Iterated Integrals

An iterated integral is simply a double integral done one variable at a time. The expression $\int_a^b\!\int_c^d f(x, y)\,dy\,dx$ looks like a stack of symbols, but read it from the inside out: $\int_c^d f(x, y)\,dy$ (freeze $x$ and add up along $y$) gives you a function of $x$ alone, and $\int_a^b \ldots\,dx$ (add those results up across $x$) finishes the job. Two single integrals, which you already know how to do. That's all there is to it.

Work from the inside out. In $\int_a^b\!\int_c^d f\,dy\,dx$, do the $dy$ integral first, treating $x$ as a constant. It's exactly like a partial derivative in reverse: there, freezing $x$ let you differentiate in $y$; here it lets you integrate in $y$.

Why it works. Think about what the inner integral measures. With $x$ frozen, $z = f(x, y)$ is an ordinary curve over the interval $c \le y \le d$, and $\int_c^d f(x, y)\,dy$ is the area under it: the area $A(x)$ of the slice of the solid at that $x$. Then $\int_a^b A(x)\,dx$ adds up the slices. You've seen this before: it's the volume by cross-sections method from Calculus 1, where a loaf of bread is the sum of its slices. There's an animation further down that sweeps a slice across a solid.

Fubini's theorem says the order doesn't matter for a continuous $f$ on a rectangle: slicing the loaf the long way or the short way gives the same volume. It makes sense, since both are adding up the same tiny boxes, just grouped into different rows. In practice, pick whichever order makes the inner integral easier.

Products split. If $f(x, y) = g(x)h(y)$ over a rectangle, the inner integral of $h$ is a constant, so $\iint f\,dA = \left(\int g\,dx\right)\left(\int h\,dy\right)$.

A common mistake. Reading left to right, it's tempting to pair the first integral sign with the first differential you see. They nest like parentheses instead: the innermost differential goes with the innermost integral sign. In $\int_0^2\!\int_0^3 \ldots\,dy\,dx$, $y$ runs from $0$ to $3$ and $x$ from $0$ to $2$.

Image processing sums pixel intensities over windows with iterated sums, the discrete version of these integrals; computing a blur is a double integral of the image against a kernel.