Question:

<div>$f(x) = \int_{0}^{x} (t-1)(t-2)^2 \, dt$<br/>$f'(x) = (x-1)(x-2)^2$<br/>From the sign scheme of $f'(x)$, it is clear that $f'(x)$ changes sign from negative to positive at $x=1$.<br/>$\Rightarrow x = 1$ is the point of minimum.<br/>$f(1) = \int_{0}^{1} (t^3 - 5t^2 + 8t - 4) \, dt = \frac{1}{4} - \frac{5}{3} + 4 - 4 = -\frac{17}{12}$.<br/>Hence $(1, -\frac{17}{12})$ is a point of minimum.</div>