The Fundamental Theorem of Calculus, in two sentences

Both halves of the theorem are stated and proved in section 5.3 of OpenStax Calculus Volume 1, a free textbook, and the net-change reading of it is in §5.4. Every statement and quoted phrase below comes from those two sections, read on 17 October 2026; the arithmetic is ours and is shown at each step so it can be checked.

The two sentences

1. Differentiate the area-so-far function and you get the curve back.

2. Integrate a rate over an interval by reading any antiderivative at the two ends and subtracting.

The first is the surprising one. The second is the one that saves the week. OpenStax puts the second this way: “It almost seems too simple that the area of an entire curved region can be calculated by just evaluating an antiderivative at the first and last endpoints of an interval.” That is the feeling worth putting in front of a class deliberately — the theorem really is that cheap to use, and that is the point.

Part 1: the slope of the area is the curve

Define the area-so-far function by fixing the left end and letting the right end move:

F(x) = ∫_a^x f(t) dt   ⇒   F′(x) = f(x)

OpenStax flags the notation first, because it is where students stall: “we have said several times that a definite integral is a number, and here it looks like it’s a function. The key here is to notice that for any particular value of x, the definite integral is a number.” Read F as a machine that swallows an x and returns the area from a to that x; nothing more.

The proof is three lines and is worth doing on the board, because it is the average value of a function doing the work. Write the derivative of F as a limit, combine the two integrals into one, and you are left with

F′(x) = lim_h→0 (1/h) ∫_x^x+h f(t) dt

That fraction is the average value of f on the shrinking interval [x, x+h]. By the Mean Value Theorem for Integrals there is a point c in that interval where the average is achieved, and as h goes to zero c is squeezed towards x, so continuity gives f(c) → f(x).

OpenStax also names the consequence that earns the theorem its adjective: it “guarantees that any integrable function has an antiderivative.” That is not a footnote. It is the reason the antiderivative-hunting of the next month of the course is not guesswork.

Part 2: the area is read off the ends

If f is continuous on [a, b] and F is any antiderivative of f, then

∫_a^b f(x) dx = F(b) − F(a)

OpenStax calls this the evaluation theorem and proves it by telescoping: write F(b) − F(a) as a sum of small changes across a partition, apply the Mean Value Theorem on each subinterval to turn each change into f(c_i)Δx, and let the number of pieces go to infinity. The sum becomes the integral. Nothing in the argument depends on which antiderivative you picked — which is exactly why the constant is harmless.

Now put numbers in, the way a class actually will. Take

∫₋₂² (t² − 4) dt

An antiderivative is t³/3 − 4t, so

[8/3 − 8] − [−8/3 + 8] = 8/3 − 8 + 8/3 − 8 = 16/3 − 16 = −32/3

Minus ten and two-thirds. Every class will ask whether that is a mistake. It is not, and the reason is the sentence to put on the board, in OpenStax’s own words: “Area is always positive, but a definite integral can still produce a negative number (a net signed area).” On [−2, 2] the parabola sits below the axis for most of the interval, so the signed total is negative even though both pieces of area are positive.

Where the two halves meet: net change

The theorem has an everyday reading that is easier to teach first, and §5.4 states it as the net change theorem: the new value of a changing quantity is the initial value plus the integral of its rate of change. OpenStax’s example is a car going north at 40 mph from 2 p.m. to 4 p.m. and then south at 30 mph for the next hour. Net displacement is 80 − 30 = 50 miles north. Total distance travelled is 80 + 30 = 110 miles. Same drive, two integrals, one answer each — and the sentence that names the difference: to get distance you integrate the absolute value of velocity. “Net change accounts for negative quantities automatically without having to write more than one integral.”

Four places a class slips

1. Signed area against total area. The −32/3 above and the 50 against 110 are the same confusion seen twice. The fix is not a rule but a question: is the answer supposed to be able to come out negative? Displacement can. Distance cannot. Area cannot. If it cannot, the integrand or the interval is wrong.

2. Chasing the + C. Students who have just learned indefinite integrals want to add a constant and then wonder what happens to it. OpenStax settles it in its own analysis of a worked example: any antiderivative works, “so, for convenience, we chose the antiderivative with C = 0. If we had chosen another antiderivative, the constant term would have canceled out. This always happens when evaluating a definite integral.” In a definite integral the constant is not lost; it is subtracted from itself.

3. The upper limit is a function, not a letter. Part 1 is stated for an upper limit of x. Make it √x and the chain rule arrives whether you invited it or not: OpenStax’s example gives F(x) = ∫₁^√x sin t dt with F′(x) = sin √x / (2√x). Make both limits functions and you split at a constant first: ∫_x^2x t³ dt differentiates to −x³ + (2x)³·2, which is 15x³. Once a student can do that second one without prompting, Part 1 is actually learned.

4. The hypothesis is continuity, and the dummy variable matters. Both parts of the theorem as OpenStax states them begin “if f is continuous”. That clause is doing work: at a jump discontinuity the area-so-far function has a corner, and the safe classroom habit is to split the integral at the jump rather than to apply the theorem across it. Separately, the t inside and the x on top are not decoration — they are what keeps the sentence readable when the top limit becomes √x or 2x.

A five-minute sequence for the board

One order that works, and needs nothing but the board and the two sentences:

Sketch f(x) = x² on [0, 2]. Shade from 0 to a movable point and label the shaded area F(x). Ask what F′(x) is — not to compute, to look at. The area gains a thin strip of height x² and width dx, so the gain per unit of x is x². That is Part 1, discovered rather than announced. Then check it numerically: ten right-hand rectangles for ∫₀¹ x² dx give 0.385, and 1/3 is 0.333 — close, and closer with more rectangles. Then use Part 2 for the exact value and let the class watch the approximation and the theorem agree.

What we are not claiming

Three limits on this page. We are not claiming this is a proof of the theorem for a calculus class — the arguments above are the sketches OpenStax gives, compressed; the full proofs are in §5.3 and are better read there in full. We are not claiming one teaching order suits every class: the board sequence is one that works, not the only one. And we are not making any claim here about how the theorem is weighted on the AP exam — that is a different page, sourced to College Board, and it is linked below. Where a number appears, we recomputed it from the source’s own stated antiderivative — −32/3, −3/2 and 41/6 for the v(t) = 3t−5 particle, and 15x³ from F(x) = ∫_x^2x t³ dt, whose closed form is 15x⁴/4.

• Sources (checked 17 October 2026): OpenStax, Calculus Volume 1 — §5.3 The Fundamental Theorem of Calculus (the Mean Value Theorem for Integrals and its proof; the notation note that a definite integral is a number; the statement and proof of Part 1, including “guarantees that any integrable function has an antiderivative”; the statement and telescoping proof of Part 2, the evaluation theorem; the worked derivative examples g(x) = ∫₁^x 1/(t³+1) dt → 1/(x³+1), F(x) = ∫₁^√x sin t dt → sin √x/(2√x) and ∫_x^2x t³ dt → 15x³; ∫₋₂²(t²−4) dt = −32/3 with the analysis that any antiderivative works, so C = 0 is chosen; and “Area is always positive, but a definite integral can still produce a negative number (a net signed area)”); §5.4 Integration Formulas and the Net Change Theorem (the net change theorem in both forms; the 40 mph north / 30 mph south car with net displacement 50 mi against total distance 110 mi; “net change accounts for negative quantities automatically”; the v(t) = 3t−5 particle with net displacement −3/2 m and total distance 41/6 m). See also our resource library and our AP Calculus 2027 changes page.