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How do I integrate the double integral of the form $|x^2-y|$ with the boundaries $-1\leq x\leq 1$ and $-1\leq y\leq 1$?

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Separate the integral into two regions, one with $x^2>y$ and one with $x^2 \leq y$. –  copper.hat Feb 11 '13 at 21:28

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You can break it into two integrals, one inside the parabola where $y \gt x^2$ and one outside where $y \lt x^2$ So $$\int_{-1}^1\int_{-1}^1 |x^2-y|\; dy \; dx=\\ \int_{-1}^1\int_{-1}^{x^2}x^2-y \;dy \; dx+\int_{-1}^1\int_{x^2}^{1}y-x^2 \;dy \; dx$$

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Sweet! Thanks for that! –  user61881 Feb 11 '13 at 22:25

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