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Question on definite integral and inflection points

  1. May 13, 2016 #1
    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. jcsd
  3. May 13, 2016 #2

    Mark44

    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.
     
  4. May 13, 2016 #3
    Ok. Thanks. :)
     
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