Let f(x)=∫0^x (t-1)(t-3)(t-2)^2 dt.

We have $f'(x)=(x-1)(x-3)(x-2)^2$. The sign is $+$ for $x<1$, $-$ on $(1,3)$ (the square factor does not change it at $2$), and $+$ for $x>3$. So $f$ has a local maximum at $1$, a local minimum at $3$, and $x=2$ is only a stationary point; (A) is true. Expanding, $f(x)=\frac{x^5}{5}-2x^4+\frac{23x^3}{3}-14x^2+12x$, giving $f(1)=\frac{58}{15}$ and $f(3)=\frac{54}{15}$, so $f(1)-f(3)=\frac{4}{15}$ and (B) is false. Next $f''(x)=2(x-2)(2x^2-8x+7)$ has three simple roots $2,\,2\pm\tfrac{1}{\sqrt2}$, each a sign change, so there are three inflection points and (C) is true. Since $f(0)=0$, $f$ is increasing up to a positive local maximum and stays positive afterwards, and the quintic's other roots are complex, so $f=0$ has only one real root and (D) is false.