Variance formula in terms of the CDF for a continuous nonnegative random variable.

Here is a derivation of a formula for $E(X^2)$. The calculation is excessively informal. For "nice" density functions it is not difficult to justify. A similar calculation gives us $E(X)$. Then the variance is $E(X^2)-(E(X))^2$.

We find $\int_0^\infty x^2f(x)\,dx$ by integration by parts. Let $u=x^2$ and $dv=f(x)\,dx$. Then $du=2x\,dx$ and we can take $v=F(x)-1$. (Here we are being a little tricky.)

Then our integral is $$\left. x^2(1-F(x))\right|_0^\infty +\int_0^\infty 2x(1-F(x))\,dx.$$ The first part vanishes at both ends. So we find that

$$E(X^2)=\int_0^\infty 2x(1-F(x))\,dx.$$

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 $Pr[X \geq x] = 1 - F_X(x)$:

$$E[X^2] = 2\int_{x=0}^{\infty}x (1 - F_X(x)) dx $$