Real Analysis 56 | Proof of the Fundamental Theorem of Calculus [dark version]

TL;DR

The Mean Value Theorem for integration guarantees an x̂ with ∫[a to b] f(x)G(x)dx = f(x̂)∫[a to b]G(x)dx when G(x)≥0 and f is continuous.

Briefing Cornell Notes

Briefing

The proof of the Fundamental Theorem of Calculus hinges on a single tool: the Mean Value Theorem for integrals, which guarantees that an integral of a product can be matched by a rectangle using some intermediate point. That “rectangle point” exists because continuity forces the function to hit every intermediate value, and nonnegativity of the weighting function keeps inequalities from flipping. Once this bridge is built, both halves of the Fundamental Theorem fall into place.

First comes the Mean Value Theorem for integration in its general form: for continuous functions f and a nonnegative function G on [a,b], there exists a point x̂ in [a,b] such that ∫[a to b] f(x)G(x) dx = f(x̂) ∫[a to b] G(x) dx. A simpler special case comes from taking G(x)=1, yielding ∫[a to b] f(x) dx = f(x̂)(b−a). Geometrically, the integral equals the area under f, and the theorem asserts there is a rectangle of width (b−a) whose height is f(x̂) that has the same area.

The proof starts by using continuity of f on the closed interval to guarantee minimum and maximum values: f(x) is trapped between m (the minimum) and M (the maximum) for every x in [a,b]. Multiplying these inequalities by G(x) preserves the order because G(x)≥0, producing bounds on the product f(x)G(x). Integrating those inequalities uses monotonicity of the integral to keep the inequalities intact, giving m∫ G ≤ ∫ fG ≤ M∫ G. Since f(x̂) must land between m and M, continuity and the Intermediate Value Theorem ensure there is some x̂ where f(x̂) equals the specific “middle” value μ that makes the equality true. With that, the Mean Value Theorem for integrals is established.

With the integral mean value tool in hand, the first Fundamental Theorem of Calculus is proved by constructing an antiderivative F(x)=∫[a to x] f(t)dt and showing F′(x)=f(x). The difference quotient F(x+H)−F(x) becomes an integral over [x,x+H]. Applying the Mean Value Theorem for integrals turns that integral into f(x̂)·H for some x̂ inside the interval. Dividing by H leaves f(x̂), and as H→0, x̂ is forced toward x; continuity then gives the limit f(x̂)→f(x). Hence F is differentiable and its derivative equals f.

The second Fundamental Theorem then follows from the fact that any antiderivative works. Since one particular antiderivative F0 already satisfies ∫[a to b] f = F0(b)−F0(a), any other antiderivative F must differ by an additive constant C. That constant cancels in the difference F(b)−F(a), leaving the same formula for ∫[a to b] f(x)dx = F(b)−F(a). The result is a complete two-part proof: integrals generate antiderivatives, and antiderivatives evaluate integrals.

Cornell Notes

The Mean Value Theorem for integration says that for continuous f and nonnegative G on [a,b], there exists x̂ such that ∫[a to b] f(x)G(x)dx = f(x̂)∫[a to b]G(x)dx. A key special case with G(x)=1 turns the integral into a rectangle area: ∫[a to b] f(x)dx = f(x̂)(b−a). The proof uses continuity to obtain minimum and maximum values of f, multiplies inequalities by G(x)≥0, integrates to bound ∫ fG, then uses the Intermediate Value Theorem to pick x̂ where f(x̂) matches the needed intermediate value. This rectangle principle then proves the Fundamental Theorem of Calculus: differentiating F(x)=∫[a to x]f gives F′(x)=f(x), and evaluating ∫[a to b]f equals F(b)−F(a) for any antiderivative F.

How does the Mean Value Theorem for integration justify replacing an integral by a rectangle?

