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Here's the problem statement of the question which I am stuck on:

Let $R_{1}$ denote the rectangle $[0, 5] \times [-4, 4]$, $R_{2}$ the rectangle $[0, 5] \times [0, 4]$, and $R_{3}$ the rectangle $[-5, 0] \times [-4, 0]$. Suppose that $\mathop{\int\!\!\! \int}_{R_{2}}f(x,y)dA=10$, that $\mathop{\int\!\!\! \int}_{R_{3}}f(x,y)dA=24$ and that $f(-x,y)=-f(x,y)$ for all $(x,y)$. Evaluate $\mathop{\int\!\!\! \int}_{R_{1}}f(x,y)dA$.

Here's my working:

$\mathop{\int\!\!\! \int}_{R_{1}}f(x,y)dA=\mathop{\int\!\!\! \int}_{R_{2}}f(x,y)dA+\mathop{\int\!\!\! \int}_{R_{3}}f(-x,y)dA=\mathop{\int\!\!\! \int}_{R_{2}}f(x,y)dA-\mathop{\int\!\!\! \int}_{R_{3}}f(x,y)dA=10-24=-14$

The solution is given as $34$. Can you please help me see where I'm going wrong? Thank you very much.

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Your work looks correct to me. –  Ron Gordon May 14 '13 at 23:08
    
Whoops! I meant to say that the solution is given as 34. This doesn't change your thoughts? :-) –  user77858 May 14 '13 at 23:37
    
When you change the region you also change the sign since your integration intervals are swapped. –  Sigur May 14 '13 at 23:58
    
Ah, I see what you're saying! Thank you very much for your help. –  user77858 May 15 '13 at 0:26
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1 Answer

Note that the area in the third quadrant is equivalent to the area in the fourth quadrant. Making use of $f(-x,y)=-f(x,y),$ we can change the region to the fourth quadrant, but the change of region results in a change of the sign back to positive giving the area of $R_1$ as $10+24=34.$

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