Fubini's Theorem

The reasoning of the previous page can be summarized in a famous theorem:

Fubini's Theorem: If $f(x,y)$ is a continuous function on a rectangle $R = [a,b] \times [c,d]$, then the double integral $\iint_R f(x,y) dA$ is equal to the iterated integral $$\int_c^d \left( \int_a^b f(x,y) dx\right ) dy$$ and also to the iterated integral $$\int_a^b \left(\int_c^d f(x,y)dy \right ) dx.$$ In both cases we can evaluate the inner integral with the fundamental theorem of calculus, treating the other variable as a constant.

Surprisingly enough, examples of unbounded and discontinuous functions $f(x,\,y)$ can be constructed where the iterated integral in one order is different from the iterated integral in the opposite order! But for any reasonable function, you can integrate in either order.

Example: Compute $\iint_R xe^y dA$, where $R$ is the rectangle $[0,2] \times [0,1]$. Solution 1: We can integrate first over $x$. \begin{eqnarray*} \iint_R xe^y \, dA & = & \int_0^1 \left (\int_0^2 xe^y dx \right ) dy \cr \cr &=& \int_0^1 \left . \frac{x^2}{2} e^y \right |_{x=0}^2 \, dy \cr \cr & = & \int_0^1 2 e^y\, dy \ = \ 2e-2 \end{eqnarray*}

Solution 2: We can integrate first over $y$. \begin{eqnarray*} \iint_R xe^y \, dA & = & \int_0^2 \left (\int_0^1 xe^y dy \right ) dx \cr \cr &=& \int_0^2 \left . x e^y \right |_{y=0}^1 \, dx \cr \cr & = & \int_0^2 (e-1)x\, dx \cr & = & \left . \frac{(e-1)x^2}{2} \right |_0^2 = 2e-2. \end{eqnarray*}