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 August 1st, 2017, 12:27 AM #11 Global Moderator   Joined: Dec 2006 Posts: 18,037 Thanks: 1394 Alternatively, substituting u = x + y, v = x - y leads to $\displaystyle \int_{-1}^1\!\int_{-1}^1 (u^2 + v^2)/4\,dudv$, which is 2/3. Thanks from SenatorArmstrong and zollen
August 1st, 2017, 12:22 PM   #12
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Quote:
 Originally Posted by romsek that's an integration of the square with corners at $(-1,-1), (-1,1), (1,1), (1,-1)$ His integration area is a square with corners $(-1,0), (0,1), (1,0), (0,-1)$
My apologies for my blunder.

 Tags double integral, wrong

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