Let f(x)=∫ e^x(x-1)(x-2) d x, then f(x) decreases in the interval

We have,

$f(x)=\int e^x(x-1)(x-2) d x \Rightarrow f'(x)=e^x(x-1)(x-2)$

For f(x) to be decreasing, we must have

$f'(x)<0$

$\Rightarrow e^x(x-1)(x-2)<0 \Rightarrow(x-1)(x-2)<0 \Rightarrow x \in(1,2)$