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?
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*}