For more information, please see full course syllabus of College Calculus: Level II

For more information, please see full course syllabus of College Calculus: Level II

## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Integration by Parts

**Main formula:**

**Hints and tips:**

Integration by parts never solves the integral completely for you. The point is to reduce a difficult integral to an easier one.

If you have a polynomial multiplied by an exponential or trigonometric function, you can use the tabular integration shortcut.

You may have to use parts more than once in the same problem.

Sometimes, after using parts one or more times, you get the original integral reappearing. Then you can set up an equation for the original integral and find the solution algebraically without doing any more integration.

Use the mnemonic LIATE (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential) to know which part of the expression to make the

*u*.Sometimes you have to make one or more substitution before you get a form that works well for integration by parts.

### Integration by Parts

_{}

^{}2xcosxdx

- Let u = 2x and dv = cosx
- du = 2dx
- dv = cosxdx
- v = sinx + C

∫

_{}

^{}2xcosxdx = 2x(sinx) − ∫

_{}

^{}sinx(2dx) + C = 2xsinx − 2∫

_{}

^{}sinxdx + C = 2xsinx + 2cosx + C

_{}

^{}xsin3xdx

- Let u = x and dv = sin3xdx
- u = x
- du = dx
- dv = sin3xdx
- v = [( − cos3x)/3] + C
- Apply Integration by Parts
- ∫
_{}^{}xsin3xdx = [( − cos3x)/3]( x ) − ∫_{}^{}[( − cos3x)/3]dx + C - = [( − xcos3x)/3] + [1/3]∫
_{}^{}cos3xdx + C - = [( − xcos3x)/3] + [1/3]( [sin3x/3] ) + C
- = [( − xcos3x)/3] + [sin3x/9] + C

_{}

^{}lnx

^{2}dx

- Use logarithm properties to simplify
- ∫
_{}^{}lnx^{2}dx = ∫_{}^{}2lnx^{}dx - = 2∫
_{}^{}lnxdx - Let u = lnx and dv = dx
- du = [dx/x]
- dv = dx
- v = x + C
- Apply Integration by Parts
- ∫
_{}^{}lnx^{2}dx = 2∫_{}^{}lnxdx

_{}

^{}dx + C ) = 2( xlnx − x + C ) = 2xlnx − 2x + C

_{}

^{}ln(x + 1)dx

- Let u = x + 1 and dv = dx
- u = ln(x + 1)
- du = [1/(x + 1)]dx
- dv = dx
- v = x + C
- Apply Integration by Parts
- ∫
_{}^{}ln(x + 1)dx = x( ln(x + 1) ) − ∫_{}^{}x( [dx/(3x + 1)] ) + C - = x( ln(x + 1) ) − ∫
_{}^{}[x/(x + 1)] dx + C - Note the alternative form of [x/(x + 1)]
- [x/(x + 1)] = 1 − [1/(x + 1)]
- x( ln(x + 1) ) − ∫
_{}^{}[x/(x + 1)] dx = x( ln(x + 1) ) − ∫_{}^{}1 − [1/(x + 1)] dx + C - = xln(x + 1) − ∫
_{}^{}dx + ∫_{}^{}[1/(x + 1)] dx + C

_{}

^{}3x

^{2}e

^{x}dx

- Let u = x
^{2}and dv = e^{x}dx - du = 2x dx
- dv = e
^{x}dx - v = e
^{x}+ C - Apply Integration by Parts
- ∫
_{}^{}3x^{2}e^{x}dx = 3∫_{}^{}x^{2}e^{x}dx - = 3( e
^{x}x^{2}− ∫_{}^{}e^{x}( 2x dx ) ) - Apply Integration by Parts again on ∫
_{}^{}e^{x}( 2x dx ) - ∫
_{}^{}e^{x}( 2x dx ) = e^{x}(2x) + ∫_{}^{}e^{x}(2dx) - = 2xe
^{x}+ 2e^{x}+ C

3( e

^{x}x

^{2}− ∫

_{}

^{}e

^{x}( 2x dx ) ) = 3( e

^{x}x

^{2}− 2xe

^{x}+ 2e

^{x}+ C )

_{0}

^{π}(x − 1)sinxdx

- Let u = x − 1 and dv = sinxdx
- du = dx
- dv = sinxdx
- v = − cosx
- Apply Integration by Parts
- ∫
_{0}^{π}(x − 1)sinxdx = − cosx(x − 1) − ∫_{0}^{π}− cosx(dx) - = (cosx − xcosx + sinx)
_{0}^{π} - = cosπ− πcos π+ sinπ− (cos0 − (0)cos0 + sin0)
- = − 1 + π+ 0 − (1)

_{0}

^{10}x e

^{x}dx

- Let u = x and dv = e
^{x}dx - du = dx
- dv = e
^{x}dx - v = e
^{x} - Apply Integration by Parts
- ∫
_{0}^{10}x e^{x}dx = [ e^{x}(x) ]_{0}^{10}+ ∫_{0}^{10}e^{x}dx

