August 1st, 2017, 12:27 AM  #11 
Global Moderator Joined: Dec 2006 Posts: 19,162 Thanks: 1638 
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.

August 1st, 2017, 12:22 PM  #12 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,821 Thanks: 1047 Math Focus: Elementary mathematics and beyond  

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double integral, wrong 
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