Help integrating Abs(x-y) dydx

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To integrate the function abs(x-y) over the square defined by 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, the integral should be split into two regions: one where x > y and another where x < y. This allows for the use of two separate integrals, with the absolute value defined as |x-y| = x-y for x ≥ y and |x-y| = y-x for x < y. The key is to calculate each integral separately and then combine the results. The final goal is to find the value of B such that B∫[0 to 1]∫[0 to 1] abs(x-y) dydx = 1. Understanding the absolute value function is crucial for solving this problem effectively.
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integrate:


0101 abs(x-y) dydx
 
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Welcome to PF!

Hi mnf! Welcome to PF! :smile:
mnf said:
integrate:


0101 abs(x-y) dydx


(no need to shout! :rolleyes:)

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


I don't know how ,please explain it
 
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.
 


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

B∫0101 abs(x-y) dydx =1
 


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


mnf said:
integrate:


0101 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.
 
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