How to integrate $4x^3e^{-2x}$ using Gamma Function

Question

Let $X$ be a random variable with PDF $f$ given by:

$$f(x)=\begin{cases}4xe^{-2x} & 0\le x\lt\infty,\\ 0 & \mathrm{otherwise}\end{cases}$$

Find $\mathrm E(X)$ and $\mathrm{Var}(X)$.

I can find $\mathrm E(X)$ using the Gamma function but I'm not sure on how to use it to find $\mathrm Var(X)$ because it's $4x^3$ instead of $4x^2$ because I make $u = 2x$

Edit : I asked this question before but with normal solution like parts by part and now I just found a new way which is Gamma function and I'm not sure on how to use it on $\mathrm Var(X)$ so I asked about this question again

Answer 1

Note that

$$\int_0^\infty x^ne^{-x}\mathrm dx=\Gamma(n+1)$$

Here,

$$E(X)=\int_0^\infty x f(x)\mathrm dx=4\int_0^\infty x^2 e^{-2x}\mathrm dx\\=4\int_0^\infty\frac{1}{4}u^2e^{-u}\frac{1}{2}\mathrm du=\frac{1}{2}\int_0^\infty u^2e^{-u}\mathrm du=\frac12\Gamma(3)$$

$$E(X^2)=\int_0^\infty x^2 f(x)\mathrm dx=4\int_0^\infty x^3 e^{-2x}\mathrm dx\\=4\int_0^\infty\frac{1}{8}u^3e^{-u}\frac{1}{2}\mathrm du=\frac{1}{4}\int_0^\infty u^3e^{-u}\mathrm du=\frac14\Gamma(4)$$

Can you take it from here?

Note also that $\Gamma(n+1)=n!$, so you can further simplify.

Answer 2

This is the one I just think of

$$E(X^2) = \int_0^\infty4x^3e^{-2x}dx = \frac{1}{2}\int_0^\infty8x^3e^{-2x}dx$$
We make the substitution $u = 2x$: $$ E(X^2) = \frac{1}{2}\int_0^\infty u^3e^{-u}\frac{du}{2} = \frac{1}{4}\Gamma(3+1) = \frac{1}{4}*3! = 1.5 $$