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I could not compute this integral. How can we get rid of the minimum function? $ \int_{x=0}^{t}\int_{y=0}^{t}\min(x,y)dydx $

Thanks a lot.

3 Answers 3

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Prove that $\min \{x,y\}=\frac{x+y-\vert x-y\vert}{2}.$

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    @DougSpoonwood,actually I mean "Prove that... and use it." I think that don't make sense to use things fallen from the sky.2012-02-05
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To evaluate this integral, observe that $\min\{y,x\}=\begin{cases}y, ~~\text{when}~~ y \leq x \\x, ~~\text{when} ~~x \leq y\end{cases}$

So, $\begin{align*}\int_o^t\int_0^t \min\{x,y\} \mathrm dy ~~\mathrm dx&=\int_0^t\left(\int_0^xy~\mathrm dy+\int_x^t x\mathrm dy\right)\mathrm d x\\&=\int_0^t\left(\dfrac{x^2}{2}+x(t-x)\right)\mathrm d x\\&=\dfrac{t^3}{6}+\dfrac{t^3}{2}-\dfrac{t^3}{3}\\&=\dfrac{t^3}{3}\end{align*}$

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How about splitting the integral into two regions, one where $x\ge y$, one where $x?

$\int_{x=0}^{t}\int_{y=0}^{t}\min(x,y)\,dy\,dx=\int_{x=0}^{t}\int_{y=0}^{x}y\,dy\,dx+\int_{x=0}^{t}\int_{y=x}^{t}x\,dy\,dx$