^{10}(10) + e

^{10}− (e

^{0}(0) + e

^{0}) = e

^{10}(10) + e

^{10}− 1 = 11e

^{10}− 1

_{0}

^{e}[x/(x + 1)]dx

- Let u = x and dv = [dx/(x + 1)]
- du = dx
- dv = [1/(x + 1)]dx
- v = ln(x + 1)
- Apply Integration by Parts
- ∫
_{0}^{e}[x/(x + 1)]dx = xln(x + 1) − ∫_{0}^{e}ln(x + 1) dx - From problem 4, we have solved ∫
_{}^{}ln(x + 1) dx - ∫
_{}^{}ln(x + 1) dx = xln(x + 1) − x + ln(x + 1) + C - xln(x + 1) − ∫
_{0}^{e}ln(x + 1) dx = xln(x + 1) − (xln(x + 1) − x + ln(x + 1)) - = [ xln(x + 1) − xln(x + 1) + x − ln(x + 1) ]
_{0}^{e} - = eln(e + 1) − eln(e + 1) + e − ln(e + 1) − ((0)ln(0 + 1) − (0)ln(0 + 1) + 0 − ln(0 + 1)
- = e − ln(e + 1) + ln1

_{2}

^{3}ln(3x − 5)

^{3}dx

- Use logarithm properties to change form
- ∫
_{2}^{3}ln(3x − 5)^{3}dx = ∫_{2}^{3}3ln(3x − 5)^{}dx - = ∫
_{0}^{2}3n(3x − 5)3dx - Let u = ln(3x − 5) and dv = 3dx
- du = [3/(3x − 5)]dx
- dv = 3dx
- v = 3x
- Apply Integration by Parts
- ∫
_{2}^{3}ln(3x − 5)dx = xln(3x − 5) − ∫_{2}^{3}[3x/(3x − 5)]dx - Find the alternative form of [3x/(3x − 5)]
- [3x/(3x − 5)] = [(3x − 5)/(3x − 5)] + [5/(3x − 5)] = 1 + [5/(3x − 5)]
- xln(3x − 5) − ∫
_{0}^{2}[3x/(3x − 5)]dx = xln(3x − 5) − ∫_{2}^{3}1 + [5/(3x − 5)] dx

_{}

^{}sinxcosx = [(sin

^{2}x)/2] + C

- Let u = sinx and dv = cosx
- du = cosx
- dv = cosx
- v = sinx
- Apply Integration by Parts
- ∫
_{}^{}sinxcosx = sinx(sinx) − ∫_{}^{}sinxcosx + C - = sin
^{2}x − ∫_{}^{}sinxcosx + C - Isolate sin
^{2}x - 2∫
_{}^{}sinxcosx = sin^{2}x + C

_{}

^{}sinxcosx = [(sin

^{2}x)/2] + C

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Integration by Parts

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Important Equation 0:07
- Where It Comes From (Product Rule)
- Why Use It?
- Lecture Example 1 1:24
- Lecture Example 2 3:30
- Shortcut: Tabular Integration 7:34
- Example
- Lecture Example 3 10:00
- Mnemonic: LIATE 14:44
- Ln, Inverse, Algebra, Trigonometry, e
- Additional Example 4
- Additional Example 5

### College Calculus 2 Online Course

I. Advanced Integration Techniques | ||
---|---|---|

Integration by Parts | 24:52 | |

Integration of Trigonometric Functions | 25:30 | |

Trigonometric Substitutions | 30:09 | |

Partial Fractions | 41:22 | |

Integration Tables | 20:00 | |

Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule | 22:36 | |

Simpson's Rule | 21:08 | |

Improper Integration | 44:18 | |

II. Applications of Integrals, part 2 | ||

Arclength | 23:20 | |

Surface Area of Revolution | 28:53 | |

Hydrostatic Pressure | 24:37 | |

Center of Mass | 25:39 | |

III. Parametric Functions | ||

Parametric Curves | 22:26 | |

Polar Coordinates | 30:59 | |

IV. Sequences and Series | ||

Sequences | 31:13 | |

Series | 31:46 | |

Integral Test | 23:26 | |

Comparison Test | 22:44 | |

Alternating Series | 25:26 | |

Ratio Test and Root Test | 33:27 | |

Power Series | 38:36 | |

V. Taylor and Maclaurin Series | ||

Taylor Series and Maclaurin Series | 30:18 | |

Taylor Polynomial Applications | 50:50 |

### Transcription: Integration by Parts

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*First natural log, then inverse trigonometry.*0184

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

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

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

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

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

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

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

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

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

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

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

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

*Which is 1/2w dx.*0059

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

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

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

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

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

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

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

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

*We have w sin(w).*0116

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

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

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

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

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

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

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

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

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

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

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

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

*That just multiplies away to nothing.*0195

*Now I will substitute back.*0198

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*Let us try a trickier example.*0208

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*Here is the secret shortcut.*0477

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*If we took derivatives of either one of these, e ^{x} or cosin(x), we would never get down to 0.*0612

*The shortcut is not available here.*0620

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*That means we can solve for i.*0820

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

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

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

*That is the answer right there.*0844

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*I stands for inverse functions.*0973

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

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

*A stands for algebra.*0994

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

*T stands for trigonometry.*1003

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

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

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

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

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

*I hope it works out for you.*1039

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

1 answer

Last reply by: Dr. William Murray

Fri Oct 7, 2016 9:56 AM

Post by selam berhe on October 5, 2016

Antibiotic pharmacokinetics An antibiotic tablet is taken and t hours later the concentration in the bloodstream

is C(t)= 3(e^-0.8t -e^-1.2t)

where C is measured in mgymL.

