Calculate $f\left(x\right)=cxe^{-\frac{x^{2}}{2}}$ with $x>0$ (a) find c, (b)the mean of X, (c) the variance of X
(a)$$E\left[X\right]=\int_{0}^{+\infty}x\cdot cxe^{-\frac{x^{2}}{2}}dx=c\int_{0}^{+\infty}x^{2}\cdot e^{-\frac{x^{2}}{2}}dx$$ Integrating by parts $$\begin{array}{cc} u=x^{2} & du=2xdx\\ dv=e^{-\frac{x^{2}}{2}}dx & v=-e^{-\frac{x^{2}}{2}} \end{array}$$ $$E\left[X\right]=c\left(\left(x^{2}\right)\left(e^{-\frac{x^{2}}{2}}dx\right)+2\int_{0}^{\infty}xe^{-\frac{x^{2}}{2}}dx\right)$$ Integrating by parts again $$\begin{array}{cc} u=x & du=dx\\ dv=e^{-\frac{x^{2}}{2}}dx & v=-e^{-\frac{x^{2}}{2}} \end{array}$$ $$E\left[X\right]=c\left(\left(x^{2}\right)\left(e^{-\frac{x^{2}}{2}}dx\right)+2\left[\left(x\right)\left(e^{-\frac{x^{2}}{2}}dx\right)+\int_{0}^{\infty}e^{-\frac{x^{2}}{2}}dx\right]\right)$$
Still not sure if this is the right procedure. (b) Missing (c) Missing
(a)
$f(x)$ is a valid pdf.
Therefore $$\int_0^\infty f(x) dx = c\int_0^\infty xe^{-\frac{x^2}{2}} dx = 1$$
Since $$\int_0^\infty xe^{-\frac{x^2}{2}} dx = 1$$ therefore $c = 1$
(b)
$$E(X) = \int_0^\infty xf(x) dx = \int_0^\infty x^2 e^{-\frac{x^2}{2}} dx = \sqrt{\frac{\pi}{2}}$$
(c) $$Var(X) = E(X^2) - (E(X))^2$$
$$E(X^2) = \int_0^\infty x^2 f(x) dx = \int_0^\infty x^3 e^{-\frac{x^2}{2}} dx = 2$$
Hence $$Var(X) = 2 - \frac{\pi}{2} = \frac{4 - \pi}{2}$$