\(\displaystyle \frac{2}{2\pi}\int^a_0xe^{-x^2}\,dx=-\frac{1}{2\pi}\int^a_0(-2x\,dx)e^{-x^2}=\) \(\displaystyle -\frac{1}{2\pi}\left[e^{-x^2}\right]^a_0=-\frac{1}{2\pi}\left[e^{-a^2}-e^0\right]\) \(\displaystyle =\frac{1-e^{-a^2}}{2\pi}\) , so there's a mistake in your notes.

Tonio

\(\displaystyle \frac{2}{2 \pi } \int_0^a xe^{-x^2} dx \)

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?

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.