I've got the integral $\int_B|\sin(x)-y|dx dy$ with $B=\{0\leq x\leq \pi, 0\leq y\leq 1\}$ My problem is how handle the abs in the integral. I'd probably go with two integrals (one when the abs is negative, one if it's positive), but I suppose there's a nicer solution. Note: I know how to deal with parametric integrals, the question is how to integrate the absolute value.
Parametric integral with abs()
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calculus
multivariable-calculus
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3First compute the integral over $0\le y\le1$ for $x$ fixed. – 2011-06-23
1 Answers
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$ \int_0^\pi \int_0^1 |\sin x-y| dy dx= $ $ \int_0^\pi\left( \int_0^{\sin x} (\sin x-y)dy +\int_{\sin x}^1 (y-\sin x)dy\right)dx=$ $ \int_0^\pi \left( \sin^2 x- \frac{\sin^2 x}{2} + \frac{1}{2}-\sin x-\frac{\sin^2 x}{2}+\sin^2 x \right)dx=$ $ \int_0^\pi \left(\frac{1}{2}+\sin^2 x-\sin x\right)dx $
From here on it's a simple one dimensional integration.