Ask most people what an integral is, and you'll get some version of "the area under the curve." It's not wrong, exactly — but it's the kind of answer that quietly ducks the actual question. What is area, precisely, for a region bounded by some wild, wiggly curve? You can't tile it with unit squares and count. You can't cut it into triangles like a geometry textbook shape. So what does the symbol ∫ actually mean, in a way that doesn't just point back at a picture?
Michael Spivak spends the opening pages of Chapter 13 of Calculus refusing to let you off the hook with a picture. Instead, he builds the integral out of two ingredients you can actually pin down: partitions and sums of rectangles, squeezed from both sides. It's one of the great "wait, that's it?" moments in undergraduate math — and it's worth walking through slowly, because the definition is doing real work.
The Trouble With "Area"
Spivak opens by pointing out something almost embarrassing: elementary geometry never actually defines area for anything but the simplest shapes. A circle of radius 1 "obviously" has area π, but try explaining what "π squares" fitting inside a circle even means. If area is shaky ground even for a circle, it's hopeless as a foundation for the area under an arbitrary function's graph.
So instead of trying to define area in general, Spivak restricts attention to a very specific kind of region: the one bounded by the x-axis, the vertical lines through (a, 0) and (b, 0), and the graph of a bounded function f on [a, b]. Call this region R(f, a, b). The number we eventually assign to it — whether or not f(x) ≥ 0 — is called the integral of f on [a, b]. The word "area" comes later, as a special case.
Squeezing From Both Sides: Partitions and Sums
Here's the actual machinery. A partition of [a, b] is just a finite set of points t₀, t₁, …, tₙ with a = t₀ < t₁ < ⋯ < tₙ = b — a way of chopping the interval into n subintervals. On each subinterval [tᵢ₋₁, tᵢ], let
- mᵢ = inf{f(x) : tᵢ₋₁ ≤ x ≤ tᵢ} — the greatest lower bound of f on that piece
- Mᵢ = sup{f(x) : tᵢ₋₁ ≤ x ≤ tᵢ} — the least upper bound
Notice these are infs and sups, not minima and maxima — f hasn't been assumed continuous, so it might not actually attain those bounds. From these, Spivak defines the lower sum and upper sum of f for the partition P:
L(f, P) = Σ mᵢ(tᵢ − tᵢ₋₁) and U(f, P) = Σ Mᵢ(tᵢ − tᵢ₋₁)
L(f, P) is the total area of rectangles guaranteed to sit inside the region; U(f, P) is the total area of rectangles guaranteed to contain it. Whatever "area" ends up meaning, it has to sit between these two numbers, for every single partition — no matter how the interval is chopped up.
A short lemma makes this rigorous: refining a partition (adding more points) can only push lower sums up and upper sums down, since finer rectangles hug the curve more tightly. From that lemma, Spivak proves Theorem 1: for any two partitions P₁ and P₂ of [a, b], L(f, P₁) ≤ U(f, P₂) — every lower sum is beneath every upper sum, even ones built from totally different partitions. That means the set of all lower sums and the set of all upper sums never cross; there's a gap (possibly zero) between sup{L(f, P)} and inf{U(f, P)}, and the integral, if it exists, has to live in that gap.
The Definition — and a Function That Breaks It
This is where Spivak finally commits: a bounded function f is integrable on [a, b] if sup{L(f, P)} = inf{U(f, P)}, taken over all partitions P. When this common value exists, it's called ∫ₐᵇ f. That's the whole definition — no limits of Riemann sums needed yet, no appeal to "infinitely thin rectangles." Just: do the best-possible lower approximations and the best-possible upper approximations converge to the same number?
They don't always. Spivak's go-to counterexample is the function equal to 0 on the irrationals and 1 on the rationals, on any interval [a, b]. Because every subinterval, no matter how small, contains both a rational and an irrational number, every mᵢ = 0 and every Mᵢ = 1, for every partition. So L(f, P) = 0 and U(f, P) = b − a always — the lower and upper sums never budge no matter how finely you partition. The gap never closes, and the function is simply not integrable. This isn't a technicality; it's the definition correctly refusing to assign an area to a region that's genuinely too pathological to have one.
Continuity to the Rescue
So which functions are integrable? Spivak proves a convenient reformulation first (Theorem 2): f is integrable on [a, b] exactly when, for every ε > 0, some partition P satisfies U(f, P) − L(f, P) < ε — you don't need to compute sups and infs over all partitions, just show the gap can be squeezed arbitrarily small.
That reformulation earns its keep immediately in Theorem 3, arguably the payoff of the whole chapter: if f is continuous on [a, b], then f is integrable on [a, b]. The proof leans on uniform continuity — since f is uniformly continuous on the closed interval, you can choose a partition fine enough that f varies by less than ε/(b−a) on each piece, which forces U(f, P) − L(f, P) < ε. Continuity, it turns out, is exactly the regularity a function needs to guarantee its lower and upper approximations converge.
Spivak makes the definition concrete with f(x) = x² on [0, b]. Partitioning [0, b] into n equal pieces and using the formula for sums of squares, he computes L(f, Pₙ) = (b³/n³) · ⅙(n−1)n(2n−1) and U(f, Pₙ) = (b³/n³) · ⅙(n+1)n(2n+1). Both expressions squeeze down on b³/3 as n grows, and since L(f, Pₙ) ≤ b³/3 ≤ U(f, Pₙ) for every n while the gap shrinks to zero, b³/3 is the only number that can be the integral. Hence ∫₀ᵇ x² dx = b³/3 — a fact known to Archimedes, now derived without a single appeal to "area," purely from the partition definition doing exactly what it was built to do.
This is the real substance behind that friendly phrase "area under the curve": a precise squeeze between two families of rectangle sums, and a theorem telling you exactly when the squeeze is guaranteed to work. If you want to see the full machinery — the proofs of the lemma, Theorem 1's uniqueness argument, and where continuity gets its uniform-continuity teeth — it's all laid out with Spivak's characteristic rigor in Spivak's Calculus, 4th Edition.
Further reading: How to self-study Spivak's Calculus · Table of contents · Index · Calculus, 4th Edition · Answer Book
Before you assumed "area under the curve" was the whole story — had you ever wondered why a function needs to be continuous (or something like it) for that area to even exist?
