now, if we integrate 1/x2, that is easy, that is just 1-x3/3.0255

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 centroid0315

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 dx0413

The integral of 1/x is ln(x).0417

Integrate that from x=1 to x=e0420

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 10460

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/x2 dx.0514

So, this is 1/2, now the integral of x2, you want to think about it as x-20520

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 - 1/2e0550

That was the y coordinate of the centroid.0555

you put those together and you get the centroid, the balance point, is0558

Put the x and the y coordinate together to get an ordered pair0565

You get (e-1, 1/2 - 1/2e)0570

The point of that problem is that we had to figure out the area first0582

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-x2)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 - x2 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-x2 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 integral0740

What you can do is say u = 1 - x2,0748

Then du is -2x, dx0752

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 centroid0795

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) = cos2(t) and y(t) = sin2(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 cos2(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 4cos2(t) sin2(t).0059

y'(t)2 is 4sin2(t) cos2(t).0070

If we add those up, x'2 + y'2,0082

Well those are the same thing.0091

We just get the sqrt(8sin2(t)cos2(t)).0094

The sqrt(8) is 2×sqrt(2).0102

The square root of sin2 and cos2, 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 u2/2.0160

Now to convert things back into t.0169

2×sqrt(2) × sin2(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 cos2(t) + sin2(t) = 1.0228

So, cos2(t) = x, sin2(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) = et + 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 = et + e-t, so the derivative of et is just et.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, 22 is just 4. 0058

Now, if we square y'(t), then we are going to follow the formula (a + b)2,0060

Is a2 + 2ab + b2.0074

Here a is et, so (et)2 is e2t, b is -e-t,0080

So this is -2ab, well ab is et × e-t,0090

Then + b2 is e-t, so that is e-2t,0102

But look at this et × e-t, is e0 which is 1.0111

This is just -2.0116

We also have a 4 here, so what we get is e2t - 2 + 4, is just + 2.0120

+ e-2t.0131

The clever thing to do here is to write that 2 as 2et × e-t again + e-2t.0134

This is (et + 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 et + 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 et + e-t dt.0181

That integral is not too bad.0192

The integral of et, is just et 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 et - e-t.0210

Then we want to evaluate that from t=0 to t=1.0213

That is e1 - e-1 - e0 + e0.0221

The e0'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'(t2 + y'(t2) 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 examples0097

The first examples is the equations are x(t) = t+1, y(t) = t2.0100

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 = t2 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 - y0, which is 1 here, is equal to the slope × x - x0 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 cos2(t) + sin2(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 circle0294

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 all0511

Let us try an example of arc length now.0519

We are given the curve x(t) = 6 - 2t3, y(t) = 8 + 3t2.0520

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 - 2t3,0549

You take its derivative and the 6 just goes away, so the derivative is -6t2.0556

y'(t), if y' is 8 + 3t2, again the 8 does not have any effect.0568

The 3t2 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 36t40584

+ y'2 is 36t2.0596

That can simplify because we can pull a 36 out and just get 6, 0600

we can pull a t2 out and get t.0610

Under the radical, what we have left is just t2 + 1.0615

The arc length = integral from t=0 to t=2 of 6t × sqrt(t2+1) dt.0620

Now, this integral is not too bad because what we can do is make a little substitution.0645

u = t2 + 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 u1/2.0683

To integrate that we get u3/2/3/2 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's0706

These 3's cancel, so we get 2 × u was (t2 + 1)3/2 evaluated from t=0 to t=2.0709

This gives us 2 now if we plug in t=2, we get 22 + 1, so that is 53/2.0727

- 02 + 1, so that is 13/2.0737

We can simplify that to a little bit.0744

53/2 is the same as 5 × 51/2.0746

13/2 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

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 42/2.0284

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 64sin2(θ) - 42 is 16.0303

Remember the 2's cancel on the outside d(θ).0313

That is the integral of sin2(θ),0319

We are going to use a half-angle formula to deal with that.0323

Remember sin2 is 1 - cos(2θ)/2.0324

So, 64sin2(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 9sin2(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) + 9cos2(θ) dθ.0168

