Expected value of $f(x)=\frac38 x^2$ with range $[0,2]$

Question

I was going through some examples of finding the expected value of continuous random variables and i found an example as below.

Could anyone please explain how the working shown works as I'm really confused on how the power value is incremented.

Answer 1

First, for the expected value of continuous random variables is definded as: \begin{equation} E(X)=\int_{-\infty}^{\infty}xf_X(x)dx \end{equation} where $f_X(x)$ is the density of $X$ and $X$ is a continuous random variable. For the example, $\displaystyle f_X(x)=\frac{3}{8}x^2,x\in[0,2],f_X(x)=0,x \notin [0,2].\rightarrow E(X)=\int_{-\infty}^{\infty}xf_X(x)dx=\int_{0}^{2}x\frac{3}{8}x^2dx=\int_{0}^{2}\frac{3}{8}x^3dx=\frac{3}{8}\frac{x^4}{4}|_0^2$.

Could you understand better?

Answer 2

In general,

$$\int x^n \, dx = \frac{x^{n+1}}{n+1}+C, n \ne -1$$

A proof of the result can be found here.

In the question $n$ was $3$.

If the doubt is where does power $3$ comes from, we get it by multiplying $x$ with $f(x)$.