The correct option is D [−1/4,2]
We have,
f(x)=∫x1(t3−3t2+2t)dt
⇒f′(x)=x3−3x2+2x=x(x−2)(x−1)
The changes in signs of f′(x) for different values of x are given in figure.
Clearly, f(x) is decreasing in [1,2] and increasing in [2,3].
∴Min.f(x)=f(2)=∫21t(t2−3t+2)dt=[t44−t3+t2]21=−14
and Max.f(x)=f(3)=∫31t(t2−3t+2)dt=[t44−t3+t2]31=2
Since f(x) is continuous on [1,3]. Therefore, range of f(x)=f[f(2),f(3)]=[−1/4,2].