How to find $E[\sqrt{X}]$ given only a distribution function? I'm given a continuous distribution function $F(x)$ and I should assume that $X$ is some random variable. My goal is to find the expected value of $\sqrt{X}$.
I'm very bad at probability but this is my attempt:
If I derivate the distribution function $F(x)$ I get the density function $f(x)$. Then $E[X] = \int_{-\infty}^{\infty}f(x)\cdot xdx$ 
But what about $\sqrt{X}$?
I'm asking this question in general terms so that I can calculate everything on my on. Though, any help would be highly appreciated!
 A: If $U$ is a nonnegative random variable with CDF $H$ then:$$\mathbb{E}U=\int_{0}^{\infty}\left(1-H\left(u\right)\right)du$$
Proof:
If function $g$ is prescribed by $\left(u,v\right)\mapsto1$ if $v>u$
and $\left(u,v\right)\mapsto0$ otherwise then:
$$\int_{0}^{\infty}\left(1-H\left(u\right)\right)du=\int_{0}^{\infty}\int_{0}^{\infty}g\left(u,v\right)dH\left(v\right)du=\int_{0}^{\infty}\int_{0}^{\infty}g\left(u,v\right)dudH\left(v\right)=\int_{0}^{\infty}vdH\left(v\right)=\mathbb{E}U$$

$\sqrt{X}$ only makes sense if we have $X\geq0$ so let us assume
that that is the case here. 
If $F$ denotes the CDF of $X$ and $G$
the CDF of $\sqrt{X}$ then $G\left(x\right)=F\left(x^{2}\right)$.
$\sqrt{X}$ is a nonnegative rv and this implies that we can find $\mathbb{E}\sqrt{X}$ by means
of the mentioned rule: $$\mathbb{E}\sqrt{X}=\int_{0}^{\infty}\left(1-G\left(x\right)\right)dx=\int_{0}^{\infty}\left(1-F\left(x^{2}\right)\right)dx$$
This allows you to find the expectation without differentiating $F$. You have in fact no guarantee that $F$ is differentiable. So this meets your needs. No PDF is involved. If a CDF is continuous then you are not allowed yet to conclude that it has a PDF. For that it must be 'absolutely continuous'.
A: We can use the law of the unconscious statistician
 to find $\operatorname E\sqrt X$. We have that
$$
\operatorname E\sqrt X=\int_0^\infty\sqrt xf(x)\mathrm dx.
$$
A: $\displaystyle E[\sqrt{X}] = \int_{-\infty}^{\infty}f(x)\cdot \sqrt{x}dx$. 
As Stefan Hansen comments, there is an issue if the is a non-zero probability of $X \lt 0$.
The law of the unconscious statistician
 gives $\displaystyle E[g(X)] = \int f(x)g(x)dx$.