That looks like it is going to be a pretty messy integral,0187

But good things happen here because we have 9sin2(θ) and 9cos2(θ) 0190

Those combine just to be 9.0196

We get 2 × integral from 0 to π of square root of,0198

Well the cos2 and the sin2 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

cos2(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 θ0374

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, cos2(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(θ2/2).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(θ2), f'(θ2).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(θ2/2) 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 θ2 is θ3/30194

This whole thing is theta3/3/2, so over 6, evaluated from θ = 0 to θ = pi/2.0196

That is just pi/23/6.0211

Now, 23 is 8 and 8 × 6 is 48 so our answer is pi3/48.0220

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(θ2)/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(θ2)/2 so our f(θ) = cos(2θ) this is cos2(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 cos2 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(θ2)/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 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'(θ2) + f(θ)2 dθ.0629

Our f'(θ) is well, r = e.0640

f'(θ) would be (5e)2 + just f(θ2) so that is (e)2.0650

We want to square root that and integrate it.0665

5 squared is 25, e)2 is e10θ.0670

Square root that and integrate it.0685

This is sqrt(26e10θ).0690

I am going to pull the sqrt of 26 all the way out of the integral.0704

Then we have the sqrt(e10θ) which is just e.0710

We want to integrate this from θ = 0 to 2pi.0718

Now the integral of e is just e/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 you get e10π.0755

If you plug 0 into e you get e0, 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(n2 + 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 = a2 - b2.0067

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(n2 + 6n) - n.0088

I want to multiply that by its conjugate, the sqrt(n2+6n)+n.0094

To pay for that multiplication, I have to do some division by the same expression,0104

sqrt(n2+6n)+n.0110

The point of that is that in the numerator we get this a2-b2 expression.0115

We get the sqrt(n2+6n)2, so that is just n2+6n.0124

Minus n2.0134

Then that same denominator.0140

sqrt(n2+6n) + n.0142

This simplifies nicely down to 6n/sqrt(n2+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(n2), that is approximately n.0186

You know there is a 6n next to it.0190

The n2 is bigger than the 6n, so the sqrt(n2) 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(n2+6n)/n + 1.0220

Sqrt(n2+6n), I can write that as,0230

Well I can make that a big square root.0236

n2+6n.0241

Now if I want to make this n part of the square root, I have to change it to n2.0244

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(n2+6n)2, that gets rid of the square root there.0324

We get n2, 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 n2, 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/n2n.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+0infinity, so that is 1infinity,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 ab,0070

Often, a very useful way to write that is,0078

Write it in the form, eln(ab).0087

The point of that is that the ln(ab),0097

Is the same as bln(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 eln(1+1/n)2n.0177

That is e2n, 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/n2.0300

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/n2 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 e2.0348

That sequence converges to e2.0357

Let us go back to the beginning.0360

If you had made this mistake of thinking, oh that is just 1infinity,0362

Which is just 1,0367

You would have gotten your answer to just be 1 which is very different from e2.0370

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 eln 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 an term is getting steadily bigger in other words the an+1 term is bigger than or equal to the an 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, a1 would be the sqrt(2+a0), so 2 + sqrt(2)0103

a1 would be the sqrt(2+a1), 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 a0 term,0165

Is certainly less than 2 because the sqrt(2) is about 1.4.0172

That is certainly less than 2.0177

If an is < 2, if one term < 2,0182

Then I am going to a do a little arithmetic here.0195

2 + an would then be less than 4.0196

So the sqrt(2+an) would be less than sqrt(4) which is 2.0202

That is saying that an + 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 an + 1 is bigger than or equal to an.0290

