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Math Help - Basic u substitution help

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

    Hello,

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

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

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

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

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

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


    \text{Substitute: }\;\int \underbrace{e^{x^{8.5}}}_{\downarrow}\underbrace{\  left(x^{16}dx\right)}_{\downarrow}
    . . . . . . . . . \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!
    \frac{2}{17}\int_{0}^{1}ue^{u} = ue^{u} -\int e^{u}
     = \frac{2}{17}*1*e^1 - e^1 - \frac{2}{17}*0*1-1 = \frac{2}{17}
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