The Fundamental Theorem, Part 2

For two thousand years, finding the exact area under a curve was a heroic act. Archimedes needed an entire treatise to pin down one parabola. Even with Riemann sums, every new curve means slicing, summing, and wrestling a limit — a fresh rectangle-counting campaign each time.

Part 1 told us that area-so-far is an antiderivative. Part 2 cashes that in. It says you can compute a definite integral without ever summing a rectangle: find any antiderivative, plug in the two endpoints, subtract. Done.

\int_{a}^{b} f(x)\,dx = F(b) - F(a), \qquad\text{where } F' = f.

Read that again, because it is the single greatest labour-saving device in mathematics. An infinite summation on the left; one function lookup and one subtraction on the right. This is the line that turned calculus from a philosophical marvel into a computational subject — the line that put the rectangle-counting industry out of business overnight.

The shorthand for "evaluate at b, evaluate at a, subtract" is the evaluation bracket you will write a thousand times:

\big[F(x)\big]_a^b := F(b) - F(a).

Three worked examples — feel the shortcut

Example 1 — a plain area. Find \int_1^3 x^2\,dx. The reverse power rule hands us an antiderivative, F(x) = \tfrac{x^3}{3}. Now evaluate at the top limit, evaluate at the bottom limit, subtract:

\int_1^3 x^2\,dx = \left[\frac{x^3}{3}\right]_1^3 = \frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3} \approx 8.67.

That exact value — \tfrac{26}{3}, not an estimate — took three substitutions and one subtraction. By Riemann sums it would have taken an algebraic siege.

Example 2 — negative contributions. Find \int_0^{2\pi} \sin x\,dx. An antiderivative of \sin x is F(x) = -\cos x (check: F'(x) = \sin x ✓). Then

\int_0^{2\pi} \sin x\,dx = \big[-\cos x\big]_0^{2\pi} = \big(-\cos 2\pi\big) - \big(-\cos 0\big) = (-1) - (-1) = 0.

Zero? Look at the graph of \sin x: from 0 to \pi it arches above the axis (area +2), and from \pi to 2\pi it dips below (area -2). The definite integral is the net signed area, and FTC2 tracks the signs for you automatically — the hill and the valley cancel to exactly nothing. If you want the total painted area, split at the crossing: \int_0^{\pi}\sin x\,dx = [-\cos x]_0^{\pi} = 1-(-1) = 2, twice, giving 4.

Example 3 — speed to distance, with units. A car accelerates from rest so that its speed after t seconds is v(t) = 12t - 3t^2 metres per second, for 0 \le t \le 4. How far does it travel? Distance is the integral of speed, and an antiderivative of v is s(t) = 6t^2 - t^3:

\int_0^4 (12t - 3t^2)\,dt = \big[\,6t^2 - t^3\,\big]_0^4 = (6\cdot 16 - 64) - (0 - 0) = 96 - 64 = 32\ \text{m}.

Notice the units book-keep themselves: metres-per-second times seconds gives metres. This is the net change theorem in action — integrate a rate of change (\text{m/s}) over a time interval and you get the total change (\text{m}). The same pattern computes water pumped from a flow rate, charge from a current, population growth from a birth rate.

Why the +C never matters

An antiderivative is only determined up to a constant — so which one should you use? Watch what happens if we redo Example 1 with a different antiderivative, say F(x) = \tfrac{x^3}{3} + 7 (also valid: F'(x) = x^2):

\left[\frac{x^3}{3} + 7\right]_1^3 = \left(\frac{27}{3} + 7\right) - \left(\frac{1}{3} + 7\right) = \frac{27}{3} - \frac{1}{3} + \underbrace{7 - 7}_{=\,0} = \frac{26}{3}.

The +7 rides in at the top limit and rides out at the bottom limit — it cancels in the subtraction, every time, whatever the constant. In general, for any constant C:

\big(F(b) + C\big) - \big(F(a) + C\big) = F(b) - F(a).

So for definite integrals you may take C = 0 and never think about it again. The +C matters for indefinite integrals (where the answer is a whole family of functions); it is invisible to definite ones (where the answer is a single number).

The proof — line by line (Part 2 from Part 1)

We are handed some antiderivative F of f (any one will do) and must show \int_a^b f = F(b) - F(a).

Step 1 — build the accumulation function. Define G as the area-so-far from a:

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

Step 2 — differentiate it (Part 1). Since f is continuous, the Fundamental Theorem Part 1 gives

G'(x) = f(x).

Step 3 — two antiderivatives of the same f. Now G' = f and (by assumption) F' = f. By the antiderivatives-differ-by-a-constant theorem, they differ by a constant:

G(x) = F(x) + C \qquad\text{for some constant } C.

Step 4 — pin down C at x = a. The area from a to a is zero, so G(a) = \int_a^a f = 0. Substituting,

0 = G(a) = F(a) + C \quad\Longrightarrow\quad C = -F(a).

Step 5 — evaluate at x = b. With C = -F(a),

G(b) = F(b) + C = F(b) - F(a).

