Integrating abs(x-y) over the unit square

[mnf]

integrate:


∫₀¹∫₀¹ abs(x-y) dydx

 

[tiny-tim]

Welcome to PF!

Hi mnf! Welcome to PF!

integrate:


∫₀¹∫₀¹ abs(x-y) dydx

(no need to shout! )

Hint: split the integral into two regions, one with x > y, and one with x < y.

 

[mnf]

I don't know how ,please explain it

 

[tiny-tim]

Instead of one integral, with limits 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 (which is a square),

split the square into two regions, one with x < y, and one with x > y,

and then use two integrals, one for each region.

 

[mnf]

i want to get answer in closed form
because question is
find B value

B∫₀¹∫₀¹ abs(x-y) dydx =1

 

[CRGreathouse]

If I didn't know better, I'd think you want us to solve it for you!

 

[HallsofIvy]

integrate:


∫₀¹∫₀¹ abs(x-y) dydx

If you honestly do not know what the absolute value of a number is, which is what you appear to be saying, you have no hope of doing this problem. Talk to your teacher about it!

If you do know the absolute value function then you know that |x- y|= x- y as long as $x\ge y$ and |x-y|= y- x if x< y. The region $0\le x\le 1$, $0\le y\le 1$ is a square. $x\ge y$ below the diagonal from (0, 0) to (1, 1) and y> x above the diagonal. Integrate those two separately and add.