Why is the expectation of $Y=X^2$ more than $\textbf{E}[X^2]$?

Question

Given $X$ is an RV with pdf $f(x)=2x$ where $0<x<1$. Then the $$\textbf{E}[X^2]=\int_0^12x^3 = \frac{1}{2}$$ Given $Y=X^2$, then the pdf of $Y$ is $g(x)=f(x)f(x)=4x^2$ giving us the expectation, $$\textbf{E}[Y] = \int_0^1 4x^3dx = 1.$$

Why is the expectation of $Y=X^2$ more than $\textbf{E}[X^2]$?

Because it seems like $\textbf{E}[Y]=\textbf{E}[X^2]$.

Answer 1

Hint: The pdf of $Y$ is not given by $f(x)\cdot f(x)$, as is evidenced by the fact that integrating that function from $0$ to $1$ does not give a total probability of $1$.