Let us plug in what an + 1 is.0300

By definition that is 2 + sqrt(2 + an), bigger than or equal to an.0305

I am going to work with this a little bit.0310

I am going to square both sides, this is saying 2 + an is bigger than or equal to an2.0312

If I move all of the terms over to the right, that is saying 0 is bigger than or equal to an2 - an - 2.0322

I can factor that into (an - 2) × (an + 1).0330

Let us remember that since an < 2.0345

Over on the left we showed that every term is < 2.0355

Since an < 2, an - 2 < 0.0360

Since all of these terms are positive, an + 1 would be bigger than 0.0370

an - 2 × an + 1 is less than or equal to 0.0377

Which means that this equation is true and this equation is true.0388

an + 1 is > or = an.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 an converges to L.0465

There is a little trick here which is that if an converges to L, then the sequence of successive terms must also converge to L.0467

But an + 1 is the sqrt(2 + an)0482

So, an + 1 = sqrt(2 + an).0491

But, an converges to L, so sqrt(2 + an) converges to 2 + L.0503

Now we have an 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

L2 = 2 + L, L2 - 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 an converging to -1, so an 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 an + 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 + 5n2/2n2 + 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 + 5n2, definitely the largest term there is 5n2.0709

In the denominator, 2n2 + 6, even though the 6 is bigger than the 2, the n2 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 n2 in the numerator and the denominator is the important thing.0740

What we are going to do is divide top and bottom by n2.0745

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/n2 + 5n2/n2)/(2n2/n2 + 6/n2).0765

(3n/n2 is 3/n + 5)/(2 + 6/n2).0783

In the limit, as n goes to infinity, the 3/n is going to go to 0.0793

The 6/n2 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 situations0810

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 n2, so we divide top and bottom by n2.0828

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 n1/2.0947

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.com1054

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, s1, 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 s1 is just the s1 term, ln(2) - ln(1),0077

+ the n = 2 term, ln(3) - ln(2).0084

s3, 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, s4.0118

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 3n/43n+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 32/43×2 + 2.0019

So that is 48.0028

Now n=3 gives us 33, over 43×3+2, so that is 411,0031

+ 34/43n+2, so 414.0041

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 × 3/43.0064

To get from the second to the third term, we multiply by 3/43.0073

So, each one of these terms, you get it by multiplying the previous term by 3/43.0082

So, what we have here is a geometric series.0090

Our common ratio is r = 3/43.0097

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 3/43 < 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 32/48.0170

The common ratio is 3/43.0176

So, that is a little bit messy.0186

We can clean it up a little bit by multiplying top and bottom by 43.0189

That will give us 32/45.0195

On 43 - 3 in the denominator.0199

That is 9/45/64-3.0205

So, that in turn becomes 9/61×45.0214

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.com0000

We are going to talk today about series0003

There are several bits of notations and definitions before we look at some examples0005

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 a0, a0 + 1, a0 + 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 a0 + a1,the third is a0 + a1 + a2 and so on.0047

The sn is a0 + a1 up to an0054

You call this the sequence of partial sums.0060

s0, s1, s2 up to sn.0064

You think of that as a sequence, not a series.0069

This is a sequence sn.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 sequence0102

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 × r2 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 value0270

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/n0513

Right away we will look at the test for divergence and see if it works. 0516

an = n-1/n.0520

We rewrite that as 1-1/n.0525

The limit of the sequence an 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 an 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

an = 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

s1 is just the first partial sum, that is just 1. 0676

s2 is 1 + 1/2.0679

s3 is 1 + 1/2 + 1/3.0684

s4 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 s3, I will not look at that.0726

s4, 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 sn.0756

sn 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

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 sn 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

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-np where p=1.0952

It is really 1/n1.0960

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

s1, 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

s2 is 1/2 × 1 - 1/3, + the n = 2 term, is 1/2 - 1/4.1127

s3 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, s4.1167

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 sn.1216

It is going to be 1/2, well after we cancel everything, the only terms left are this 11218

