Imagine totalling the rainfall collected across a rectangular field from an array of sensors. Does it matter whether you first add up each row of sensors, then add the row totals — or first add up each column, then add those? Fubini's Theorem says no. For a well-behaved function over a region like a rectangle, a double integral gives the same answer no matter which variable you integrate first.
Slice the same region two different ways — add up the strips either way, and you get the same total. That's Fubini's Theorem.
What the theorem actually says
For a function f(x, y) that's continuous over a rectangle where x runs from a to b and y runs from c to d:
∬R f(x,y) dA = ∫ab ∫cd f(x,y) dy dx = ∫cd ∫ab f(x,y) dx dy
In words: hold one variable constant, integrate over the other, get an ordinary function of one variable — then integrate that. You can do the two steps in either order and land on the same number.
Why it matters
It turns a genuinely two-dimensional problem — the volume under a surface — into two ordinary one-variable integrals, done one after another, which is something you already know how to do. And because the order is your choice, you can pick whichever one is actually easy: some regions make integrating over x first nearly impossible while y first is routine, or the reverse.
This quietly sits underneath a lot of real calculations: finding the mass or centre of mass of a flat plate, computing moments of inertia in engineering, working with joint probability distributions in statistics, and the convolution operations used in image processing and machine learning.
One condition worth remembering: Fubini's Theorem needs the function to be well-behaved (technically, absolutely integrable) over the region. For the continuous functions you'll meet at university, order genuinely never matters.
A quick worked example
Take f(x, y) = xy over the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1.
- Integrate x first: ∫₀¹ [∫₀² xy dx] dy = ∫₀¹ [x²y/2]₀² dy = ∫₀¹ 2y dy = [y²]₀¹ = 1.
- Integrate y first: ∫₀² [∫₀¹ xy dy] dx = ∫₀² [xy²/2]₀¹ dx = ∫₀² (x/2) dx = [x²/4]₀² = 1.
Same answer, either order — exactly what Fubini's Theorem promises. Once this clicks, double integrals stop feeling like a new topic and start feeling like two integrals you already know, done back to back. If multivariable calculus like this is something you or your child wants properly explained rather than just memorised, that's exactly what our undergraduate maths tuition is for, and you can see the full learning pathway here.
Frequently asked questions
Does Fubini's Theorem work for any region, not just rectangles?
Yes — as long as the region can be described with sensible bounds in either order (what textbooks call a 'type I' or 'type II' region), and the function is integrable over it. You do have to rewrite the bounds correctly when you swap the order, which is usually the fiddly part in practice.
What's the difference between Fubini's Theorem and Tonelli's Theorem?
Fubini's Theorem needs the function to be absolutely integrable; Tonelli's Theorem covers non-negative functions even when that's not guaranteed in advance. For continuous functions on a closed, bounded region — the situation in most undergraduate courses — Fubini applies without extra checks.
Where does this show up outside a maths degree?
Anywhere you're summing something over two dimensions at once: computing a centre of mass, finding expected values of joint probability distributions, image-processing convolution kernels, and numerical integration routines all lean on being able to integrate one variable at a time.
About the author
Sudershan Soni
Founder & Lead Tutor at Mostak Services — an MSc-qualified Mathematics, Science, Computer Science & STEM tutor with 20+ years of professional experience, teaching students from 11+ and GCSE to A-Level and beyond, online worldwide.
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