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Integration by Parts Evaluate the integral

  1. Jul 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral. (Use C for the constant of integration.)

    ∫te ^ (-9t) dt


    2. Relevant equations

    ∫udv = uv - ∫vdu

    u=t dv= e ^ (-9t) dt
    du=dt v=(-1/9) e ^(-9t)

    3. The attempt at a solution

    = -1/9 te^(-9t) - ∫-1/9 e ^(-9t) dt

    Second Integral:
    w=-9t
    dw=-9dt
    -81∫-1/9 * -81 e ^(-9t) dt
    -81∫e^w * w
    -81 * e^(-9t) +C

    Final Answer:
    = -1/9 te^(-9t) + 81 e ^(-9t) +C

    This answer isn't right and I'm not sure where I'm going wrong, so any help would be appreciated. Thanks!!
     
    Last edited: Jul 10, 2014
  2. jcsd
  3. Jul 10, 2014 #2
    A tip which has helped me is to always extract as much as possible before doing the integral. Doing that, you can see that it is actually the same integral as before, e^-9t. The error is simply that you multiplied with 81,while you should have divided.
     
  4. Jul 10, 2014 #3

    HallsofIvy

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    Where did that "-1/9" come from in "dv"?
    The integral you are given is [itex]\int te^{-9t}dt[/itex]. Writing that as [itex]\int u dv[/itex], you could take u= t, [itex]dv= e^{-9t}[/itex].

    Or you could write the integral as [itex]-9\int t(-(1/9)e^{-9t})dt[/itex] and then take [itex]dv= (-1/9)e^{-9t}dt[/itex] but you don't seem to have done that.

     
  5. Jul 10, 2014 #4
    You're completely right - I accidentally switched v and dv, thanks for pointing that out!
     
  6. Jul 10, 2014 #5
    So is the final answer -1/9 t e ^(-9t) + 1/81 e ^(-9t) + C ?
     
  7. Jul 10, 2014 #6

    LCKurtz

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    Differentiate it and see if you get your integrand.
     
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