# Thread: integral inequalities

1. ## integral inequalities

Hi
it would be really great if someone could help me with the following problem. I seem to have a great thick fog descending and cannot find my way out!!

Prove that:

(pi^3)/48 <= integral [0,pi/2] x^2/(2-sinx) dx <= (pi^3)/24

2. To get the upper bound, consider where the integrand reaches its maximum value - what would be the value of the integral if the integrand were constantly equal to this maximum value?

Can you use a similar idea for the lower bound?

3. Originally Posted by horse2152
Hi
it would be really great if someone could help me with the following problem. I seem to have a great thick fog descending and cannot find my way out!!

Prove that:

(pi^3)/48 <= integral [0,pi/2] x^2/(2-sinx) dx <= (pi^3)/24
On $\left[0,\frac{\pi}{2}\right] 0 \le \sin x \le 1$ so

$
-1 \le - \sin x \le 0
$

$
1 \le 2 - \sin x \le 2
$

$
\frac{1}{2} \le \frac{1}{2- \sin x} \le 1
$

$
\frac{x^2}{2} \le \frac{x^2}{2- \sin x} \le x^2
$

Now integrate over the interval.

4. ## integral inequalities

Hi
Thank you both very much, the fog is lifting!!