integration by substitution xe^x^2

**Craka** · October 5th 2008, 09:44 PM

Question is to calculate the integral $\int\limits_0^2 {xe^{x^2 } dx} =$

This is what I have done, however looks to be incorrect.

I let $u = e^{x^2 }$

and so $\begin{array}{l} \frac{{du}}{{dx}} = 2xe^{x^2 } \\ du = 2xe^{x^2 } dx \\ \end{array}$

and $ \begin{array}{l} \frac{1}{2}\int {udu} \\ = \frac{1}{2}\left[ {\frac{{u^2 }}{2}} \right]_0^2 \\ = \frac{1}{2}\left[ {\frac{{(e^{x^2 } )^2 }}{2}} \right]_0^2 \\ = \frac{1}{2}\left[ {\left( {\frac{{(e^{2^2 } )^2 }}{2}} \right) - \left( {\frac{{(e^{0^2 } )^2 }}{2}} \right)} \right] \\ = \frac{1}{2}\left( {\frac{1}{2}e^{16} - \frac{1}{2}} \right) \\ \end{array} $
However answer is given as $ \frac{1}{2}\left( {e^4 - 1} \right) $

Can someone tell me the errors of my way. Thanks

**o_O** · October 5th 2008, 10:48 PM

You don't get $\int u du$ with that substitution.

Use: $u = x^2 \ \Rightarrow \ du = 2x \ dx \iff {\color{red}\frac{1}{2}du} = {\color{red}x \ dx}$

Note that: $u(x) = x^2 \ \Rightarrow u(2) = (2)^2 = 4 \ \text{and} \ u(0) = (0)^2 = 0$

So: $\int_{0}^{2} {\color{red}x}e^{x^2} {\color{red}dx} = \int_{u(0)}^{u(2)} e^{u} \left({\color{red}\frac{1}{2} du}\right) = \frac{1}{2} \int_{0}^{4} e^u \ du$

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