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 s1, s2, s3, s4.1338

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)/x2 × log(x)2.0137

If we look at that, 1 + log(x) > 0.0151

x2 × 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/n2 + 1.0005

We would set up f(x) = 1/n2 + 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/x2+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/x2+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/n2+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.com0286

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/np 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/xp converges.0106

What we figured out was that the integral from 1 to infinity of dx/xp 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/n1/2.0148

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 n1/2 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/np.0231

A geometric series, if you had something like 1/2n, 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 rn,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/n3/2.0309

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 n2/en.0402

This one is a little more complicated and is not just a simple p series.0405

Let us look at f(x) = x2/xn.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 x2e-x.0424

I want to look at the improper integral from 1 to infinity of x2e-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 x2, 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 x2e-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 x2/e-x.0543

As we plug in infinite values there, larger and larger values of x,0551

The e-x dominates the x2.0560

When we take the limit as t goes to -infinity,0568

Since the e-x dominates the x2, 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 n2/en converges by the integral test.0613

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 en/n!.0005

That one is a little tricky.0011

I want to write that one as en/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 = en/1×2×3, and there were n-2 factors there.0044

So 3n-2.0062

Just to make the algebra a little bit cleaner, I am going to multiply the top and bottom by 32.0064

32 × en/1 × 2, now I can say that is 3n.0071

I just multiplied in 32 × 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, 32 is 9/2 × en/3n.0085

This is (e/3)n.0094

That is the series that I am going to use as my bn.0100

Remember an is always the series that you are given.0102

The point here is that an is < or = to bn.0109

The sum of bn converges.0114

Why does it converge?0118

Because, bn is just a constant 9/2 × e/3n.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

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 bn converges.0171

So, our smaller series an, 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 n4+3n-8/3n7+6n5.0006

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, n4, in the numerator, and n7 in the denominator.0024

It looks like if you strip away the extraneous stuff there,0030

It looks like the sum of n4/n7.0038

That in turn would simplify down to 1/n3.0041

OK, I am going to call my 1/n3, that is going to be the bn that I create.0051

This entire given series, that is an.0057

We will look at an vs. bn,0062

We will divide them together and try to go for something using the limit comparison test.0065

an/bn is n4+3n-8/3n7+6n5.0069

bn is 1/n3,0081

But we can flip that up into the numerator so we can write that as n3/1.0084

That turns into n7+3n4-8n3/3n7+6n5.0094

Now you look at something like this and identify the biggest term anywhere,0107

Which is an n7 in the top and the bottom and you divide by it.0113

So, you get 1+3/n3-8/n4/3+6/n2.0116

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 bn,0164

Well that is 1/n3, 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/an 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.com0259

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 an,0006

And you create your own series that we will call bn.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 an.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 bn.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 bn.0149

So, the an is less than the bn.0155

The sum of bn, 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

3n/2n - 1, 0233

That looks like the sum of 3n/2n.0240

So an is the given series, for our bn, we will use the sum of 3n/2n.0249

Then we will compare those to each other.0255

So, 3n/2n-1 vs. 3n/2n.0260

Well 2n-1 is a smaller denominator which means that 3n/2n-1 is actually a bigger number.0268

What that is saying is that an is bigger than bn.0278

We know that the sum of the bn'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/2n is just 3/2n, 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 bn.0398

This series is our an.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 aN/bn,0411

And that is the sqrt(2n+17)/n.0414

All of that divided by bn, 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(2n2) + 17n/n.0444

If we look at the top and bottom there, the biggest terms we have in the top, we have an n2 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 n2.0472

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 bn diverges, well how do we know that,0498

That is because we can think of 1/sqrt(n) as 1/n1/2.0509

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 bn is a divergent series.0527

The limit comparison test says that the given series an does the same thing that your series bn does.0530

So, we can conclude that the sum of an 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 bn.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 bn diverges, because it is a p series,0593

