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Math Help - Squeezing integrals

  1. #1
    Senior Member I-Think's Avatar
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    Squeezing integrals

    Use the fact that
    x-\frac{x^3}{6}\leq{sin{}x}\leq{x-\frac{x^3}{6}+\frac{x^5}{120}}

    to prove that \int_0^{\frac{1}{2}} sin(x^2)=\frac{223}{5376} with error no greater than \frac{1}{2703360}

    I attempted to use the inequality by replacing x with x^2 and integrating both sides to get bounds on sin (x^2), but I got nothing close to the actual answer

    Hints please, and thank you
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  2. #2
    MHF Contributor
    Prove It's Avatar
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    If \displaystyle X - \frac{X^3}{6} \leq \sin{X} \leq X - \frac{X^3}{6} + \frac{X^5}{120}

    then by letting \displaystyle X = x^2 we have

    \displaystyle x^2 - \frac{x^6}{6} \leq \sin{(x^2)} \leq x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}

    \displaystyle \int_0^{\frac{1}{2}}{x^2-\frac{x^6}{6}\,dx} \leq \int_0^{\frac{1}{2}}{\sin{(x^2)}\,dx} \leq \int_0^{\frac{1}{2}}{x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}\,dx}

    \displaystyle \left[\frac{x^3}{3} - \frac{x^7}{42}\right]_0^{\frac{1}{2}} \leq \int_0^{\frac{1}{2}}{\sin{(x^2)}\,dx} \leq \left[\frac{x^3}{3} - \frac{x^7}{42} + \frac{x^{11}}{1320}\right]_0^{\frac{1}{2}}

    \displaystyle \frac{223}{5376} \leq \int_0^{\frac{1}{2}}{\sin{(x^2)}} \leq \frac{784967}{18923520}.


    The difference between the endpoints is \displaystyle \frac{1}{2703360}.

    Therefore \displaystyle \int_0^{\frac{1}{2}}{\sin{(x^2)}\,dx} = \frac{223}{5376} with an error no greater than \displaystyle \frac{1}{2703360}.
    Last edited by Prove It; January 27th 2011 at 08:34 PM.
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  3. #3
    Senior Member I-Think's Avatar
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    I integrated wrong... Shame on me.

    Thanks
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