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Math Help - Substitution then integration by parts

  1. #1
    Senior Member mollymcf2009's Avatar
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    Substitution then integration by parts

    Hey y'all,

    The instructions for this problem are:

    First make a substitution then evaluate the integral using integration by parts.

    Here is what I did, can someone let me know if I'm on the right track?

    \int 2t^{9}e^{-t^5} dt

    2 \int t^{9}e^{-t^5} dt

    u = -t^5
    du = -5t^4 dt

    -\frac{1}{5} du = t^4 dt

    -\frac{1}{5} \cdot 2 \int -u \cdot e^u du

    * It that what I'm supposed to do for the substitution part?
    * If this is correct, do I plug my original numbers back in and then do my integration by parts the usual way?

    Thanks!!
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  2. #2
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    Hello, mollymcf2009!

    First make a substitution.
    then evaluate the integral using integration by parts.

    . . \int 2t^9e^{-t^5}\,dt

    Here is what I did, can someone let me know if I'm on the right track?

    \int 2t^9e^{-t^5}\,dt \;=\;2 \int t^9e^{-t^5}\,dt

    . . w \,=\, \text{-}t^5 \quad\Rightarrow\quad dw \,=\,\text{-}5t^4 dt \quad\Rightarrow\quad t^4 dt\,=\,\text{-}\frac{dw}{5}

    -\frac{1}{5} \cdot 2 \int -w \cdot e^w dw . . . . correct, but tidy up!

    You have: . \frac{2}{5}\int w\!\cdot\!e^w\,dw


    By parts: . \begin{array}{ccccccc}u &=& w & & dv &=& e^w\,dw \\ du &=& dw & & v &=& e^w \end{array}


    Then we have: . \frac{2}{5}\left[w\!\cdot\!e^w - \int e^w\,dw\right] . . . etc.

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