Is the cumulative distribution function evaluated on the expected value $\frac{1}{2}$ Let $X$ be a random variable with density function $f(x)$, so
$$
\int_{-\infty}^{+\infty}f(x)dx=1
$$
and
$$
E(X)=\int_{-\infty}^{+\infty}{xf(x)dx}.
$$
Then, my question is if
$$
\int_{-\infty}^{E(X)}{f(x)dx}=\frac{1}{2} ?
$$
and if it is true, how can be proved? Otherwise, which would be a contra-example of this?
Thanks!
 A: Since you're writing what look to be Riemann integrals, I presume we're keeping to the case where $X$ is a continuous random variable.
If $m$ is a value for which $\int_{-\infty}^m f_X(x)\, dx =\frac12$ then $m$ is a median of $X$ (if the density is $>0$ in a neighborhood of $m$, then $m$ will be unique; the median). 
Since here the upper limit of the integral is given as $m=E(X)$, this amounts - as Nick Cox pointed out in comments - to saying "is there a distribution for which the mean differs from the median", and the obvious thing to do when searching for a counterexample is to try some distributions that are skew*. 
Here's an obvious example: the exponential distribution, which has its median at $\ln 2$ times the mean (i.e. about 70% of the mean). 
You might like to also consider the density 
$$f(x) = \cases{ 
       \begin{array}{lr}
        0, & \text{for } x\leq 0\\
        2x, & \text{for } 0< x\leq 1\\
        0, & \text{for } x> 1
        \end{array}}
         $$
as a simple one to try for yourself (since the integrations are particularly simple).

*(not every asymmetric distribution has mean different from median, however)
