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