Then we can say the an 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 an might behave like the series bn = 1/n.0650

Because sin(x) behaves like x.0660

Let us try that out.0662

We will take the series bn = 1/n, and then we will look at an/bn.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/-n2.0707

The derivative of sin(1/n) is cos(1/n).0710

× derivative of 1/n by the chain rule which is -1/n2.0718

Those -1/n2'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

The sum of bn does what?0763

Well bn 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 an 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 -1n+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 bn to be 1/n.0018

We do have that that is decreasing, bn+1 is less than bn.0025

We do have that the limit of bn=0.0033

Our two conditions are satisfied.0038

That means the alternating series converges.0040

So, the sum of -1n+1 bn,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 an+1.0079

By the first term that you cut off.0086

We want that to be less than 0.01.0090

We want an to be less than 0.01.0103

Well an+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 s99 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 -xn/n.0208

If you plug in x=-1, then we get ln(2)=the sum of -1n/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 -1n/en 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 bn, we just take everything except the -1n.0018

That is 1/en.0024

Then we check our two conditions, one is that it is decreasing.0026

bN+1 should be less than bn.0030

That should not be an en that should be an en.0037

Certainly that is decreasing.0040

en+1 is bigger than en, so 1/en+1 is less than 1/en.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

-1n/en 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, an+1, if we take the absolute value, that knocks away any possible negative signs.0117

So this is 1/en < or = 0.001 is 1/1000.0124

If we take reciprocals, remember that reverses the inequality.0135

So we get en > 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 en 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 an+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 210 = 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, e10 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 en+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 s6, 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 s6 for this series,0340

Add up terms and find the partial sum, you get s6 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 s6 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 s6 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 -1n, or -1n+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 bn 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 bn 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(an+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, -1n/n!.0200

Here the bn is just 1/n!.0205

We have to check our two conditions, whether the bn is decreasing, and whether the limit is 0.0210

If we look at bn+1 vs. bn, bn should be less than bn, 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 bn 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 -1n × sin(n)/n2.0287

So, I sort of strip off -1n.0294

We look at this part,0296

bn is going to be sin(n)/n2.0298

Now, right away we have a problem because remember I said that we want to have bn > 0.0306

Here, the n2, 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 bn 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 -1n/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

-1n/n!, the n = 0 term, gives us -10, that is 1.0407

0! by definition is 1.0413

The n=1 term is -11, 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 an + 1, the abs(an+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 ex is the sum of xn/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 -1n/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, -100n n4/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 an+1/an.0022

I am going to work out this whole expression, but substituting an n+1 for n.0028

We get -100n+1 (n+1)4/n+1! × ln(n+1).0036

All of that divided by the same thing just with n, so -100n n4/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 100n+1/100n.0084

By the way, I am getting rid of the negative signs because we are taking absolute values there.0089

n+1/4/n4.0096

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

100n+1/100n is 100.0126

Now (n+1)4 is going to be some big complicated polynomial.0130

The key thing here is that n4 plus some smaller terms.0136

We are dividing it by n4,0145

So both in top and bottom there, the biggest term is n4.0148

We can divide top and bottom by n4,0154

So, that will make the n4 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

n4, if you look at n4 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 n4 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

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 n4, 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 -100n,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 100n, 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 n4,0450

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/nn2.0004

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 an1/n,0050

That is the nth root of the terms.0055

This is 1-2/n raised to the n2, 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 n2 raised to the 1/n,0090

Remember you multiply those exponents.0092

That is n2/n, that is just n.0097

Now this is a tricky expression, 1-2/nn.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 eln(ab), the point of doing that is you can pull the exponent b out of the natural log.0121

You can write it as eb, ln(a).0132

Here the a is the 1-2/n and the b is n.0138

We can write this as en × 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/n2,0250

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/n2, and the negative of that turns that into a positive.0267

