If `P(x)` is a polynomial of the least degree that has a maximum equal to 6 at `x=1`, and a minimum equalto 2 at `x= 3`,then `int_0^1 P(x)dx` equals:

Options

A. `17/4`
B. `13/4`
C. `19/4`
D. `5/4`

Answer

C

Explanation

The polynomial function is differentiable everywhere. Therefore the points of extremum can only be the roots of the derivative. Further, the derivative of a polynomial is a polynomial. The polynomial of the least degree with roots `x=1` and `x=3` has the form `a(x-1)(x-3)`.
Hence, `P'(x)=a(x-1)(x-3)`
Since at `x=1`, we must have `P(1)=6`.
`P(x)=int_(1)^(x) P'(x)dx+6`
`=aint_(1)^(x)(x^(2)-4x+3)dx+6`
`=a((x^(3))/3-2x^(2)+3x-4/3)+6`
Also, `P(3)=2`. So `a=3`. Hence `P(x)=x^(3)-6x^(2) +9x+2`.
Thus, `int_(0)^(1) P(x)=1/4-2+9/2+2=19/4`