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Homework Help: Integration by parts

  1. Jun 3, 2006 #1
    hi..im new to this topic..can someone check to see if this is right?

    [tex]\int (xe^-^x)dx = \int udV = uV - \int Vdu[/tex]
    [tex]=x(-e^-^x)- \int -e^-^x[/tex]
    [tex]=-xe^-^x-e^-^x+C[/tex]

    thanks
     
  2. jcsd
  3. Jun 3, 2006 #2

    arildno

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    Why not check it yourself?
    Differentiate your answer and see if you get back your integrand! :smile:
     
    Last edited: Jun 3, 2006
  4. Jun 3, 2006 #3
    wow..why didnt i think of that haha

    THanks
     
  5. Jun 3, 2006 #4

    arildno

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    Spoiler alert!!




    Yeah, it's right
     
  6. Jun 3, 2006 #5
    Ok, I have another one im trying to figure out...but i keep getting 0!

    [tex] \int_{0}^{\pi} tsin3t dt = udV \right]_{0}^{\pi} - \int_{0}^{\pi}vdu[/tex]

    work...

    [tex]=t(sin3t) \\right]_{0}^{\pi} - \int_{0}^{\pi}vdu[/tex]

    [tex]=\pi (sin3\pi) - \int_{0}^{\pi} \frac{-1}{3} cos3t(1)dt[/tex]

    [tex]=0 - \frac{1}{sin3t} \int_{0}^{\pi}z dz[/tex]..
    let z = cos3t
    and then use substitution to get [tex]\frac{dz}{sin3t} = -3dt[/tex]

    [tex]0-\frac{1}{sin3t} \frac{z^2}{2}[/tex]

    [tex]=\frac{-(cos3t)^2}{2sin3t}[/tex]
    this must be wrong..i solved it out a bit more..but keep getting zero, where did i go wrong?
     
    Last edited: Jun 3, 2006
  7. Jun 3, 2006 #6

    arildno

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    Well, what you have written is just nonsense.
    As a novice, it is safer for you to do these problems like this:
    [tex]u(t)=t\to\frac{du}{dt}=1,\frac{dv}{dt}=\sin(3t)\to{v}(t)=-\frac{1}{3}\cos(3t)[/tex]

    Use these relations in the integration by parts formula.
     
  8. Jun 3, 2006 #7
    hrmm..i already knew these relations..i still dont see what i did wrong..i followed it just how they do it in the book, my LaTeX skills arent very good, is it just hard to follow?
     
  9. Jun 3, 2006 #8

    arildno

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    well, you didn't compute the uv-term correctly.
     
  10. Jun 3, 2006 #9
    o wait..it is :[tex]uv - \int_{0}^{\pi}vdu[/tex]
    i have [tex]udv - \int_{0}^{\pi}vdu[/tex] in my work..my bad
     
  11. Jun 3, 2006 #10
    giving me [tex]\frac{\pi}{3}[/tex] which i think is right..thanks for the help
     
  12. Jun 4, 2006 #11

    HallsofIvy

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    I would strongly recommend that you write out exactly what "u", "v", "du", and "dv" are!
     
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