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Please bear with me because I have only little experience in using codes to construct the symbols for the equations. The question is:

Determine the average value of $f(x,y) = x^2 y^2$, in the region $$R: a\le x\le b, c\le y\le d,$$ where $a+b=5, ab=13, c+d=4, cd=7.$

The formula for average value is

$$\dfrac{\displaystyle\iint f(x,y)dA}{(b-a)(d-c)}.$$

Please use only elementary calculus, and no complex analysis. Much appreciated!

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I don't think so - the expression $a \le x \le b$ does not make sense with $a,b$ complex. –  Johannes Kloos Mar 27 '12 at 8:40
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As stated the problem simply makes no sense; it must be defective. –  Brian M. Scott Mar 27 '12 at 8:49
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Even if you use the fact that all of $a,b,c,d$ are imaginary, the question still does not make any sense. On the other hand, if you evaluate the integrals as line integrals over $\mathbb C$, you do get a (negative) real solution since the real parts cancel out, and the product of the imaginary parts is real. –  Johannes Kloos Mar 27 '12 at 9:13
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Even if it doesn't make sense I've computed formally an average value of 12. –  Américo Tavares Mar 27 '12 at 9:24
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This is also a pretty idiotic way to present a region of integration, even if the solution to the system of equations had $a,b,c,d \in \mathbb{R}$. –  Michael Joyce Mar 27 '12 at 18:25
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1 Answer 1

I guess this example could be a weird way to show possible transformation formalities during (double-)integration. Let's just do it formally:

$$\dfrac{\displaystyle\iint f(x,y)dA}{(b-a)(d-c)} = \frac{\int\limits_a^b\int\limits_c^d x^2y^2 dxdy}{(b-a)(d-c)} = \frac{\int\limits_a^b x^2 dx}{b-a}\times\frac{\int\limits_c^d y^2 dy}{d-c} = \frac{\frac{x^3}{3}\big|_a^b}{b-a}\times\frac{\frac{y^3}{3}\big|_c^d}{d-c} = \frac{1}{9}\frac{b^3-a^3}{b-a}\frac{d^3-c^3}{d-c} = \frac{((a+b)^2-ab)((c+d)^2-cd)}{9} = \frac{12\cdot 9}{9} = 12$$

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