Given distribution function $F(x)$:
Determine the interval boundary coefficient $c$, the density function and the mathematical expectation.
In my opinion, to get $c$ you have to differentiate $F(x)$ by that getting $f(x)$. Then we would get an integrated $(4x)$ within (from $0$ to $c$) which should be equal to $1$.
As a result $4c = 1\Rightarrow c=1/4$. Can someone confirm if I am correct?
$F$ satisfes all the properties of a CDF as long as $c>0, 2c^{2}\le 1$, so we need $0<c\leq \frac 1 {\sqrt 2 }$.
This CDF need not have a density. It can have jump at $c$. If you assume that there is no jump at $c$ then we need $2c^{2}=1$ or $c=\frac 1 {\sqrt 2}$.
The density function in this case is $f(x)=4x, 0<x<c$ and the mean is $\int_0^{c}4x^{2}dx=\frac 4 3 c^{3}=\frac 2 {3 \sqrt 2}$.