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Math Help - integration problem

  1. #1
    Senior Member Sambit's Avatar
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    Question integration problem

    \iint\limits_R \cos [\frac{\pi x^2}{2}]dxdy=?, where R is a region bounded by y=0, x=1, y=x.
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  2. #2
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    integration problem-img.jpg
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  3. #3
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    mr fantastic's Avatar
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    Quote Originally Posted by Sambit View Post
    \iint\limits_R \cos [\frac{\pi x^2}{2}]dxdy=?, where R is a region bounded by y=0, x=1, y=x.
    The first step is to reverse the order of integration (Fubini's Theorem).
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  4. #4
    Senior Member Sambit's Avatar
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    @ Pandevil1990,

    what you have done is absolutely right...thanks for that. but here the "[ ]" means box-function....so what will be the answer in that case?
    Last edited by Sambit; October 20th 2010 at 10:53 PM.
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  5. #5
    Senior Member Sambit's Avatar
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    ok...so here is what i have done when "[ ]" means box-function....

     0 < x < 1 ==> 0 < \frac{\pi x^2}{2} < \frac{\pi}{2} = 1.57 (approx.)
    ==> [\frac{\pi x^2}{2}]=0, if  \frac{\pi x^2}{2} < 1, ie,  x < \sqrt\frac{2}{\pi}

    so the integration becomes  \int_0^{\sqrt{\frac{2}{\pi}}}\int_0^x cos [\frac{\pi x^2}{2}]dy dx + \int_{\sqrt{\frac{2}{\pi}}}^1\int_0^x cos [\frac{\pi x^2}{2}]dy dx<br /> <br />
= \int_0^{\sqrt{\frac{2}{\pi}}}\int_0^x cos 0 dy dx + \int_{\sqrt{\frac{2}{\pi}}}^1\int_0^x cos 1 dy dx <br />
= \frac{1}{\pi} + cos 1(\frac{1}{2} - \frac{1}{\pi})

    this is all i could do. but this answer does not agree with any of the the given possible options.
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