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Thread: integration by parts

  1. #1
    Junior Member
    Sep 2009

    integration by parts

    Use integration by parts to find the integral.

    $\displaystyle \int t^2*\exp{3t} dt$
    $\displaystyle u=t^2$ $\displaystyle v'=\exp{3t}$
    $\displaystyle u'=2t$ $\displaystyle v=\frac{\exp{3t}}{3}$
    $\displaystyle \frac{1}{3}t^2\exp{3t}-\int\frac{2t\exp{3t}}{3}dt =\frac{1}{3}t^2\exp{3t}-\frac{2}{3}\int t\exp{3t} dt$

    $\displaystyle u=t$ $\displaystyle v'=\exp{3t}$
    $\displaystyle u'=1$ $\displaystyle v=\frac{\exp{3t}}{3}$

    $\displaystyle \frac{1}{3}t^2\exp{3t}-\frac{2}{3}(\frac{1}{3}t\exp{3t}-\int\frac{1}{3}\exp{3t} dt)$
    $\displaystyle \frac{1}{3}t^2\exp{3t}-\frac{2}{3}(\frac{1}{3}t\exp{3t}-\frac{1}{3}*\frac{1}{3}\exp{3t})+C$

    To clean this up a bit...
    $\displaystyle \frac{1}{3}t^2\exp{3t}-\frac{2}{3}(\frac{1}{3}t\exp{3t}-\frac{1}{3}*\frac{1}{3}\exp{3t})+C$

    $\displaystyle \frac{1}{3}t^2\exp{3t}-\frac{2}{9}t\exp{3t}-\frac{1}{3}\exp{3t}+C=$

    $\displaystyle \frac{\exp{3t}}{3}(t^2-\frac{2}{9}t-1)$

    Did I do all of that alright? It was not very easy to check with differentiation.
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  2. #2
    MHF Contributor Calculus26's Avatar
    Mar 2009
    The last term should be 2/29 e^(3t)

    not that hard to differentiate
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