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.

This is basically proving the fact that the variance of a zero mean Gaussian random variable X with probability density function of $f_X(x)=\frac{1}{s\sqrt{2\pi}}e^{-\frac{x^2}{2s^2}}$ is equal to $s^2$:
$$[\text{Var}X]=\mathbb{E}[X^2]-\mathbb{E}^2[X]=\mathbb{E}[X^2]=\int_{-\infty}^\infty \frac{1}{s\sqrt{2\pi}}x^2 e^{-\frac{x^2}{2s^2}}dx$$
Noting that the integrand is an even function of $x$ and using integral by parts, we have:
$$ [\text{Var}X]=\sqrt{\frac{2}{\pi}}\frac{1}{s} \int_{0}^\infty \underbrace{x}_{f(x)}.\underbrace{x e^{-\frac{x^2}{2s^2}}}_{g'(x)}dx$$
$$=\sqrt{\frac{2}{\pi}}\frac{1}{s}\bigg[\underbrace{x}_{f(x)}.\underbrace{(-s^2)e^{-\frac{x^2}{2s^2}}}_{g(x)}|_0 ^\infty- \int_{0}^\infty \underbrace{1}_{f'(x)}.\underbrace{(-s^2)e^{-\frac{x^2}{2s^2}}}_{g(x)}dx\bigg]$$
$$=\sqrt{\frac{2}{\pi}}\frac{1}{s}\bigg[0+s^2 \int_{0}^\infty e^{-\frac{x^2}{2s^2}}dx\bigg]=\sqrt{\frac{2}{\pi}}s\int_{0}^\infty e^{-\frac{x^2}{2s^2}}dx$$
$$=\frac{2}{\sqrt{2\pi}s}s^2\int_{0}^\infty e^{-\frac{x^2}{2s^2}}dx$$
Again the integrand is an even function so we can modify the range by incorporating the factor of $2$:
$$=s^2\underbrace{\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}s}e^{-\frac{x^2}{2s^2}}dx}_{=1 \text{, since it's the integran of Gaussian pdf over all } x}=s^2 \ \square$$
