Let f(x)=∫0^x^2(t-1)(t-4) dt for real x.

Here $f(x)=\frac{x^6}{3}-\frac{5x^4}{2}+4x^2$ and $f'(x)=2x(x^2-1)(x^2-4)$, which changes sign at $x=0,\pm1,\pm2$. Signs: negative for $x<-2$, positive on $(-2,-1)$, negative on $(-1,0)$, positive on $(0,1)$, negative on $(1,2)$, positive for $x>2$. So minima occur at $-2,0,2$ and maxima at $\pm1$: (A) is true. $f(\pm2)=-\frac83$ and $f\to\infty$, and the other local minimum $f(0)=0$ is larger, so the global minimum is $-\frac83$: (B) is true. $f(x)=x^2(2x^4-15x^2+24)/6$ has the double root $0$ and four more roots from $2u^2-15u+24=0$ ($u=x^2>0$, discriminant $33>0$, both roots positive), so there are $5$ distinct real zeros, not $3$: (C) is false. On $(1,2)$ we found $f'<0$, so $f$ is decreasing there: (D) is false.