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Math Help - Simple integration problem.

  1. #1
    Junior Member
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    Simple integration problem.

    What is the cost of a swimming pool cover. the pool is bounded by y=x^2 and y=2 and it costs $2 / foot for the cover. One unit of the x and y axes are 50 feet.

    I have tried the following:
    Area under the graph y=x^2
    \int\limits_{ - \sqrt 2 }^{\sqrt 2 } {{x^2}dx}

    Which is
    \left. {\frac{{{x^3}}}{3}} \right|_{ - \sqrt 2 }^{\sqrt 2 }

    Which equals
    \frac{4\sqrt 2}{3}.

    Area of cover is rectangle less area under the graph.
    4\sqrt{2} -\frac{4\sqrt 2}{3}

    That multiplied by 50, and then by 2
    \frac{8\sqrt 2}{3}(50)(2)

    Equals \approx{377.123}

    My textbook has $18,856. I am not even in the same ball park.

    Thanks
    Regards
    Craig.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Simply you forgot that tha area is in foot square... that means you have to multiply by 50 again the result and you obtain 50 x 377.123 = 18,586... just as in your textbook ...

    Kind regards

    \chi \sigma
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