# integration by substitution xe^x^2

• Oct 5th 2008, 09:44 PM
Craka
integration by substitution xe^x^2
Question is to calculate the integral $\displaystyle \int\limits_0^2 {xe^{x^2 } dx} =$

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

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

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

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

Can someone tell me the errors of my way. Thanks
• Oct 5th 2008, 10:48 PM
o_O
You don't get $\displaystyle \int u du$ with that substitution.

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

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

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