For real x define F(x)=∫-1^1|t-x| dt.

For $|x|\le1$ split at $t=x$: $\int_{-1}^{x}(x-t)\,dt+\int_x^1(t-x)\,dt=\frac{(x+1)^2}2+\frac{(1-x)^2}2=x^2+1$. For $x>1$ every $t\in[-1,1]$ has $t<x$, giving $\int_{-1}^{1}(x-t)\,dt=2x$, and similarly $2|x|$ for $x<-1$: (A) correct. At $x=1$ the left derivative of $x^2+1$ is $2$ and the right derivative of $2x$ is $2$, and the same holds at $x=-1$, so $F$ is differentiable everywhere (B correct). But $F''(1^-)=2$ while $F''(1^+)=0$, so $F''(1)$ does not exist (C incorrect). Using evenness, $\int_{-2}^{2}F=\int_{-1}^{1}(x^2+1)\,dx+2\int_1^2 2x\,dx=\frac83+6=\frac{26}3$ (D correct). Answer: A, B and D.