Integral Inequality

diorific

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 Thumbnails

 

uart

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

BTW. Your second inequality is the wrong way around.

diorific

I can't get this one. What bounding function. I'm lost....

SammyS

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]

diorific

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.

Millennial

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.

diorific

thank you so much.

I finally managed to resolve this!!!