The Fundamental Theorem, Part 1

Watch rain collect in a gauge during a storm. The total caught so far climbs fast when the rain is heavy, slows when it eases, and holds steady when it stops — the rate the total rises is simply how hard it is raining at that instant. That everyday link between a running total and its current rate is exactly the theorem on this page, and it is the hinge the whole of calculus swings on.

Calculus has two halves, and for most of its history they were built as separate subjects. One half measures slope — the derivative, born from tangent lines and instantaneous speed. The other measures area — the definite integral, born from chopping regions into slivers and adding them up. Tangents and areas: two problems that look about as related as a speedometer and a warehouse inventory.

The Fundamental Theorem of Calculus throws open the connecting door. Part 1 says: accumulating area, then differentiating, returns the integrand you started with. Differentiation undoes integration. The two halves are one subject running in opposite directions — and that single sentence is what turned a pile of clever tricks into the machine called calculus.

Here is the object the theorem is about. Fix a continuous f and define the accumulation function — the area so far, swept from a base point a up to a moving right edge x:

F(x) = \int_{a}^{x} f(t)\,dt.

Notice the cast of characters: x is the input — where the right edge of the shading currently sits — and t is a private, internal dummy variable that runs across the shaded region and is used up by the integral. (We write it as t precisely so it isn't confused with the limit x.) Slide x right and F grows; slide it back and F shrinks. F is a perfectly ordinary function of x — it just happens to be defined by an area.

The claim — the shock — is that this area function has a derivative, and that derivative is exactly the function under the integral sign:

F'(x) = f(x).

Watch the theorem emerge: build F by hand

Before any proof, let's catch the theorem in the act. For simple integrands we can compute the accumulated area with nothing but school geometry — and then differentiate the result and see what comes out.

Example 1 — a constant. Take f(t) = 3 and base point a = 0. The region from 0 to x under a flat line of height 3 is a rectangle: width x, height 3. So

F(x) = \int_0^x 3\,dt = 3x, \qquad\text{and}\qquad F'(x) = 3 = f(x).\;\checkmark

The derivative of the area function is the integrand. Once might be a coincidence.

Example 2 — a linear integrand. Take f(t) = t. The region from 0 to x under the line y = t is a right triangle with base x and height x:

F(x) = \int_0^x t\,dt = \tfrac12 x^2, \qquad\text{and}\qquad F'(x) = x = f(x).\;\checkmark

Twice is a pattern. Note what happened to the degree: accumulating pushed t up to \tfrac12 x^2, and differentiating pulled it straight back down. The two operations are running the same staircase in opposite directions.

Example 3 — where the theorem starts paying rent. Now take f(t) = \cos t. There is no rectangle-and-triangle trick for

F(x) = \int_0^x \cos t\,dt,

and if you only knew Riemann sums, computing F'(x) would look hopeless: a limit of a limit. But the theorem answers instantly, without computing the area at all:

\frac{d}{dx}\int_0^x \cos t\,dt = \cos x.

No sums, no geometry — just read the integrand off and evaluate it at the upper limit. That is the power move: FTC Part 1 turns "differentiate this area" from a research project into a lookup.

Why it's true: the sliver argument

The honest heart of the theorem fits in one picture. Ask how much new area F gains when the right edge nudges from x to x + h. The gain, F(x+h) - F(x), is a skinny sliver of width h standing under the curve. For tiny h the curve barely moves across the sliver, so the sliver is almost a rectangle of height f(x):

F(x+h) - F(x) \;\approx\; f(x)\cdot h \qquad\Longrightarrow\qquad \frac{F(x+h)-F(x)}{h} \;\approx\; f(x).

Let h \to 0 and the "approximately" hardens into "equals": F'(x) = f(x). Everything below is just this picture made airtight — the word "almost a rectangle" replaced by a squeeze.

The proof — line by line

We compute F'(x) straight from the limit definition of the derivative. The whole argument is a squeeze.

Step 1 — form the difference quotient. By definition,

F'(x) = \lim_{h\to 0}\frac{F(x+h) - F(x)}{h}.

Step 2 — turn the numerator into one integral. Using additivity of the definite integral, the area up to x+h minus the area up to x is just the area of the thin sliver [x, x+h]:

F(x+h) - F(x) = \int_a^{x+h} f(t)\,dt - \int_a^{x} f(t)\,dt = \int_{x}^{x+h} f(t)\,dt.

Step 3 — write the difference quotient as an average. Divide that sliver by its width:

\frac{F(x+h) - F(x)}{h} = \frac{1}{h}\int_{x}^{x+h} f(t)\,dt.

The right side is the average value of f over the little interval [x, x+h].

Step 4 — bound that sliver. Let m_h and M_h be the smallest and largest values of f on [x, x+h] (they exist because f is continuous on a closed interval). The bounding inequality m_h\, h \le \int_x^{x+h} f \le M_h\, h, divided by h > 0, gives

m_h \;\le\; \frac{1}{h}\int_{x}^{x+h} f(t)\,dt \;\le\; M_h.

