Find $\mathbb{E}(X^2)$ given the pdf of $X$.

Question

I have that a continuous random variable $X$ has the pdf $$ f_X(x) = \frac{3}{2}\sqrt{x} \text{ if }x\in[0,1] $$ $$ f_X(x) = 0 \text{ if }x\notin[0,1] $$ I wish to find $\mathbb{E}(X^2)$ via determining $X^2$'s pdf. Here's what I've tried $$ F_X(x) = \sqrt{x^3} = \mathbb{P}(X\leq x) \text{ for }x\in[0,1] $$ This implies that $$ F_{X^2}(x) = \mathbb{P}(X^2\leq x) = \mathbb{P}(X\leq \sqrt{x}) = F_X(\sqrt{x}) $$ Is this correct?

Answer 1

You just need to calculate

$$ \mathbb{E}[X^2] = \int_{-\infty}^{+\infty} {\rm d}x~ x^2 f_X(x) = \frac{3}{2}\int_0^1 {\rm d}x~ x^2 x^{1/2} = \frac{3}{7} $$