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Calculus I / The Integral

Derivation — Proof of the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is stated and applied in the preceding lesson. This derivation note supplies the rigorous proofs of both parts. Both proofs rely on the comparison properties of the integral and the continuity of ff: Part I uses the squeeze theorem applied to [g(x+h)−g(x)]/h[g(x+h) - g(x)]/h; Part II follows from Part I by the corollary that two functions with the same derivative differ by a constant.
Proof of FTC Part I. Let g(x)=∫axf(t) dtg(x) = \int_a^x f(t)\,dt. We want to show g′(x)=f(x)g'(x) = f(x).

By the definition of the derivative:

g′(x)=lim⁡h→0g(x+h)−g(x)h.g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h}.

By additivity of the integral (Property 6):

g(x+h)−g(x)=∫ax+hf(t) dt−∫axf(t) dt=∫xx+hf(t) dt.g(x+h) - g(x) = \int_a^{x+h} f(t)\,dt - \int_a^x f(t)\,dt = \int_x^{x+h} f(t)\,dt.

So:

g(x+h)−g(x)h=1h∫xx+hf(t) dt.\frac{g(x+h) - g(x)}{h} = \frac{1}{h}\int_x^{x+h} f(t)\,dt.
Since ff is continuous on [a,b][a, b], by the Extreme Value Theorem it attains its minimum mm and maximum MM on [x,x+h][x, x+h] (for h>0h > 0). By the comparison property:
m⋅h≤∫xx+hf(t) dt≤M⋅h.m \cdot h \le \int_x^{x+h} f(t)\,dt \le M \cdot h.

Dividing by h>0h > 0:

m≤1h∫xx+hf(t) dt≤M.m \le \frac{1}{h}\int_x^{x+h} f(t)\,dt \le M.
The squeeze bound on [x, x+h]: the true area under f (blue curve) between x and x+h is trapped between the inscribed rectangle m·h (gray, using the minimum m of f on the sub-interval) and the circumscribed rectangle M·h (amber, using the maximum M)
The squeeze bound on [x, x+h]: the true area under f (blue curve) between x and x+h is trapped between the inscribed rectangle m·h (gray, using the minimum m of f on the sub-interval) and the circumscribed rectangle M·h (amber, using the maximum M)

Now m=f(u)m = f(u) and M=f(v)M = f(v) for some u,v∈[x,x+h]u, v \in [x, x+h]. As h→0+h \to 0^+, both u→xu \to x and v→xv \to x, so by continuity of ff:

lim⁡h→0+f(u)=f(x)andlim⁡h→0+f(v)=f(x).\lim_{h \to 0^+} f(u) = f(x) \quad \text{and} \quad \lim_{h \to 0^+} f(v) = f(x).

By the Squeeze Theorem:

g′(x)=lim⁡h→0+1h∫xx+hf(t) dt=f(x).g'(x) = \lim_{h \to 0^+} \frac{1}{h}\int_x^{x+h} f(t)\,dt = f(x).

An analogous argument for h→0−h \to 0^- completes the proof. □\square

Proof of FTC Part II. Let FF be any antiderivative of ff on [a,b][a, b]: F′=fF' = f. By FTC Part I, the function g(x)=∫axf(t) dtg(x) = \int_a^x f(t)\,dt also satisfies g′=fg' = f. Since F′=g′F' = g' on (a,b)(a, b), by the corollary of the Mean Value Theorem (functions with identical derivatives differ by a constant):
F(x)=g(x)+Cfor all x∈[a,b].F(x) = g(x) + C \quad \text{for all } x \in [a, b].

Evaluating at x=ax = a: g(a)=∫aaf(t) dt=0g(a) = \int_a^a f(t)\,dt = 0, so F(a)=0+C=CF(a) = 0 + C = C.

Evaluating at x=bx = b: F(b)=g(b)+C=∫abf(t) dt+F(a)F(b) = g(b) + C = \int_a^b f(t)\,dt + F(a).

Rearranging:

∫abf(t) dt=F(b)−F(a).□\int_a^b f(t)\,dt = F(b) - F(a). \quad\square
Remark 1 (Why the constant cancels). The proof shows explicitly why the choice of FF does not matter: the constant CC appears in both F(b)F(b) and F(a)F(a) and cancels in the difference. Any antiderivative works.
Remark 2 (Role of continuity). Continuity of ff is used in two places: to guarantee that gg is well-defined (integrability of continuous functions), and to pass the limit through the squeeze (f(u)→f(x)f(u) \to f(x) as u→xu \to x). The theorem fails for discontinuous integrands where the EVT does not apply on the tiny sub-interval [x,x+h][x, x+h].

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