Evaluate $f(x)= {\int_{-\infty}^0\ }\frac{x^2}{e^{x^2}}\operatorname d\!x$ How to evaluate this integral:
$$f(x)= {\int\limits_{-\infty}^0\ }\frac{x^2}{e^{x^2}}\operatorname  d\!x$$
 A: As stated in the comments integrate by parts. Since $(e^{-{x^2}})'=-2xe^{-{x^2}}$ you have that $$\begin{align*}f(x)&=\int_{-\infty}^o x^2e^{-{x^2}}dx=-\frac{1}{2}\int_{-\infty}^0 x(e^{-{x^2}})'dx=-\frac{1}{2}\left[x(e^{-{x^2}})\right]_{-\infty}^0+\frac{1}{2}\int_{-\infty}^0 (x)'e^{-{x^2}}dx=\\&=0+\frac{1}{2}\int_{-\infty}^0 e^{-{x^2}}dx\end{align*}$$
Now the last can be evaluated in many ways. One probabilistic way is to observe that this is the normal distribution function with $\mu=0$ and $\sigma=\frac{1}{\sqrt{2}}$ that is to be integrated (missing a constant). The integral limits are the half of the normal distribution's domain (which is $(-\infty,+\infty)$ so the integral will be equal to $\frac{1}{2}$ instead of $1$ (due to the symmetry of the standard normal distribution to zero) after adding the constant. So by adding this constant you have that:
 $$\int_{-\infty}^0 e^{-{x^2}}dx=\frac{\sqrt{2\pi}}{\sqrt{2}}\int_{-\infty}^0\frac{1}{\sqrt{2\pi}\frac{1}{\sqrt{2}}}e^{-\frac{x^2}{2\frac{1}{\sqrt{2}^2}}}dx=\frac{\sqrt{2\pi}}{\sqrt{2}}\frac{1}{2}=\frac{\sqrt{\pi}}{2}$$
So, in sum $$f(x)=\int_{-\infty}^o x^2e^{-{x^2}}dx=\frac{1}{2}\frac{\sqrt{\pi}}{2}=\frac{\sqrt{\pi}}{4}$$
