Fubini's Theorem

Compute a double integral one variable at a time, in either order.

The idea

Theorem (Fubini's theorem).

Let $R = [a,b] \times [c,d]$ and let $f$ be integrable on $R$, meaning $\iint_R |f| < \infty$; a function continuous on $R$ always qualifies. Then both iterated integrals exist and $\iint_R f = \int_a^b \int_c^d f(x,y) \, dy \, dx = \int_c^d \int_a^b f(x,y) \, dx \, dy.$

The double integral on the left is defined as a single limit over ever finer grids covering $R$, a definition that offers no method of evaluation. The theorem replaces it with ordinary one-variable integration done twice: in the middle expression the inner integral holds $x$ fixed and integrates the function of $y$ alone, producing a number that depends on $x$, and the outer integral then integrates that function of $x$. Since the two orders give the same value, we may choose whichever is shorter.

The hypothesis $\iint_R |f| < \infty$ carries real content. When it fails, both iterated integrals can still exist and yet disagree: the positive and negative parts of $f$ then each contribute infinite mass, and the order of integration decides how the two are balanced against each other. For a continuous function on a closed bounded rectangle this cannot happen, so there the swap is always valid.

Ways to work on it

• Walkthrough. Why a double integral equals either iterated integral, with both orders worked out.
• Practice. Integrate a simple function over a rectangle.
• Hardest. A non-separable integrand, then when Fubini's order swap can fail.

Not sure where to start? Take the ten-question placement test.