Another potential derivation for $E[X^2]$ uses the tail-probability equality.
$Y = X^2$ is a nonnegative random variable, so the tail-probability equality applies:
$$E[Y] = \int_{y=0}^{\infty} Pr[Y \geq y] dy $$
Replace $Y = X^2$ and $y$ with appropriate expressions of $X$ and $x$:
$$E[X^2] = \int_{x^2=0}^{\infty} Pr[X^2 \geq x^2] dx^2 $$
Because we are assuming that $X$ is nonnegative, $Pr[X^2 \geq x^2] = Pr[X \geq x]$. We also change our variable of integration from $x^2$ to $x$ (essentially $u$-substitution) by noting that $dx^2 = 2xdx$. Making these changes, we get
$$E[X^2] = 2\int_{x=0}^{\infty}x Pr[X \geq x] dx $$
If you want it in CDF form, make the replacement $Pr[X \geq x] = 1 - F_X(x)$:
$$E[X^2] = 2\int_{x=0}^{\infty}x (1 - F_X(x)) dx $$