Proof of FTC Part II

A Partition-Based Proof

The Fundamental Theorem of Calculus, Part II, says that if \(F\) is an antiderivative of a continuous function \(f\), then the integral of \(f\) is the change in \(F\) between the endpoints. The previous tutorial proved this by comparing \(F\) with the accumulation function from Part I. Here we give a different proof: divide the interval into small pieces and apply the mean value theorem on each one.

The key observation is local. On each subinterval, the mean value theorem identifies the change in \(F\) with the value of its derivative at some point, multiplied by the subinterval’s length. Adding these local changes makes the interior endpoint values cancel. The resulting sum is a Riemann sum for \(f\).

We will make the approximation precise before taking a limit. This yields an error bound that explains exactly why the partition method works. In what follows, a partition of \([a,b]\) is written \(P=\{x_0,\ldots,x_n\}\), where \(a=x_0<x_1<\cdots<x_n=b\), and its mesh is \(\|P\|=\max_i(x_i-x_{i-1})\).

The Mean Value Theorem on Each Subinterval

Proof. Fix a subinterval \([x_{i-1},x_i]\). The function \(F\) is continuous on this closed subinterval and differentiable in its interior, so the mean value theorem gives a point \(c_i\in(x_{i-1},x_i)\) with

Add these equalities for \(i=1,\ldots,n\). On the left, every value \(F(x_i)\) at an interior partition point occurs once with a positive sign and once with a negative sign. Thus

Substituting the mean value theorem expression for each difference proves the identity. \(\square\)

The points \(c_i\) are selected by the theorem; they need not be specified in advance. The sum in the proposition is a tagged Riemann sum, with each tag chosen from the interior of its subinterval. The identity itself is exact for every partition, even though the sum is not yet identified with the integral.

A Quantitative Error Estimate

Continuity of \(f\) on the closed bounded interval \([a,b]\) implies uniform continuity. Measure its variation over distances at most \(\delta\) by

This quantity is finite, and uniform continuity says that \(\omega_f(\delta)\) tends to zero as \(\delta\) tends to zero. If \(x\) and \(c_i\) lie in the same partition subinterval, their distance is at most the mesh, so the difference between \(f(x)\) and \(f(c_i)\) is bounded by \(\omega_f(\|P\|)\).

Proof. Write \(\Delta x_i=x_i-x_{i-1}\). For any \(x\in[x_{i-1},x_i]\), the distance from \(x\) to \(c_i\) is at most \(\Delta x_i\), and \(\Delta x_i\leq\|P\|\). Therefore

Integrating the resulting pointwise bounds on this subinterval and using order preservation of the Riemann integral gives

The integral of the constant function \(f(c_i)\) on this interval is \(f(c_i)\Delta x_i\). Subtracting that quantity from the bounds shows

Sum over all subintervals. Finite additivity of the integral across a partition gives \(\sum_i\int_{x_{i-1}}^{x_i}f=\int_a^b f\), while the Mean-Value Telescoping Identity gives \(\sum_i f(c_i)\Delta x_i=F(b)-F(a)\). The triangle inequality now yields

This proves the stated bound. \(\square\)

Proof of the Endpoint Formula

We now recover the Fundamental Theorem of Calculus, Part II, by making the mesh small. This is a second proof of the theorem stated in the previous tutorial; it does not use the accumulation function or the result about antiderivatives differing by a constant.

Proof. For each positive integer \(n\), take the uniform partition \(P_n\) with \(n\) subintervals of length \((b-a)/n\). Its mesh is \(\|P_n\|=(b-a)/n\), which tends to zero. Apply the Partition Error Bound for an Antiderivative:

Uniform continuity of \(f\) implies that the right-hand side tends to zero as \(n\) tends to infinity. The expression inside the absolute value on the left does not depend on \(n\). Its absolute value is bounded by quantities tending to zero, so it must be zero. Hence

This proves FTC Part II. \(\square\)

The proof has two distinct ingredients. The mean value theorem produces an exact sum for the change in \(F\), and uniform continuity ensures that this sum is close to the integral of \(f\). Telescoping connects the local changes to the endpoint difference; the shrinking mesh connects the sum to integration.

Worked Applications

What the Proof Does—and Does Not—Assume

A common pitfall is to use the mean value theorem on every partition interval without checking its hypotheses. For each closed subinterval, \(F\) must be continuous up to both endpoints and differentiable in the interior. The assumptions that \(F\) is continuous on \([a,b]\) and differentiable on \((a,b)\) guarantee this for every subinterval, including those touching \(a\) or \(b\). No derivative at either endpoint is required.

Another important point is the role of continuity of \(f\). It ensures uniform continuity on \([a,b]\), so the error bound tends to zero as the mesh shrinks. The argument is not claiming that every bounded function with an antiderivative can be treated in this way without further hypotheses. Continuity is part of the stated theorem and is what controls the approximation here.

The partition estimate also gives more than a limiting argument. If the variation of \(f\) on distances up to a chosen mesh is known, it supplies a numerical bound for the discrepancy between the endpoint change and the integral. For example, when \(\omega_f(\delta)\leq C\delta\), the error is at most \(C(b-a)\|P\|\). Thus the proof explains both why the formula holds and how partition size affects the approximation.