# Basic u substitution help

• Aug 10th 2009, 04:10 AM
Jones
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?
• Aug 10th 2009, 05:01 AM
Soroban
Hello, Jones!

Your substitution is incomplete . . .

Quote:

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 . . .

• Aug 10th 2009, 06:21 AM
Jones
Thanks!
$\displaystyle \frac{2}{17}\int_{0}^{1}ue^{u} = ue^{u} -\int e^{u}$
$\displaystyle = \frac{2}{17}*1*e^1 - e^1 - \frac{2}{17}*0*1-1 = \frac{2}{17}$