Let `P(x)` be a polynomial of least degree whose graph has three points of inflection `(-1,-1),(1,1)` and a point with abscrissa `0` at which the curve is inclined to the axis of abscissa at an angle of `60^0dot` Then find the value of `int_0^1P(x)dx`

Explanation

Polynomial `P(x)` has point of inflection at `x=-1,1,` and `0`.
`:. P"(x)=ax(x-1)(x+1)=a(x^(3)-x)`
Also, given that `P'(0)=tan 60^(@)=sqrt(3)`
`P'(x)=int_(0)^(x)P''(x)dx+sqrt(3)=a((x^(4))/4-(x^(2))/2)+sqrt(3)`
`P(1)=1`
`:.P(x)=int_(1)^(x)P'(x)dx+1`
`=a((x^(5))/20-(x^(3))/6+7/60)+sqrt(3)(x-1)+1`
`P(-1)=-1`
`:.a=(60(sqrt(3)-1))/7`
`:.int_(0)^(1)P(x)dx=int_(0)^(1)dx=int_(0)^(1)[(sqrt(3)-1)/7(3x^(5)-10x^(3))+xsqrt(3)]dx`
`=[(sqrt(3)-1)/7((x^(6))/2-5/2x^(4))+(sqrt(3)x^(2))/2]_(0)^(1)`
`=(sqrt(3)-1)/7(1/2-5/2)+(sqrt(3))/2`
`=(3sqrt(3))/14+2/7`