For continuous f on [a,b] and nonnegative G, the theorem guarantees some x̂ in [a,b] with ∫[a to b] f(x)G(x)dx = f(x̂)∫[a to b]G(x)dx. In the special case G(x)=1, this becomes ∫[a to b] f(x)dx = f(x̂)(b−a). Geometrically, the area under f equals the area of a rectangle whose width is (b−a) and whose height is f(x̂), meaning the graph of f hits exactly the height needed to match the integral’s area.

Why does nonnegativity of G matter in the proof?

The proof begins with m ≤ f(x) ≤ M for all x in [a,b], where m and M are the minimum and maximum of f. Multiplying by G(x) preserves inequalities only because G(x)≥0. If G could be negative, the inequality directions could flip, breaking the later bounds on ∫ fG.

What role do continuity and the Intermediate Value Theorem play in choosing x̂?

Continuity on a closed interval ensures f attains a minimum m and maximum M. After integrating the inequality bounds, the integral of fG is squeezed between m∫G and M∫G. The proof then defines a value μ between m and M so that μ∫G equals ∫ fG. Since μ lies between the minimum and maximum, continuity plus the Intermediate Value Theorem guarantees there exists x̂ with f(x̂)=μ.

How does the Mean Value Theorem for integration prove that F′(x)=f(x) for F(x)=∫[a to x]f?

Consider the difference quotient: (F(x+H)−F(x))/H. The numerator equals ∫[x to x+H] f(t)dt. Applying the special mean value theorem gives ∫[x to x+H] f(t)dt = f(x̂)·H for some x̂ in [x,x+H]. Dividing by H yields f(x̂). As H→0, the interval collapses, forcing x̂→x; continuity then gives f(x̂)→f(x), so the derivative exists and equals f(x).

Why does the second Fundamental Theorem work for any antiderivative, not just one specific F0?

If F is any antiderivative of f, then F differs from a known antiderivative F0 by an additive constant C: F = F0 + C. In the evaluation formula, F(b)−F(a) = (F0(b)+C)−(F0(a)+C) = F0(b)−F0(a), so the constant cancels. Since the formula already holds for F0, it holds for every antiderivative.

Review Questions

In the special case G(x)=1, what equation shows the integral equals a rectangle area, and what are the rectangle’s width and height?
Where exactly does the proof use continuity of f: to get min/max values, to apply the Intermediate Value Theorem, or both? Explain.
In the derivative proof, why does the difference quotient reduce to f(x̂) and why does f(x̂) approach f(x) as H→0?

Key Points

1
The Mean Value Theorem for integration guarantees an x̂ with ∫[a to b] f(x)G(x)dx = f(x̂)∫[a to b]G(x)dx when G(x)≥0 and f is continuous.
2
The special case G(x)=1 turns ∫[a to b] f(x)dx into f(x̂)(b−a), matching the integral’s area with a rectangle’s area.
3
Continuity on [a,b] ensures f attains a minimum m and maximum M, enabling tight inequality bounds on f(x)G(x).
4
Multiplying inequalities by G(x) preserves order only because G(x) is nonnegative.
5
After bounding the integral, the Intermediate Value Theorem supplies x̂ such that f(x̂) equals the required intermediate value μ.
6
For F(x)=∫[a to x]f(t)dt, the difference quotient becomes ∫[x to x+H]f(t)dt, which the mean value theorem turns into f(x̂)·H; dividing by H yields f(x̂).
7
Any two antiderivatives differ by an additive constant, and that constant cancels in F(b)−F(a), making the integral evaluation formula universal.

Highlights

The integral mean value theorem turns ∫[a to b] f(x)dx into f(x̂)(b−a), asserting the graph of f hits the exact height needed to match the area.

The Fundamental Theorem’s first half follows by applying the integral mean value theorem to ∫[x to x+H]f(t)dt inside the difference quotient.

The second half becomes automatic once it’s known that antiderivatives differ only by a constant, which cancels in F(b)−F(a).