The Fundamental Theorem, Part 2

Part 1 told us area-so-far is an antiderivative. Part 2 cashes that in. It says you can compute a definite integral without ever summing a rectangle — just find any antiderivative and subtract its values at the endpoints:

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

This is the line that makes calculus a computational subject. The shorthand \big[F(x)\big]_a^b := F(b) - F(a) is the evaluation bracket you will write a thousand times.

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).\;}

And it did not matter which antiderivative F we chose: a different one shifts both F(b) and F(a) by the same constant, which cancels in the subtraction. The +C is irrelevant to a definite integral.

Let us actually use it. To integrate x^2 we need an antiderivative, which the reverse power rule supplies: F(x) = \tfrac{x^3}{3} (take C = 0). Then

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

Compare the labour: in Riemann sums we fought a limit of growing sums to get such a number. Here it is two substitutions and a subtraction. That is the power Part 2 hands you.

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

The definite integral is the net change of the antiderivative: \int_{a}^{b} f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a).
The answer is independent of which antiderivative is used — the constant of integration cancels in the subtraction.
Equivalently (the net change theorem): the integral of a rate of change F' over [a,b] is the total change F(b) - F(a) it produces.

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

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.