Help integrating Abs(x-y) dydx

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The discussion focuses on the integration of the function abs(x-y) over the square defined by the limits 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. Participants suggest splitting the integral into two regions: one where x > y and another where x < y. This approach allows for the application of the absolute value function, leading to two separate integrals that can be solved and combined to find the closed form of the integral. The goal is to determine the value of B in the equation B∫[0,1]∫[0,1] abs(x-y) dydx = 1.

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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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