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Integration by parts review

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.E (LO)
,
FUN‑6.E.1 (EK)
Review your integration by parts skills.

What is integration by parts?

Integration by parts is a method to find integrals of products:
integral, u, left parenthesis, x, right parenthesis, v, prime, left parenthesis, x, right parenthesis, d, x, equals, u, left parenthesis, x, right parenthesis, v, left parenthesis, x, right parenthesis, minus, integral, u, prime, left parenthesis, x, right parenthesis, v, left parenthesis, x, right parenthesis, d, x
or more compactly:
integral, u, space, d, v, equals, u, v, minus, integral, v, space, d, u
We can use this method, which can be considered as the "reverse product rule," by considering one of the two factors as the derivative of another function.
Want to learn more about integration by parts? Check out this video.

Practice set 1: Integration by parts of indefinite integrals

Let's find, for example, the indefinite integral integral, x, cosine, x, d, x. To do that, we let u, equals, x and d, v, equals, cosine, left parenthesis, x, right parenthesis, d, x:
integral, x, cosine, left parenthesis, x, right parenthesis, d, x, equals, integral, u, d, v
u, equals, x means that d, u, equals, d, x.
d, v, equals, cosine, left parenthesis, x, right parenthesis, d, x means that v, equals, sine, left parenthesis, x, right parenthesis.
Now we integrate by parts!
xcos(x)dx=udv=uvvdu=xsin(x)sin(x)dx=xsin(x)+cos(x)+C\begin{aligned} \displaystyle\int x\cos(x)\,dx &=\displaystyle\int u\,dv \\\\ &=uv-\displaystyle\int v\,du \\\\ &=\displaystyle x\sin(x)-\int\sin(x)\,dx \\\\ &=x\sin(x)+\cos(x)+C \end{aligned}
Remember you can always check your work by differentiating your result!
Problem 1.1
integral, x, e, start superscript, 5, x, end superscript, d, x, equals, question mark
Choose 1 answer:
Choose 1 answer:

Want to try more problems like this? Check out this exercise.

Practice set 2: Integration by parts of definite integrals

Let's find, for example, the definite integral integral, start subscript, 0, end subscript, start superscript, 5, end superscript, x, e, start superscript, minus, x, end superscript, d, x. To do that, we let u, equals, x and d, v, equals, e, start superscript, minus, x, end superscript, d, x:
u, equals, x means that d, u, equals, d, x.
d, v, equals, e, start superscript, minus, x, end superscript, d, x means that v, equals, minus, e, start superscript, minus, x, end superscript.
Now we integrate by parts:
=05xexdx=05udv=[uv]0505vdu=[xex]0505exdx=[xexex]05=[ex(x+1)]05=e5(6)+e0(1)=6e5+1\begin{aligned} &\phantom{=}\displaystyle\int_0^5 xe^{-x}\,dx \\\\ &=\displaystyle\int_0^5 u\,dv \\\\ &=\Big[uv\Big]_0^5-\displaystyle\int_0^5 v\,du \\\\ &=\displaystyle\Big[ -xe^{-x}\Big]_0^5-\int_0^5-e^{-x}\,dx \\\\ &=\Big[-xe^{-x}-e^{-x}\Big]_0^5 \\\\ &=\Big[-e^{-x}(x+1)\Big]_0^5 \\\\ &=-e^{-5}(6)+e^0(1) \\\\ &=-6e^{-5}+1 \end{aligned}
Problem 2.1
integral, start subscript, 1, end subscript, start superscript, e, end superscript, x, cubed, natural log, x, space, d, x, equals, question mark
Choose 1 answer:
Choose 1 answer:

Want to try more problems like this? Check out this exercise.

Want to join the conversation?

  • leafers tree style avatar for user Austin.Connell
    in the int (0 -> pi) of xsin(2x)dx problem, in the solution, the third to last line, shouldn't that be (sin(2x)/4) not (sin(4x)/4)? or am I missing something?
    (11 votes)
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  • blobby green style avatar for user earl kraft
    Like Bhoovesh I am also fuzzy about the compact notation. It seems that the confusion is not with Leibniz notation vs Newton's, but rather I am concerned about falling in a pit as a consequence of having only one letter in an expression for which I am accustomed to two. The dropping of the x's and dx's makes me nervous about getting fouled up as a consequence of x not being the only variable. Where is y, and how does one keep track of it with the more compact notation? I would like to know the conventions and rules for this.
    (7 votes)
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  • starky ultimate style avatar for user Bhoovesh Waran
    Why hasn't Sal explained about the compact form of Integration by parts??i don't understand it!! It contradicts to what Sal said about differentials earlier that the differentials are not numbers or function which can't cancelled or algebraically manipulated!!
    (5 votes)
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    • leafers tree style avatar for user dragon purls
      The "compact form" is just a different way to write the form used in the videos. Basically, the only difference is that the "video form" uses prime notation (f'(x)), and the "compact form" uses Leibniz notation (dy/dx). If you are used to the prime notation form for integration by parts, a good way to learn Leibniz form is to set up the problem in the prime form, then do the substitutions f(x) = u, g'(x)dx = dv, f'(x) = v, g(x)dx = du. At least, that's how it clicked for me.

      As far as the manipulating differentials goes, it's true that you can't just treat differentials like they are normal terms in an equation (as if dx were the variable d times the variable x), but it is legal to split up the dy/dx when differentiating both sides of an equation. The concept here is exactly the same as what is used when doing u-substitution (URL to video below if you need it).
      https://www.khanacademy.org/math/calculus-home/integration-techniques-calc/u-substitution-calc/v/u-substitution
      Hope this helps, and good luck with your work!
      (3 votes)
  • blobby green style avatar for user Mahnoor Aamer
    What is the use of integration? When do we use it in our practical lives?
    (1 vote)
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  • aqualine ultimate style avatar for user Hafsa Kaja Moinudeen
    Why does the integral of e^5x dx = 1/5 e^5x? Is it an application of the reverse chain rule?
    Thanks very much!
    (4 votes)
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    • piceratops ultimate style avatar for user Fai
      That's one way of thinking about it, yes. As you continue on in math, this will become almost second-nature and you won't even think about the chain rule when integrating simple exponential functions.
      (2 votes)
  • aqualine seed style avatar for user Varun
    How would you integrate something in parentheses, like (x^2 +1)^1/2?
    (3 votes)
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  • male robot hal style avatar for user Leo
    Is there a reverse division rule that can sometimes serve as a substitute for this? An example where this would be useful is (ln(5x))/(x^2)
    (3 votes)
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    • female robot grace style avatar for user loumast17
      the quotient rule for derivatives is just a special case of the product rule. f(x)/g(x) = f(x)*(g(x))^(-1) or in other words f or x divided by g of x equals f or x times g or x to the negative one power. so it becomes a product rule then a chain rule.

      So when you have two functions being divided you would use integration by parts likely, or perhaps u sub depending.

      Really though it all depends. finding the derivative of one function may need the chain rule, but the next one would only need the power rule or something. It's kinda hard to predict if two functions being divided need integration by parts or what to integrate them.
      (2 votes)
  • blobby green style avatar for user mbalikhosta
    Thanks Sal. I am the master of integration by parts. Do not test me. I am now the best, I think I should do the testing.
    (3 votes)
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  • blobby green style avatar for user Bhargava Sivateja
    ILATE technique will be useful
    (3 votes)
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  • leaf green style avatar for user Daniel Arges
    Hi. There is a gap of explanation about the compact notation. I didn't catch why 'dx' becomes 'dv' and 'du'. Thank you.
    (2 votes)
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