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

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

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    Since $f$ is a PDF, $\int f=1$ – 2017-01-01

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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 $C=1/(\sigma^2 + \mu^2)$.

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    I don't see why your answer is downvoted. We are not supposed to prove things from scratch. (+1) – 2017-01-01
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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, $$ \int_{-\infty}^\infty {1 \over \sqrt{2\pi\sigma^2}} exp(-{x^2 \over 2 \sigma^2})\:dx=1 $$ $$ \int_{-\infty}^\infty {1 \over \sqrt{2\pi\sigma^2}} x^2 exp(-{x^2 \over 2 \sigma^2})\:dx=\sigma^2. $$

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    Yes, they are correct now. – 2017-01-01
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Hint. Integrate over all $x$ and equal it to 1.