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Integral Inequality

  • Thread starter diorific
  • Start date
  • #1
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Homework Statement


Prove


Homework Equations



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

Also look at atachment

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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Attachments

Answers and Replies

  • #2
uart
Science Advisor
2,776
9
Hint. Try replacing the sin(x) with a simpler "bounding" function.

BTW. Your second inequality is the wrong way around.
 
  • #3
19
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I can't get this one. What bounding function. I'm lost....
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
I can't get this one. What bounding function. I'm lost....
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]
 
  • #5
19
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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.
 
  • #6
296
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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.
 
  • #7
19
0
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.
thank you so much.

I finally managed to resolve this!!!
 

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