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Thread: Basic u substitution help

  1. #1
    Member Jones's Avatar
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    Basic u substitution help

    Hello,

    I need to evaluate $\displaystyle \int_{0}^{1}x^{16}\times e^{x^{8.5}} $

    So i figured i let $\displaystyle \left \{\begin{array}{rcll}u = x^{8.5} & \\
    \text{then}~u^2 = x^{17} & \\
    du^2 = 17x^{16}\end{array} \right.$

    $\displaystyle \frac{1}{17}\int_{0}^{1}du^2\times e^u$

    Hmm, do i substitute my limits at this point?
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  2. #2
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    Hello, Jones!

    Your substitution is incomplete . . .


    Evaluate: .$\displaystyle \int_{0}^{1}x^{16} e^{x^{8.5}} $

    Let: .$\displaystyle u \:=\:x^{8.5}$

    Then: .$\displaystyle u^2 \:=\:x^{17}$

    And: .$\displaystyle 2u\,du \:=\:17x^{16}dx \quad\Rightarrow\quad x^{16}dx \:=\:\tfrac{2}{17}u\,du$


    $\displaystyle \text{Substitute: }\;\int \underbrace{e^{x^{8.5}}}_{\downarrow}\underbrace{\ left(x^{16}dx\right)}_{\downarrow} $
    . . . . . . . . .$\displaystyle \int e^u\left(\tfrac{2}{17}u\,du\right) \;=\;\tfrac{2}{17}\int ue^u\,du$


    Now integrate by parts . . .

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  3. #3
    Member Jones's Avatar
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    Thanks!
    $\displaystyle \frac{2}{17}\int_{0}^{1}ue^{u} = ue^{u} -\int e^{u}$
    $\displaystyle = \frac{2}{17}*1*e^1 - e^1 - \frac{2}{17}*0*1-1 = \frac{2}{17}$
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