The derivative of 2/n is -2/n2, derivative of -2/n is positive 2/n2.0283

Let us clean this up a little.0294

The 1/n2 cancels with that 1/n2 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, an 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/e2.0347

We have to figure out approximately how big that is.0350

e is approximately 2.7.0354

1/e2, 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 eln rule.0439

You sort it out using this eln 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

The case that is missing is what if an diverges,0127

But the absolute series converges.0137

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

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 an 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 -en/n!.0347

So we are going to go for the ratio test.0351

We are going to look at, remember an +1/an , 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 -en+1/n+1!.0369

an just means the original term, -en/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 en+1/en × n!/n+1!.0393

There are a couple things to explain here.0403

Notice with the n! and the en,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

en+1/en 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 an .0535

Ratio test says you divide an +1/an .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 n2 + 2.0564

Let us go for the ratio test again.0568

so an +1/an .0573

Is n+1!/n+12 + 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 n2 + 2/n+12 + 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, n2 + 2/n+12 + 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 n2 + some other stuff divided by,0656

Now if you work out n2, you can work it out, you are going to get n2 + some smaller terms, guaranteed.0664

If you divide top and bottom by n2, 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 an = n+1n/9n.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 an , well the nth root of that,0829

That is the same as an 1/n,0832

So that is n-1n/9n, absolute value of that to the 1/n,0836

The absolute values do not do anything here because the terms are positive.0847

n1/n is just 1, so we get n-11/9.0855

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-1n, we know that we want to use the root test.0936

The root test says you look at an 1/n, or the nth root of an ,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-2n/6n × 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-2n+1/6n+1, and then this big polynomial n+12 + 3 × n+1 - 7.0027

All of that divided by the an term, so x-2n.0048

Divided by 6n, and n2 + 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-2n+1/x-2n.0078

6n in the numerator, 6n+1 in the denominator.0089

In the numerator I get n2 + smaller stuff.0102

In the denominator I get n+12 + 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

6n/6n+1 cancels down to 6 in the denominator.0134

n2/n+12, that is a big polynomial, but both of the big polynomials,0142

Their leading term is n2, so if you divide both terms by n2,0150

What happens there is that the n2 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-2n/6n.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 limit0184

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 - 2n.0278

So that is -6n/6n.0287

Then, the n2 term.0293

The -6n and 6n gives you -1n/n2 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 8-2n is 6n/6n × the n2 term.0332

That simplifies down to 1/the n2 term.0351

That is a series that is very similar to 1/n2.0362

We can use the limit comparison test with the sum of 1/n2.0369

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/n2 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 series0000

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+x2.0012

Then, that is something that we can write as 1/1-(-x2).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)n0042

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 -x2 in absolute value less than 1.0063

Motivated by that, we will write 1/1+x2.0071

I am going to expand out this series in two individual terms.0079

That n=0 gives you 1, - x2 + (x2)2, so that is x4 - x6, and so on.0085

This is for x2 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+x2.0118

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 - -x3/3 - x5/5 - x7/7,0128

We still have the problem of adding on that constant, so I will add it on at the beginning this time.0155

C + x + x3, 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 - x3/3 + x5/5 - x7/7 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 x2n+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, -x3, that is - -13/3 + -15/3 - -17/3,0301

So that is -13 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+x2,0424

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.com0464

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 cn,0007

Then it has a power of either xn or x-an.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-an, 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

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 xn/2n × 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 an+1/an in absolute value.0231

That in this case is xn+1/2n+1 × n+1/xn/2n × 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 xn+1/xn.0264

2n/2n+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 2n/2n+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 -2n/2n × n, which simplifies down into -1n/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 2n/2n × 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

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 -1n × xn/n!.0656

We will look at an+1/an.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 xn+1/n+1! divided by the an term, xn/n!.0680

Still need my absolute values here.0694

If I flip my denominator and pull it up, we get xn+1/xn × 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

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 xn, 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+x2, and so on, dx.0995

