Let f:mathbb R→mathbb R be defined by f(x)=∫0^x t |t-1| dt.

Since $t|t-1|$ is continuous, the fundamental theorem gives $f'(x)=x|x-1|$ for all $x$, so (A) is true. Now $f'(x)<0$ for $x<0$ and $f'(x)>0$ for $0<x<1$, so $f'$ changes sign from negative to positive at $0$: a local minimum, and (B) is true. At $x=1$, $f'(1)=0$ but $f'(x)=x|x-1|\ge0$ on both sides, so $f'$ does not change sign and there is no extremum: (C) is false (although $f''$ does not exist there). For (D), $f(3)=\int_0^1t(1-t)\,dt+\int_1^3t(t-1)\,dt=\tfrac16+\left[\tfrac{t^3}{3}-\tfrac{t^2}{2}\right]_1^3=\tfrac16+\left(\tfrac92-\left(-\tfrac16\right)\right)=\tfrac16+\tfrac{14}{3}=\tfrac{29}{6}$, so (D) is true.