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Math Help - integral inequalities

  1. #1
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    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
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    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?
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  3. #3
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    Quote Originally Posted by horse2152 View Post
    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

     <br />
-1 \le - \sin x \le 0<br />

     <br />
1 \le 2 - \sin x \le 2<br />

     <br />
\frac{1}{2} \le \frac{1}{2- \sin x} \le 1<br />

     <br />
\frac{x^2}{2} \le \frac{x^2}{2- \sin x} \le x^2<br />
    Now integrate over the interval.
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  4. #4
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    integral inequalities

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