If you integrate that, you get negative, integral of 1 is x, integral of x is x2/2,1009

Integral of x2, is x3/3, 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 - x2/2 - x3/3, and so on.1067

We can write that as the sum from n=1 to infinity, of -xn/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 xn, we would have gotten,1111

Well, xn+1/n+1, then shifting the indices from n=0 to n=1,1114

Would have converted that into xn/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 -1n/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 1n/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 ex.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 fn(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

fn(x), well ex, that is very easy, that is ex all the way down.0038

So, fn(0), that is just plugging 0 into ex, 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 fn(a)/n!/x-an.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! × x2 × 1/3! x3, and so on.0099

So, we get the McLauren series for ex is the sum from 0 to infinity of just xn/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 ex, is just xx/n!.0164

OK, for this example we are asked to find the McLauren series for f(x) = (ex)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) = ex. We are going to write down n fn(x). Oops, f(x) = (ex)2.0012

So f0(x) is just (ex)2.0031

First derivative is just (ex)2 × 2x.0036

The second derivative is 2x × derivative of (ex)2, so that is 2x × 2x × (ex)2.0041

Plus (ex)2 × derivative of 2x that is 2 (ex)2.0053

That can simplify down to 4x2 (ex)2,0062

I guess 4x2 + 2, we could combine those terms (ex)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 ex.0114

ex was the sum from 0 to infinity of xn/n!.0121

A much easier way than writing down all of these derivatives is to substitute in x2 for x.0136

So (ex)2 is the sum form n=0 to infinity of x2n/n!.0145

You could write that as the sum from n=0 to infinity of x2n/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 +x2/1.0176

n=2 gives you x4/2!.0182

n=3 gives you x6/3!, and so on.0187

Notice that that is just what you would have gotten if you had looked at the series for ex.0193

1+x+x2/2!+x3/3!.0199

If you had taken that series and just changed x into x2, 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 (ex)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 ex, and when you have to find (ex)2 you just do a quick substitution, and we get our series for (ex)20271

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 nth derivative there.0023

Of the function, you plug in a, you divide it by n!, and then you multiply by x-an.0030

We will call that cn, that is the nth coefficient of the Taylor Series.0047

Then the x-an 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 tk(x), and what that means is, essentially that you get this thing that only runs up to xk 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 xk 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 nth derivative of f at a.0163

Well a=0 here, because it is a Maclaurin Series.0171

fN(0)/n! × normally x-an, but since a=0 it is just xn.0174

I want to figure out what these coefficients are, the cn's.0184

I will make a little chart here.0188

n and the nth derivative of x, then I will plug in the nth derivative of 0.0191

Then I will figure out what the coefficient n is.0198

That is the same as fn(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 × x0 + 0 × x1.0302

So I will leave that out, - 1/2, 1/2! × x2, + 0 x3, + 1/4! × x4.0307

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 x5 - 1/6! × x6,0333

0 x7 + 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 x2n/2n!.0358

I am going to write -1n 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 - x2/2! + x4/4! - x6/6! 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 x2/2! is just x. Sorry 2x/2!0488

+ now the derivative of x4 + 4x3/4!.0503

- derivative of x6 is - x5/6! 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, 4x3/4!.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 x5, 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 x5/5! 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

x7/7!, 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 x2n+1/(2n+1)!.0601

And, -1n 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 ones0645

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 t4(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

n fn(x), the nth derivative, fn(pi/3).0718

Then the full coefficient cn, which is fn(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 nth derivative x, of a, you plug in x=a,0742

Divided by n! multiplied by the power part x-an.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 cn 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-a0 which is just 0.0924

n=1 says we take the 1/2 × x-a is pi/3, so pi/31.0931

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/33 + sqrt(3)/48 × x-pi/34.0967

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-an,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 nth derivative of f at x over n! × x-an.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-x2/2+x4/4! 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 ex 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-x2/2 + x4/24. 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 + 0x2 + 0x4, 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 - x2/2 + x4/24,1214

