Determine constant in PDF

Question

How can I determine constant C to be $f(x)$ a PDF in this case?

$$f(x)=Cx^2 {1 \over \sqrt{2\pi\sigma^2}} exp(-{(x-\mu)^2 \over 2 \sigma^2})$$

Answer 1

If you integrate over $x$, you obtain $E(CX^2)$ where $X$ follows a normal distribution with parameters $\mu$ and $\sigma^2$. Since $\sigma^2 = E(X^2) - \mu^2$, we obtain $E(CX^2)=C E(X^2) = C(\sigma^2 + \mu^2)$. Since this has to equal 1, we obtain $C=1/(\sigma^2 + \mu^2)$.

Answer 2

Hint. Integrate over all $x$ and equal it to 1.

Answer 3

Hint. One is looking for $C$ such that $$ \int_{-\infty}^\infty Cx^2 {1 \over \sqrt{2\pi\sigma^2}} exp(-{(x-\mu)^2 \over 2 \sigma^2})dx=1 $$ make the change of variable $u=x-\mu$, expand $(x+\mu)^2$ then use the gaussian results, $$ \int_{-\infty}^\infty {1 \over \sqrt{2\pi\sigma^2}} exp(-{x^2 \over 2 \sigma^2})\:dx=1 $$ $$ \int_{-\infty}^\infty {1 \over \sqrt{2\pi\sigma^2}} x^2 exp(-{x^2 \over 2 \sigma^2})\:dx=\sigma^2. $$