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Thread: 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 $\displaystyle y=x^2$ and $\displaystyle y=2$ 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 $\displaystyle y=x^2$
    $\displaystyle \int\limits_{ - \sqrt 2 }^{\sqrt 2 } {{x^2}dx}$

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

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

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

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

    Equals $\displaystyle \approx{377.123}$

    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

    $\displaystyle \chi$ $\displaystyle \sigma$
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