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Integral Inequality

  1. Aug 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove


    2. Relevant equations

    (π^3)/12≤∫_0^(π/2)▒〖(4x^2)/(2-sinx) dx≥(π^3)/6〗

    Also look at atachment

    3. The attempt at a solution

    I can't get round this one, since when you substitute x by 0 is always 0 and I don't know how to get ∏^3/12
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Aug 1, 2012 #2

    uart

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    Hint. Try replacing the sin(x) with a simpler "bounding" function.

    BTW. Your second inequality is the wrong way around.
     
  4. Aug 21, 2012 #3
    I can't get this one. What bounding function. I'm lost....
     
  5. Aug 21, 2012 #4

    SammyS

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    What are the minimum and maximum values of [itex]\displaystyle \frac{1}{2-\sin(x)}[/itex] for [itex]\displaystyle 0\le x\le \frac{\pi}{2}\ ?[/itex]
     
  6. Aug 22, 2012 #5
    Ok, that is

    [itex]\displaystyle \frac{1}{2}[/itex] ≤ [itex]\frac{1}{2-\sin(x)}[/itex] ≤ 1

    But then 0 ≤ 4x2 ≤ π2

    So then

    0 ≤ [itex]\displaystyle \frac{4x^2}{2-sinx}[/itex] ≤ π2

    I know this might be wrong, but I don't really know how to continue.
     
  7. Aug 22, 2012 #6
    Don't take the maximum (or the minimum) of the polynomial. By the mean value theorem, there exists C such that
    [tex]\int_{0}^{\pi/2}\frac{4x^2}{2-\sin(x)}\,dx=\frac{1}{2-\sin(c)}\int_{0}^{\pi/2}4x^2\,dx[/tex]
    and [itex]0\leq c\leq \pi/2[/itex].
    Take the integral and maximize/minimize the factor by adjusting C appropriately.
     
  8. Aug 22, 2012 #7
    thank you so much.

    I finally managed to resolve this!!!
     
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