Given the distirbution function as follows:
$F(x) = x^2$ for $ 0 \leq x < \frac{1}{2}$, $F(x) = x$ for $\frac{1}{2} \leq x < 1$, $F(x) = 1$ for $x > 1$ and $F(x) = 0$ for $x<0$.
Is it true that the density then equals $f(x) = 2x$ for $0 \leq x < \frac{1}{2}$ and $f(x) = 1$ for $ \frac{1}{2} < x <1$ and $0$ else?
Because the integral doesn't sum up to 1, i.e. $\int_{-\infty}^{\infty} f(x)dx=\frac{3}{4} \neq 1$, or did I make a mistake somewhere?
$$ F_X(x)= \begin{cases} 0&\text{ if }& x<0\\ x^2&\text{ if }& 0\le x\le \frac12\\ x&\text{ if }&\frac12<x\le 1\\ 1&\text{ if }& x>1 \end{cases}$$
The graph of this cdf looks like this
$F$ does have a discontinuity at $\frac12$. The limit is $\frac14$ from the left and it is $\frac12$ from the right. The point is that there is a jump of $\frac14$ at $\frac12$.
So, there is no density because there is a single point $\left(\frac12\right)$ whose probability is $\frac14$.
The distribution consists of an absolute continuous part, the density mentioned by the OP (its integral is $\frac34$) and an atom. The total probabilty is $1$.