$\displaystyle \frac{2}{2 \pi } \int_0^a xe^{-x^2} dx $
Let $\displaystyle u=x^2 $ and $\displaystyle du = 2xdx $
$\displaystyle \frac{1}{2 \pi } \int_0^{a^2} e^{-u} du = \frac{1}{2 \pi } [ e^{-a^2} - 1 ] $
I think you dont carry through your negative on the third step?
$\displaystyle \int e^{-u}\,du=-e^{-u}$ , and not only $\displaystyle e^{-u}$ . You forgot the minus sign here. Tonio
But the above is NOT equal to
$\displaystyle \frac{2}{ \sqrt{ \pi }} [ e^{-a^2} - 1 ] $
Your professor has messed up the constants somewhere along the way.