Let f (x) be a function defined by `f(x)=int_(0)^(x) t(t^(2)-3t+2)dt, 1 le, x le3`.
Then, the range of f(x) is

Options

A. [0,2]
B. [-1/4,4]
C. [-1/4,2]
D. none of these

Answer

C

Explanation

We have ,
`f(x)=underset(1)overset(x)int (t^(3)-3t^(2)+2t)dt`
`rArr f'(x) =x^(3)-3x^(2)+2x=x(x-2)(x-1)`
The changes in sings of f'(x) for different values of x are given in Fig. 11

Clearly, f(x) is decreasing in [1,2] and increasing in [2,3]
`:. "Min" f(x)=f(2)=underset(1)overset(2)intt(t^(2)-3t+2)dt=[(t^(4))/(4)=t^(3)+t^(2)]_(1)^(2)=-(1)/(4)` and
`:. "Max" f(x)=f(3)=underset(1)overset(2)intt(t^(2)-3t+2)dt=[(t^(4))/(4)=t^(3)+t^(2)]_(1)^(3)=2`
Since f(x) is continuous on [1,3]. Therefore, range of `f(x)=[f(2),f(3)]=[-1//4,2]`