Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

up vote 1 down vote accepted

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

share|improve this answer
    
Um, don't you mean to use that formula, instead of proving it? –  Doug Spoonwood Feb 5 '12 at 3:45
    
@DougSpoonwood,actually I mean "Prove that... and use it." I think that don't make sense to use things fallen from the sky. –  leo Feb 5 '12 at 3:50

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

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

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.