- #1

- 65

- 15

## Homework Statement

Let P(x) be a polynomial of least degree whose graph has three points of inflection (-1,-1) ; (1,1) and a point with abscissa 0 at which the curve is inclined to the axis of abscissa at an angle of ## \frac {\pi}{3} , Then \int_0^1 P(x) \,dx = ? ##

## Homework Equations

No equations Relevant.

## The Attempt at a Solution

Since P(x) has three inflection points at x=-1,x=0,x=1

=>## P''(x)= a(x-1)(x)(x+1) = a(x^3 - x) ##

=> ## P'(x) = a( \frac {x^4}{4} - \frac {x^2}{2}) + c1 ##

## now~ it~ is ~given ~that~ P'(0) = \sqrt 3 ##

## => c1= \sqrt 3 ##

=> ## P(x)= a( \frac {x^5}{20} - \frac {x^3}{6} ) + \sqrt 3 x +c2 ##

now P(-1) = -1 and P(1)= 1

solving for a and c2

## P(x)= \frac {60(\sqrt 3 - 1)}{7} ( \frac {x^5}{20} - \frac {x^3}{6} ) + \sqrt 3 x ##

## \int_0^1 P(x)dx = \frac {3 \sqrt 3 + 4 }{14} ##

**The question is solved but it is a very lengthy method, can there be a quicker method??**
Last edited: