Let $F(x)$ be a cumulative distribution function where,
$F(x) = \begin{cases} 0, & \mbox{if } x\leq\mbox{-1} \\ \frac{1}{2}(x+1)^2, & \mbox{if } -1<\mbox{x}\leq0 \\ 1-\frac{1}{2}(x-1)^2, & \mbox{if } 0<\mbox{x}\leq1 \\ c, & \mbox{if } 1<\mbox{x} \end{cases}$
I found by definition $c=1$ with the limits at $-\infty$ and $+\infty$. The excersice tells you that the PDF function is continuous so i took the derivative of CDF and found
$f(x) = \begin{cases} 0, & \mbox{if } x\leq\mbox{-1} \\ x+1, & \mbox{if } -1<\mbox{x}\leq0 \\ 1-x, & \mbox{if } 0<\mbox{x}\leq1 \\ 0, & \mbox{if } 1<\mbox{x} \end{cases}$
Then it asks you to find the mean value $E[X]$. I took the integral and found $0$. I would appreciate it if someone wants to help me correct my solution if I am wrong.
The function $f$ is even
$$ f(-x) = f(x) $$
That means that $xf(x)$ is odd, and when you integrate it over an interval of the form $[-a,a]$ it will yield zero, in other words $\mathbb{E}[X] = 0$. In case you're interested in higher moments this is the result
$$ \begin{array}{c|c}\hline n & \mathbb{E}[X^n] \\\hline 0 & 1 \\ 1 & 0 \\ 2 & 1/6 \\ 3 & 0 \\ 4 & 1/15 \\ \hline \end{array} $$
$$ \mathbb{E}[X^n] = \frac{1+(-1)^n}{(1+n)(2+n)} $$