Basic u substitution help

**Jones** · August 10th 2009, 04:10 AM

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?

**Soroban** · August 10th 2009, 05:01 AM

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 . . .

**Jones** · August 10th 2009, 06:21 AM

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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