Step 6 — identify G(b) as the integral we wanted. By its definition, G(b) = \int_a^b f(t)\,dt. Therefore

\boxed{\;\int_{a}^{b} f(x)\,dx = F(b) - F(a).\;}

Six lines, and the engine of Part 1 does all the heavy lifting. Notice where the constant went: the whole proof is really the observation that the area function G and your antiderivative F can only disagree by a constant, and evaluating at a tells you exactly which constant. The subtraction F(b) - F(a) is what silently removes it.

Let f be continuous on [a, b] and let F be any antiderivative of f (so F' = f). Then:

Trap 1 — the integrand must be continuous on the whole interval. FTC2 comes with a licence condition, and ignoring it produces confident nonsense. Try it on \int_{-1}^{1} \tfrac{1}{x^2}\,dx. An antiderivative of \tfrac{1}{x^2} is F(x) = -\tfrac{1}{x}, so the machine happily grinds out

\left[-\frac{1}{x}\right]_{-1}^{1} = \left(-\frac{1}{1}\right) - \left(-\frac{1}{-1}\right) = -1 - 1 = -2. \quad\text{(!?)}

A negative "area" — for a function that is strictly positive everywhere it is defined. That is impossible, so something has gone badly wrong. The culprit: \tfrac{1}{x^2} blows up to infinity at x = 0, which sits right inside [-1, 1]. The integrand is not continuous on the interval, the hypothesis of the theorem fails, and the formula's output is garbage. (The true integral diverges — the area under that spike is infinite.) Before you reach for the bracket, glance along [a, b] for holes, jumps and vertical asymptotes.

Trap 2 — actually subtract F(a), with its sign. In exam-hall haste, two errors account for most lost marks: evaluating only F(b) and forgetting the bottom limit entirely, and fumbling the signs when F(a) is negative. Compute \int_{-2}^{1} x^2\,dx carefully:

\left[\frac{x^3}{3}\right]_{-2}^{1} = \frac{1}{3} - \frac{(-2)^3}{3} = \frac{1}{3} - \left(-\frac{8}{3}\right) = \frac{1}{3} + \frac{8}{3} = 3.

The hasty version, \tfrac13 - \tfrac83 = -\tfrac73, is negative — and x^2 \ge 0, so a negative answer is a flashing red light. A sanity-check on the sign of your answer catches this error almost every time. Wrap F(a) in brackets, minus sign and all, before simplifying.

Around 240 BC, Archimedes proved that the area of a parabolic segment is \tfrac43 the area of an inscribed triangle. His method — the method of exhaustion — packed the region with infinitely many ever-smaller triangles, summed the geometric series 1 + \tfrac14 + \tfrac1{16} + \cdots, then sealed the result with a double proof by contradiction (the area can be neither more nor less than \tfrac43). It is one of the greatest arguments of antiquity, and it fills a treatise: Quadrature of the Parabola, twenty-four propositions long.

Here is that result today:

\int_0^1 x^2\,dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3} - 0 = \frac{1}{3}.

Two lines. When Newton and Leibniz found this bridge between slopes and areas in the 1660s–70s, they didn't just solve one problem — they made every such problem routine. Areas that had each demanded a bespoke stroke of genius became homework. (The tidy evaluation bracket \big[F(x)\big]_a^b came later still — 19th-century notational housekeeping for a theorem that had already changed the world.) It is worth pausing on how rare this is: mathematics does not often hand you a machine that converts "lifetime achievement" into "exercise 3(b)".

The thunderclap

Put the two halves of the theorem side by side and the whole of calculus snaps into focus.

Part 1 says: take the area-so-far, then differentiate, and you recover the integrand —

\frac{d}{dx}\int_a^x f(t)\,dt = f(x).

Part 2 says: take an antiderivative, then integrate its derivative across an interval, and you recover the net change —

\int_a^b F'(x)\,dx = F(b) - F(a).

One reads "differentiate ∘ integrate = identity"; the other reads "integrate ∘ differentiate = identity (up to the endpoints)." Together they declare that differentiation and integration are inverse operations — two doors into the same room. The unbounded labour of summing infinitely many rectangles and the local act of measuring a slope turn out to be the very same arithmetic, run forwards and backwards. That is the thunderclap at the centre of the subject.

Evaluate by subtracting endpoints

Now watch the theorem breathe. The curve is f(x) = x^2, whose antiderivative is F(x) = \tfrac{x^3}{3}. Slide the limits a and b; the region under the curve between them shades, and the readout shows the area computed two ways that always agree: as the definite integral, and as F(b) - F(a) = \tfrac{b^3}{3} - \tfrac{a^3}{3}. No rectangles required.

Two things worth noticing as you play. First, drag a up towards b: the shaded sliver shrinks and the readout glides towards zero — that is \int_a^a f = 0, the fact the proof leaned on in Step 4. Second, nudge b a little to the right where the curve is tall: the area jumps by more than the same nudge does where the curve is low. The rate at which the readout grows is the height of the curve itself — which is exactly Part 1, living inside the Part 2 picture.

See it explained