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Use Fubini's theorem to show that for continuous functions f and g, and a rectangle R, $\iint_Rf(x)g(y)dA$ =$\int_a^bf(x)dx$$\int_c^dg(y)dy$. Use this property to evaluate the integral $\iint_Rxe^{x^2-y}dA$ where R is (1,2) x (-1,0).

I know that Fubini's theorem means we can interchange the variables inside the double integral but how can we seperate the double integral to two integrals?

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    $\begingroup$ The point is that the two variables have both constant bounds, and the function to integrate is the product of a function of only $x$ and one of only $y$. $\endgroup$ – alex Mar 4 '14 at 16:17
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The key to showing this is, like Alex said, the given functions are single variable and so they become constant when integrating over the other variable:

By Fubini,

$\int\int_Rf(x)g(y)d(A) = \int_x(\int_yf(x)g(y)dy)dx$ Since we're integrating over y, $f(x)$ acts as a constant, say $C$. We have $\int_yf(x)g(y)dy = \int_yCg(y)dy = C\int_yg(y)dy$. Using this substitution, we have, $\int\int_Rf(x)g(y)d(A) = \int_xC(\int_yg(y)dy)dx = \int_xf(x)(\int_yg(y)d(y))dx$ Using the same type of substitution, let $B=\int_yg(y)d(y)$, then

\begin{align*} \int_xf(x)(\int_yg(y)d(y))dx &= \int_xf(x)Bdx \\ &= B\int_xf(x)dx \\ &= \int_yg(y)dy\int_xf(x)dx \\ &= \int_xf(x)dx\int_yg(y)dy \end{align*}

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