1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration by parts

  1. May 4, 2009 #1
    Formula for integration by parts:

    [tex]\int f(x)dx = \int u dv = uv - \int v du[/tex]



    Use integration by parts to find the following integrals:

    a) [tex]\int x e^{1-x} dx [/tex]

    b) [tex]\int_1^4 \frac {ln \sqrt x} {\sqrt x} dx[/tex]

    c) [tex]\int_{-2}^1 (2x+1)(x+3)^{3/2} dx[/tex]

    d) [tex]\int x^3 \sqrt{3x^2+2} dx[/tex]

    Answers in back of the book:

    a) [tex]-e^{1-x}(x+1)+C[/tex]

    b) 4 ln 2 - 2

    c) [tex]\frac{74}{4}[/tex]

    d) [tex]\frac {1}{9}x^2(3x^2+2)^{\frac {3}{2}}-\frac{2}{135}(3x^2+2)^{\frac {5}{2}} +C[/tex]

    My attempts:

    a) [tex]\int xe^{(1-x)} dx[/tex]

    [tex]u = x[/tex]

    [tex]dv = e^{1-x} dx[/tex]

    [tex]du = 1 [/tex]

    [tex]v = -e^{1-x}[/tex]

    which gives:

    [tex]-xe^{1-x} - \int -e^{1-x}(1)[/tex]

    [tex]-xe^{1-x} - e^{1-x}[/tex]

    [tex]-e^{1-x}(x+1) +C[/tex]

    GOT IT! Thanks Melawrghk and dx!


    b) [tex]\int_1^4 \frac {ln \sqrt x} {\sqrt x} dx[/tex]

    [tex]u = ln x^{{1}{2}}[/tex]

    [tex]dv = x^{\frac {-1}{2}}[/tex]

    [tex]du = \frac { \frac {1}{2}\sqrt x}{\sqrt x} = \frac {1}{2x}[/tex]

    [tex]v = 2x^{\frac{1}{2}}[/tex]

    which gives:

    [tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int 2x^{\frac{1}{2}} \frac {1}{2x}[/tex]

    [tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int \frac{x^{\frac {1}{2}}}{x}[/tex]

    [tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int x^{\frac {-1}{2}}[/tex]

    [tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - 2x^{\frac{1}{2}}[/tex]

    [tex] 2x^{\frac{1}{2}}(ln x^{\frac {1}{2}}-1)[/tex]

    Plug in 4 and 1:

    [tex]4(ln2-1)-2(ln1-1)[/tex]

    [tex] = 4ln2-4+2[/tex]

    [tex] = 4ln2-2[/tex]

    GOT IT! Thanks again dx!!

    I'm looking at the other two and I'm blanking.

    I've been out of school for three years, and jumped right into business calculus since it's a requirement. So I apologize if this is elementary, but I've been banging my head on this for hours now... and of course the teacher is nowhere to be found. :cry:

    Thanks in advance!

    EDIT: Got a correct now, thanks to Melawrghk and dx.
     
    Last edited: May 4, 2009
  2. jcsd
  3. May 4, 2009 #2
    For a), switch your u and dv and try again :)

    EDIT: and for b), check your derivative of ln(sqrt(x)). sqrt(x) can be written as x^(1/2), so when you take the derivative you'll get (1/2)x^(3/2). Coupled with the square root of x on the bottom it will cancel nicely.
     
  4. May 4, 2009 #3

    dx

    User Avatar
    Homework Helper
    Gold Member

    For part (a), v = e1-x would be a better choice.
     
  5. May 4, 2009 #4
    Is [tex] \int e^{1-x} = e^{1-x}[/tex] ?

    Edit: I'm only assuming this because

    [tex] \int e^{x} = e^{x}[/tex]

    ... but now that I think about it...

    [tex] \int e^{1-x} = \frac {1}{1-x} e^{1-x+1} = \frac {1}{1-x} e^{x}[/tex]

    right?
     
    Last edited: May 4, 2009
  6. May 4, 2009 #5
    Wouldn't the derivative of x^(1/2) = (1/2)x^(-1/2)?
     
  7. May 4, 2009 #6

    dx

    User Avatar
    Homework Helper
    Gold Member

    Nope. ∫ e1-x dx = -e1-x. Use the substitution t = 1-x to see why.
     
  8. May 4, 2009 #7

    dx

    User Avatar
    Homework Helper
    Gold Member

    Yes, that's correct.
     
  9. May 4, 2009 #8
    Aha! Great, now I got part a and edited up there. Thanks :)
     
  10. May 4, 2009 #9

    dx

    User Avatar
    Homework Helper
    Gold Member

    For part (b), use v = √x.
     
  11. May 4, 2009 #10
    Could you walk me through this method, please?

    I'm taught to look at the equation and assign u and dv. Then derive u to get du and integrate dv to get v.

    The book says:

    Step 1: Choose functions u and v so that f(x)dx = u dv. Try to pick u so that du is simpler than u and a dv that is easy to integrate.

    Step 2: Organize the computation... (they put it in a neat box :tongue:)

    Step 3: Complete the integration by finding ∫f(x) dx = ∫ u dv = uv - ∫ v du

    So I'm not sure how I assign v = √x directly.

    Here's the problem again:

    [tex]
    \int_1^4 \frac {ln \sqrt x} {\sqrt x} dx
    [/tex]
     
  12. May 4, 2009 #11

    dx

    User Avatar
    Homework Helper
    Gold Member

    Hint: If v = √x, then

    [tex] dv = \frac{dx}{2\sqrt{x}}[/tex]
     
  13. May 4, 2009 #12
    Got part b! Dx, you rock!
     
  14. May 4, 2009 #13
    OK so for part d I'm running into a seemingly basic problem. I'm not sure how to integrate √(3x2+2)

    I'm just fried....
     
  15. May 4, 2009 #14
    You have a direct formula for that.
     
  16. May 4, 2009 #15
    What is the formula? I'm trying to study for tomorrow morning's final, doing these practice questions and papers are everywhere.
     
  17. May 5, 2009 #16
    http://planetmath.org/encyclopedia/IntegrationFormulas.html [Broken]
     
    Last edited by a moderator: May 4, 2017
  18. May 5, 2009 #17

    Cyosis

    User Avatar
    Homework Helper

    Don't use a formula, splitting it up in x^3 and the root is a poor way to split things. Do it like this instead.

    [tex]x^3 \sqrt{3x^2+2}=x^2*x\sqrt{3x^2+2}[/tex]. Now use partial integration and note that [itex]x\sqrt{3x^2+2}[/itex] is really easy to integrate by substituting u=3x^2+2, or even "guessing" the primitive.
     
    Last edited: May 5, 2009
  19. May 5, 2009 #18
    Oops.. My apologies... I just read the post no 13 and not the original post. So i just thought its [itex]sqrt{3x^2+2}[/itex] and didnt notice the x cube
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integration by parts
  1. Integration by parts (Replies: 6)

  2. Integration by parts (Replies: 8)

  3. Integration by Parts (Replies: 8)

Loading...