(a) What is the maximum concentration of the antibiotic

and when does it occur?

(b) Calculate integral of 0 to 2 C(t) dt and interpret your result.

(c) Calculate integral of 0 to infinity of C(t) and explain its meaning.

`

0 Cstd dt and explain its meaning.

3 answers

Last reply by: Dr. William Murray

Mon Jun 8, 2015 5:08 PM

Post by Joshua Bowen on May 11, 2015

How do we recognize a Intergration by part quickly?

1 answer

Last reply by: Dr. William Murray

Mon Aug 18, 2014 11:04 AM

Post by ahmed ahmed on August 16, 2014

Thank you from the bottom of my heart ... you are of a great help and i got A on my calculus-2 summer class... i am so happy... thank you again and again and forever...

1 answer

Last reply by: Dr. William Murray

Tue Feb 11, 2014 4:16 PM

Post by Silvia Gonzalez on February 3, 2014

Hi, thank you for taking the time to prepare such good lectures.

I am doing the practice exercise integral from 0 to n of (x-1)sinx.dx and I do not understand how you go from step 8 to 9, can you help me?

1 answer

Last reply by: Dr. William Murray

Sun Nov 3, 2013 6:23 PM

Post by Christian Fischer on November 3, 2013

Hi Professor, Thank you for a great video. There is one thing I don't understand: If u=x then du=dx but if u=x^2 then du=2x

To me it looks like we differentiated x^2 to get du=2x but did nothing to x to get du=dx? if we had differentiated x we would get du=d*1 What is the explanation for that?

Have a great day

Christian

2 answers

Last reply by: Dr. William Murray

Fri Aug 23, 2013 11:48 AM

Post by Maximillian Lanander on August 17, 2013

I think you made a boo boo. Not sure. If u is x^2 then du surely = x^2dx ?

3 answers

Last reply by: Dr. William Murray

Fri Mar 11, 2016 9:41 AM

Post by Ramez Hajelsawi on July 7, 2013

That shortcut blew my mind, I went back and redid all the problems that it works for again just to prove it to myself. Fantastic lecture!

1 answer

Last reply by: Dr. William Murray

Tue Apr 23, 2013 7:49 PM

Post by Hye Goo Yoon on April 23, 2013

integration of [(3X^2)/7^3]e^-(x/7)^3 dx 0<X<âˆž

At first I have used by parts method but it was a never ending integration.

I have found a formula from the table of integrals, it still gave me unfinished integral. How can I effectively solve this kind of integration by parts? Is there any ways for me to easily approach this kind of integration? Since the range is given,

Somehow I have ended up learning about Gamma function. I am very confused now.

Thank you

1 answer

Last reply by: Dr. William Murray

Fri Jan 18, 2013 5:30 PM

Post by Abdelrahman Megahed on January 17, 2013

How come LIATE doesn't work for example 3?

1 answer

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Fri Jan 18, 2013 5:30 PM

Post by Shehan Gunasekara on May 16, 2012

WOW!! This is so goood!! you are better than my lecturer

1 answer

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Fri Jan 18, 2013 5:33 PM

Post by Emudiare Sowho on May 7, 2012

How do we get the practice problems?

1 answer

Last reply by: Dr. William Murray

Fri Jan 18, 2013 5:34 PM

Post by Kathy Hodge on November 7, 2011

Thanks. Helped me understand and apply this independently, without having to look at aids, for the first time, despite some previous attempts.

3 answers

Last reply by: Dr. William Murray

Fri Jan 18, 2013 5:36 PM

Post by Maimouna Louche on June 7, 2011

Why does it stop and start over? It stops at the "Shortcut: Tabular Integration".

3 answers

Last reply by: Dr. William Murray

Fri Jan 18, 2013 5:37 PM

Post by Maimouna Louche on June 7, 2011

what is the difference between derivatives and integrals?

1 answer

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Fri Jan 18, 2013 5:38 PM

Post by Anthony Simon on April 20, 2010

This is a great video, i can't wait for those practice problems.

1 answer

Last reply by: Dr. William Murray

Fri Jan 18, 2013 5:41 PM

Post by Stephan Maric on March 13, 2010

Excellent video instruction.

My only question is whether or not you have access to supplemental problems. This would really help me practice out the problems.