And then we subtract just like we do long division of numbers,1225

So that is + x2/2, then we multiply across there, x2/2.1228

x2/2 × -x2/2 is -x4/4 + x6/48.1237

We subtract again and the x2/2's cancel.1258

x4/4, I can write that as 6/24.1263

So, we have 6/24 - 1/24, that gives us 5/24 × x4 - x6/481269

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 x4.1292

sec(x) = 1+x2/2 + 5x4/24.1305

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, t4(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 - x2/2! + x4/4! - x6/6! and so on.0031

We have been told to use the Taylor Polynomial t4(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, t4(x) is 1 - x2/2! + x4/4!.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 x6/6!.0128

Remember, in the alternating series error bound, that was an+1.0143

An easier way to think of that is just the cutoff term, so that is x6/6!.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 x6/6!, 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

x6, 6! is 720, so 0.01 × 720, we got that from doing 6!.0181

So, we get abs(x6) < 7.2,0195

If we solve that for x we get x < 7.21/6.0205

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 t4(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 4th 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 x6/6! 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, t4(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-ak+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

1/205, is, 205, well 25=32 and 105 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 t4(x), t4(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 t4(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 t4(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

And technically it is a bound on the 5th 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 tk(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-ak+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 t2(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 - x2/2! + x4/4! - x6/6! ...0230

So, t2(x), that means we cut off the Taylor series at the degree 2 term.0246

Because that is the degree 2 term, t2(x) says we just look at 1 - x2/2!.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 t2(1/2).0277

That is 1 - 1/22/2, 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 x4/4!, and we plug in 1/2 there.0375

1/24/4!, so that is the cutoff term there, that is aN+1.0384

That in turn, that is 4!, so this is 1/24 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+x5.0561

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+x5. 0575

Instead, we are going to notice that this is the same as 1/1-(-x5).0585

That is the sum of the geometric series.0593

We can write that as the sum from n=0 to infinity of (-x5)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 + -x5, so -x5.0625

+ (-x5)2, so + x10 - x15 and so on.0635

That was just the function 1/1+x5.0645

What we really want is the integral of all that.0650

The integral of 1/1+x5 dx is the integral of 1-x5 + x10 - x15 and so on dx. 0655

I can integrate that just term by term so I get x - x6/6 + x11/11 - x16/16.0672

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+x5.0707

That is 1/2 - (1/2)6/6 + 1/211/11 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 1/211/6, which is the same as 1/ 26 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, 1/211/11.0834

Now, that is 1 over 211 × 11.0842

I know that 210 is about 1000, if you multiply out 210 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 - 1/26/6 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, nth derivative of x,1079

The nth derivative, now I am going to plug in the value a=4,1084

The whole coefficient is obtained by taking that nth 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 0th derivative just means you take the original function, sqrt(x).1110

First derivative, if you think of that as x1/2 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 43/2, that is 1/8 × -1/4 that is -1/32.1165

x-5/2, that is 45/2, sqrt(4) is 2, 25 is 32, so we get 3/8 × 32, 1177

Which is 25 × 23, 28 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-ak+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/x5/2.1338

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 3/8/3.85/2.1421

OK, so I can write that as 3/8, now I do not want to work out 3.85/2.1430

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 15/2, that is not 1 and 5/2, that is 15/2.1453

This is because 3.8 is way bigger than 1, so in the denominator, 3.85/2 will be less than 15/2.1467

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 3.8-4k+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 0.23.1547

0.23 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 cn, 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

t2(x) means we read off these coefficients and we attach each one to a power of x. 1681

So, 2 × x0 is just 1.1690

+ 1/4 ×, sorry not a power of x, but x-a, so that is x-41.1694

We go up to the 2 term, so that is -1/64 × x-42, 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 × -0.22.1743

We can simplify that a little bit, this is 2 - 0.2/4, 0.22 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