Prove that the area of the region on plane bounded by $\max (|x|, |y|)\leq 1$ and $xy \leq \frac{1}{2}$ is $3+\ln(2)$
How to plot $\max (|x|, |y|)\leq 1$. Could someone help me with this?
Area bounded by $\max (|x|, |y|)\leq 1$ and $xy \leq \frac{1}{2}$
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calculus
integration
definite-integrals
area
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0In general, $\max(a,b)\le L$ if and only if $a\le L$ and $b\le L$. – 2017-01-30
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0Can you see that $max (|x|, |y|)\leq 1$ is the square centered in $0$ with sides of lenght $2$ – 2017-01-30
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0[Here's a graph - the blue is $\max(|x|,|y|)\le1$, and the red is $xy\le\frac12$.](https://i.imgur.com/eRGYYME.jpg) – 2017-01-30
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0@Tryss Could you elaborate how side would be 2 units? – 2017-01-30
1 Answers
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so we can do the integral to calculate the region between $$x\in(\frac{1}{2},1) \quad y\in(\frac{1}{2},\frac{1}{2x})$$ that it is the same region between $$x\in(-1,-\frac{1}{2})\quad y\in(\frac{1}{2x},-\frac{1}{2})$$ and the last region have measure $3,5$ so: $$2\int_\frac{1}{2}^1\int_\frac{1}{2}^{\frac{1}{2x}}dydx=\int_\frac{1}{2}^1\frac{1}{x}-1dx=(\ln[x]-x)_\frac{1}{2}^1=\ln[2]-\frac{1}{2}$$ so we have finish