Iterated Integrals and Fubini's Theorem

Riemann sums define a double integral, but nobody computes one that way. Instead, integrate one variable at a time, the way you found volumes by slicing in Calculus 2.

Key idea

Fubini's theorem: for continuous $f$ on $R = [a, b] \times [c, d]$, $$\iint_R f\,dA = \int_a^b \left(\int_c^d f(x, y)\,dy\right) dx = \int_c^d \left(\int_a^b f(x, y)\,dx\right) dy.$$ Work from the inside out; in the inner integral, the outer variable is a constant.

Why it works

Slice the solid with a plane $x = c$ (a constant). The slice is a region under the curve $z = f(x, y)$, $c \le y \le d$, with area $A(x) = \int_c^d f(x, y)\,dy$. A slab of thickness $\Delta x$ has volume about $A(x)\,\Delta x$, and adding slabs gives $\int_a^b A(x)\,dx$. Slicing with planes $y = c$ instead gives the other order. Both compute the same volume, which is why the order does not change the answer.

The red slice at $x = c$ has area $A(c) = \int_0^2 f(c, y)\,dy$, the inner integral. Slide it from $x = 0$ to $x = 2$: adding up those areas gives the volume.

When the integrand is a product $g(x)h(y)$ over a rectangle, the iterated integral splits into two single integrals: $$\int_a^b \int_c^d g(x)h(y)\,dy\,dx = \left(\int_a^b g(x)\,dx\right)\left(\int_c^d h(y)\,dy\right),$$ because in the inner integral $g(x)$ is a constant that can come outside. For example, $\int_0^1 \int_0^\pi x\sin y\,dy\,dx = \frac{1}{2} \cdot 2 = 1$. Fubini's theorem also allows switching the order, which matters when one order leads to an antiderivative you cannot find; the third example below shows this.

Worked examples

Worked example: Inside out

Evaluate $\displaystyle\int_0^2\int_1^3 x^2y\,dy\,dx$.

Inner, in $y$ ($x$ is a constant): $\int_1^3 x^2y\,dy = x^2\left[\frac{y^2}{2}\right]_1^3 = x^2 \cdot 4 = 4x^2$.
Outer: $\int_0^2 4x^2\,dx = \frac{32}{3}$.

Answer: $\frac{32}{3}$

Worked example: The other order

Evaluate $\displaystyle\int_1^3\int_0^2 x^2y\,dx\,dy$.

Inner, in $x$: $\int_0^2 x^2y\,dx = \frac{8}{3}y$.
Outer: $\int_1^3 \frac{8}{3}y\,dy = \frac{8}{3} \cdot 4 = \frac{32}{3}$, the same, as Fubini promises.

Answer: $\frac{32}{3}$

Worked example: Choosing the easy order

Evaluate $\displaystyle\int_0^1\int_0^2 xe^{xy}\,dy\,dx$.

Inner in $y$: $\int_0^2 xe^{xy}\,dy = \left[e^{xy}\right]_{y=0}^{y=2} = e^{2x} - 1$ (the $x$ in front is exactly the inner derivative).
Outer: $$\int_0^1 (e^{2x} - 1)\,dx = \left[\frac{e^{2x}}{2} - x\right]_0^1 = \frac{e^2}{2} - 1 - \frac{1}{2} = \frac{e^2 - 3}{2}.$$
Integrating $x$ first would need integration by parts: both orders are correct, but one is much easier.

Answer: $\frac{e^2 - 3}{2}$

Common mistake

Mismatching limits and differentials. In $\int_a^b\int_c^d f\,dy\,dx$, the inner limits $c$ to $d$ belong to $y$ (the inner differential $dy$) and the outer limits to $x$. Writing $dx\,dy$ while keeping the $x$-limits inside integrates over a different rectangle. Read from the inside out: innermost integral sign, innermost differential.

Iterated Integrals and Fubini's Theorem

Worked examples

Practice

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