• Support PF! Buy your school textbooks, materials and every day products Here!

Question on definite integral and inflection points

  • #1

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:

Answers and Replies

  • #2
33,310
5,004

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??
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
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.
Ok. Thanks. :)
 

Related Threads on Question on definite integral and inflection points

Replies
1
Views
1K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
3
Views
17K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
9
Views
6K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
944
Top