# Question on definite integral and inflection points

1. May 13, 2016

### Sahil Kukreja

1. The problem statement, all variables and given/known data

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 $\frac {\pi}{3} , Then \int_0^1 P(x) \,dx = ?$

2. Relevant equations

No equations Relevant.

3. 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: May 13, 2016
2. May 13, 2016

### Staff: Mentor

I doubt that there is a quicker method. I wouldn't call your method especially lengthy -- you took the given information and worked out the consequences of that information.

3. May 13, 2016

### Sahil Kukreja

Ok. Thanks. :)