Step 5 — squeeze as h \to 0. As the sliver shrinks to the single point x, continuity of f forces both the minimum and the maximum on it to converge to f(x):

\lim_{h\to 0} m_h = f(x) \qquad\text{and}\qquad \lim_{h\to 0} M_h = f(x).

Step 6 — apply the squeeze theorem. The difference quotient is trapped between two quantities that both tend to f(x), so it has no choice but to do the same:

\lim_{h\to 0}\frac{F(x+h) - F(x)}{h} = f(x).

Step 7 — read off the conclusion. The left side is F'(x). Hence

\boxed{\;F'(x) = f(x).\;}

Differentiation undoes accumulation. (For h < 0 the same argument runs with the inequalities and the average reversed, giving the identical limit, so the two-sided derivative exists.)

Think of F(x) as a tank filling with water: at time x, F is the total volume collected so far. The instantaneous rate the tank fills is the height of the inflow at this instant — which is exactly f(x). So F'(x), the rate of change of accumulated area, equals the current height of the curve. Where f is tall, area piles up fast and F rises steeply; where f = 0, the area momentarily stalls and F is flat; where f < 0, area is being removed and F falls.

Step 5 is the only place we used continuity, and the proof leans its whole weight there. If f had a jump at x, the slivers approaching from the left and the right would average to different values, so m_h and M_h would not both close on a single number — the squeeze collapses, and F'(x) need not equal f(x) there. (The accumulation function F is still continuous — area can't jump — but it develops a corner, so it isn't differentiable at the jump.) Continuity of the integrand is exactly what guarantees a clean derivative.

Let f be continuous on [a, b], and define the accumulation function

F(x) = \int_{a}^{x} f(t)\,dt \qquad (a \le x \le b).

Then F is differentiable on (a, b) and

F'(x) = f(x) \qquad\text{for every } x \in (a, b).

In words: the accumulation function is an antiderivative of its integrand — every continuous function has an antiderivative, namely its own area-so-far function.

The area filling, the slope matching

On top is f(t) = \tfrac12\,t with the area from a = 0 up to the moving edge x shaded. Below is the accumulation function F(x) = \int_0^{x}\tfrac12 t\,dt = \tfrac14 x^2, with a dot riding along it at the current x. Slide x: the shaded area on top equals the height of the dot below, and the slope of F at the dot equals the height of f on top — that is F'(x) = f(x), made visible.

Two things worth hunting for as you drag. First, near x = 0 the integrand is short, so area trickles in and the lower curve is nearly flat; out at x = 4 the integrand is tall, area floods in, and F climbs steeply — the lower graph's steepness is a live readout of the upper graph's height. Second, watch the two printed numbers: the tangent's slope always agrees with f(x), at every position, to the last decimal. The theorem isn't an end-of-chapter fact; it's happening continuously, everywhere along the curve.

Moving the goalposts: a chain-rule upper limit

The theorem hands you the derivative when the upper limit is plain x. But limits of integration love to be sneakier — say G(x) = \int_0^{x^2} \cos t\,dt. This is a composition in disguise: it is the accumulation function F(u) = \int_0^{u} \cos t\,dt with u = x^2 plugged into it, so G(x) = F(x^2). Differentiate with the chain rule and use F'(u) = \cos u from FTC Part 1:

G'(x) = F'(x^2)\cdot \frac{d}{dx}\!\left(x^2\right) = \cos(x^2)\cdot 2x.

The recipe for any differentiable upper limit g(x): evaluate the integrand at the limit, then multiply by the limit's derivative —

\frac{d}{dx}\int_a^{g(x)} f(t)\,dt \;=\; f\!\big(g(x)\big)\cdot g'(x).

And if x sits at the bottom? Flip the limits (which flips the sign of the integral) and apply the theorem: \frac{d}{dx}\int_x^{b} f(t)\,dt = -f(x). A limit at both ends splits into two pieces through any convenient midpoint — one term per moving endpoint, each with its own sign and chain factor.

Nobody woke up one morning and proved FTC. The pieces assembled across a generation. Isaac Barrow — Newton's teacher at Cambridge, and the man who gave up his professorship so the young Newton could have it — proved a geometric theorem in the 1660s that is essentially FTC Part 1: the tangent to an area-curve is governed by the original curve's height. But he stated it in the heavy costume of classical geometry, and seems not to have grasped that he was holding a master key.

Newton and Leibniz, working independently (and later feuding bitterly about it), each saw what Barrow hadn't: that "areas undo tangents" is not a curiosity about one curve but an engine. Instead of inventing a bespoke trick for every area problem — the state of the art since Archimedes — you could solve them all at once by running differentiation backwards. That reversal is why the subject is called calculus, "a method of calculating": the theorem turned two thousand years of one-off ingenuity into an algorithm. It is the reason this page's theorem gets the adjective fundamental.

See it explained