# Thread: Simple integration problem.

1. ## 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.

2. 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

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