Question

Find the point of extremum of $f\left(x\right)={\int }_0^x(t-2{)}^2(t-1)dt$

Answer 1

$f\left(x\right)={\int }_0^x(t-2{)}^2(t-1)dt$

Differentiating both sides

$f^{\prime }\left(x\right)=(x-2{)}^2(x-1)$

[using Liebinitz Rule)

$f^{\prime }\left(x\right)=0$

$\Rightarrow x=2,1$

$\Rightarrow 1$ is point of extremum.