How to calculate this integral. Var(x)=$sigma^{2}$ I have to calculate this  . I must basically prove that $Var(x)$ $=$ $s^{2}$ in terms of probability/statistics and this is $E^{2}$ (x) if it helps. Any help is appreciated. I think that it has something to do with the parity of the function. Thank you.

Your integral is
$$ \dfrac{1}{s \sqrt{2\pi}} \int_{-\infty}^\infty x^2 e^{-(x-m)^2/(2s^2)}\; dx$$
Change variables $x = m + st$:
$$ \dfrac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty (st+m)^2 e^{-t^2/2}\; dt$$
Expand the square.  
$\int_{-\infty}^\infty t e^{-t^2/2}\; dt = 0$ by symmetry. 
$\int_{-\infty}^\infty e^{-t^2/2}\; dt$ you probably already know.
$\int_{-\infty}^\infty t^2 e^{-t^2/2}\; dt$ can be done by parts: $u = t$, $dv = t e^{-t^2/2}\; dt$.
