Let f(x) be a function defined by

$f(x)=\int\limits_1^x t\left(t^2-3 t+2\right) d t, x \in[1,3]$

Then the range of f(x), is

$[-1 / 4,2]$

We have,

$f(x)=\int\limits_1^x t\left(t^2-3 t+2\right) d t, x \in[1,3]$

$\Rightarrow f(x)=x\left(x^2-3 x+2\right)$

$\Rightarrow f(x)=x(x-1)(x-2)$

The changes in signs of f(x) are shown in Figure.

We observe that

$f'(x)<0$ for all $x \in(1,2)$ and, $f^{\prime}(x)>0$ for all $x \in(2,3)$

$\Rightarrow f(x)$ is decreasing on $[1,2]$ and increasing on $[2,3]$

∴ Minimum value of f(x)

$=f(2)=\int\limits_1^2 t\left(t^2-3 t+2\right) d t=\left[\frac{t^4}{4}-t^3+t^2\right]_1^2=-\frac{1}{4}$

and, $f(3)=\int\limits_1^3 t\left(t^2-3 t+2\right) d t=\left[\frac{t^4}{4}-t^3+t^2\right]_1^3=2$

∴ Maximum value of f(x) = 2

Since f(x) is continuous on $[1,3]$. So, it attains every value between its minimum and maximum values.

Hence, range (f) = [-1/4, 2]