Proof of Fundamental Theorem of Calculus, Part II: Evaluating an Integral by an Antiderivative

Let $\Pi_{a,b}$ be the set of partitions of $[a,b]$, and let $P\in\Pi_{a,b}$; by Partitions of a Closed Interval, the Set of Partitions, and Refinements (ii), $P$ is a partition of $[a,b]$. Let $n\in\mathbb{N}$ and the increasing listing $a=x_{1}<x_{2}<\dots<x_{n+1}=b$ of $P$ be as in The Increasing Listing of a Partition: Endpoints, Lengths, Covering, Unions, Inserting a Point, Concatenation and Uniform Partitions (i), and for $i\in[n]$, where $x_{i}$ and $x_{i+1}$ are defined by The Increasing Listing of a Partition: Endpoints, Lengths, Covering, Unions, Inserting a Point, Concatenation and Uniform Partitions (ii), let $S_{i}=\{f(t):t\in[x_{i},x_{i+1}]\}$ be as in Upper and Lower Darboux Sums of a Bounded Function on a Partition (i), and write $M_{i}:=M(i)=\sup S_{i}$ and $m_{i}:=m(i)=\inf S_{i}$ for the values of the maps $M,m:[n]\to\mathbb{R}$ of Upper and Lower Darboux Sums of a Bounded Function on a Partition (i).

Step 1: one subinterval. Let $i\in[n]$. Then $x_{i},x_{i+1}\in P\subseteq[a,b]$ by The Increasing Listing of a Partition: Endpoints, Lengths, Covering, Unions, Inserting a Point, Concatenation and Uniform Partitions (ii) and Partitions of a Closed Interval, the Set of Partitions, and Refinements (i), and $x_{i}<x_{i+1}$ by The Increasing Listing of a Partition: Endpoints, Lengths, Covering, Unions, Inserting a Point, Concatenation and Uniform Partitions (iii). By Basic Facts about Intervals of the Real Line (v), $[a,b]$ is an interval. If $x\in[a,b]$ and $u<x<v$ for some $u,v\in[a,b]$, then $a\le u<x<v\le b$, so $x\in(a,b)$ and $F$ is differentiable at $x$. Thus Mean Value Theorem on a Closed Real Interval and on Any Interval (ii), applied with $I=[a,b]$ to $F$, derivatives being taken relative to $[a,b]$, and to the points $x_{i}<x_{i+1}$ of $[a,b]$, gives $\xi\in[a,b]$ with $x_{i}<\xi<x_{i+1}$ and $F(x_{i+1})-F(x_{i})=F'(\xi)\,(x_{i+1}-x_{i})$. As $a\le x_{i}<\xi<x_{i+1}\le b$, $\xi\in(a,b)$, so $F'(\xi)=f(\xi)$ and

$$F(x_{i+1})-F(x_{i})=f(\xi)\,(x_{i+1}-x_{i}).$$

As $\xi\in[x_{i},x_{i+1}]$, $f(\xi)\in S_{i}$ by Upper and Lower Darboux Sums of a Bounded Function on a Partition (i), so $m_{i}\le f(\xi)\le M_{i}$, an infimum being a lower bound and a supremum an upper bound of $S_{i}$. Multiplying by $x_{i+1}-x_{i}>0$,

$$m_{i}\,(x_{i+1}-x_{i})\le F(x_{i+1})-F(x_{i})\le M_{i}\,(x_{i+1}-x_{i}).\qquad\text{(1)}$$

The point $\xi$ does not occur in (1), so (1) holds for every $i\in[n]$; no choice of a family of points is made.

Step 2: summing. The map $d:[n+1]\to\mathbb{R}$, $k\mapsto F(x_{k})$, is defined as $x_{k}\in[a,b]$ for $k\in[n+1]$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (v)); for $i\in[n]$, both $i$ and $i+1$ lie in $[n+1]$ by The Increasing Listing of a Partition: Endpoints, Lengths, Covering, Unions, Inserting a Point, Concatenation and Uniform Partitions (ii), so $d(i)=F(x_{i})$ and $d(i+1)=F(x_{i+1})$. By The Map of Consecutive Differences of a Finite List of Reals Exists, and Its Sum Telescopes (i), applied to $d$, the map $[n]\to\mathbb{R}$, $i\mapsto F(x_{i+1})-F(x_{i})$, exists and

$$\sum_{i=1}^{n}\bigl(F(x_{i+1})-F(x_{i})\bigr)=F(x_{n+1})-F(x_{1})=F(b)-F(a).$$

Summing (1) over $i\in[n]$ by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (v), and recalling the lower sum and the upper sum of Upper and Lower Darboux Sums of a Bounded Function on a Partition (iii) and Upper and Lower Darboux Sums of a Bounded Function on a Partition (ii), we obtain

$$L(f,P)\le F(b)-F(a)\le U(f,P).$$

Step 3: conclusion. As $P\in\Pi_{a,b}$ was arbitrary, $F(b)-F(a)$ is an upper bound of the set of lower sums $\mathcal{L}(f)=\{L(f,P):P\in\Pi_{a,b}\}$ and a lower bound of the set of upper sums $\mathcal{U}(f)=\{U(f,P):P\in\Pi_{a,b}\}$, by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (i) and The Riemann Integral on a Closed Interval via Lower and Upper Integrals (ii). By The Real Line: Standing Notation and Background for Calculus on Intervals (iii), the supremum of the first set is at most $F(b)-F(a)$ and the infimum of the second set is at least $F(b)-F(a)$, so by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (iii)

$$\underline{\int_{a}^{b}}f\le F(b)-F(a)\le\overline{\int_{a}^{b}}f .$$

Since $f$ is integrable on $[a,b]$, both the lower and the upper integral equal $\int_{a}^{b}f$, by The Riemann Integral on a Closed Interval via Lower and Upper Integrals (v). Hence $\int_{a}^{b}f=F(b)-F(a)$, which is (i).