Teaching the Fundamental Theorem of Calculus: A Historical Reflection - The Question of Existence of an Integral for a Continuous Function on a Closed Bounded Interval

One of the most attractive features of the didactic approach to integration presented here is that it leads naturally to the discussion of more abstract analysis, specifically to Cauchy's limit-sum definition of the integral and to Darboux sums. In fact, early in his exposition, Gillman [20, p. 17] established the relationship between his axiomatic approach to the integral and the Darboux integral. His reasoning leads also to the Cauchy limit-sum definition of the integral. Of course, both these topics are, cognitively speaking, of a higher order of mathematical abstraction, and thus harder for students to understand. One advantage of our presentation, which differs a little from the presentation found in Lang [28] and in Gillman and McDowell [21], is that the question of the existence of an integral comes up as a natural consideration, thus leading to more advanced mathematics.

Of course, the question of the existence of an integral would have been completely anachronistic in the setting of the mathematics of the seventeenth century, when the fact that all curves had areas associated with them was something that was evident and not to be questioned. This “escalation” of the level of abstraction is consistent with the fact that, historically, this discussion belongs to the 18th century, a time in the history of mathematics when such existence questions were natural (and expected). The question of which functions (not necessarily continuous) can have integrals with properties analogous to the ones we have just discussed seems impossible to formulate within the mathematical mind frame of the 17th century.

Is there an integral for a continuous \(f\) on \([a,b]\)? If so, how can we find it? In this section we point out the connection between the axiomatic formulation of the integral presented here and the integral as a limit of Riemann sums. For the discussion we shall assume the existence of at least one such integral (as in Definition 1) in order to show that it must be a limit of Riemann sums. The deliberations that ensue from pursuing the question of the existence of an integral for a continuous function on a closed bounded interval are, as expected, of a higher level of mathematical abstraction. For this reason, this discussion is also harder for the student to sort out. We present it here to show that it fits nicely into our teaching proposal.

The next result opens the door to explore Darboux integration and introduces topics and issues characteristic of mathematical analysis. The point at hand is, of course, the existence of an integral for \(f\) on \([a,b]\). The result also introduces Riemann sums.

Theorem 4. Let \([u,v]\mapsto I_{u}^{v}(f)\) \((a\leq u\leq v\leq b)\) be an integral for a continuous function \(f:[a,b]\rightarrow\mathbb{R}\) on \([a,b]\). Then \[I_{a}^{b}(f)=\lim_{n\to\infty}\sum_{k=1}^{n}f(x_{k-1})\Delta x_{k},\quad\quad\quad\quad(7)\] where for each \(n\geq 1\), \(\{ a=x_0\,{\rm{<}}\,x_1\,{\rm{<}}\,\cdots\,{\rm{<}}\,x_n = b\}\) is the uniform subdivision of \([a,b]\) into \(n\) subintervals, with \[x_{k}={a+\frac{k(b-a)}{n}}\,\,{\rm{for}}\,\,{\rm{each}}\,\,k=0,1,\dots ,n\] and \[\Delta x_{k}=x_{k}-x_{k-1}=\frac{b-a}{n}\,\,{\rm{for}}\,\,{\rm{each}}\,\,k=1,2,\dots , n.\] In fact, if for the uniform subdivision \(\{ a=x_0\,{\rm{<}}\,x_1 < \cdots\,{\rm{<}}\,x_n = b\}\) of \([a,b]\) into \(n\) subintervals, we choose any \(z_{k-1}\in [x_{k-1},x_{k}]\) for each \(k=1,\dots ,n\), then \[I_{a}^{b}(f)=\lim_{n\to\infty}\sum_{k=1}^{n}f(z_{k-1})\Delta x_{k},\] also holds.

Note that this result does not settle the issue of the existence of an integral for a continuous function on a closed bounded interval, but it gives us a glimpse into the beginnings of Darboux integration.

Proof. 

We shall assume in this proof that there is an integral for \(f\) on \([a,b]\). Let \(F\) be an antiderivative of \(f\) on \([a,b]\). Let \(\{a=x_0\,{\rm{<}}\,x_1\,{\rm{<}}\,\cdots\,{\rm{<}}\,x_n=b\}\) be the uniform partition of \(n\) subintervals \((n\geq 1)\). Then \[F(b)-F(a)=\sum_{k=1}^{n}[F(x_{k})-F(x_{k-1})].\] By the Mean Value Theorem [22, p. 167], it is possible to choose \(x_{k}^{*}\in [x_{k-1},x_{k}]\) such that \[F(b)-F(a)=\sum_{k=1}^{n}[F(x_{k})-F(x_{k-1})]\] \[=\sum_{k=1}^{n}F^{\,\prime}(x_{k}^{*})\Delta x_{k}=\sum_{k=1}^{n}f(x_{k}^{*})\Delta x_{k}.\] As every continuous function on a closed bounded interval is uniformly continuous, given \(\epsilon\,{\rm{>}}\,0\), there is a \(\delta\,{\rm{<}}\,0\) so that \(|f(u)-f(v)|\,{\rm{<}}\,\epsilon/(b-a)\) whenever \(u,v\in [a,b]\) and \(|u-v|\,{\rm{<}}\,\delta\). Choose an integer \(n_{0}\geq 1\) such that \((b a)/n_{0}\,{\rm{<}}\,\delta\). Then, for any \(n\geq n_{0}\), \[\big|F(b)-F(a)-\sum_{k=1}^{n}f(x_{k-1})\Delta x_{k}\big|\] \[=\bigg|\sum_{k=1}^{n}f(x_{k}^{*})\Delta x_{k}-\sum_{k=1}^{n}f(x_{k-1})\Delta x_{k}\bigg|\] \[\leq\sum_{k=1}^{n}|f(x_{k}^{*})-f(x_{k-1})|\Delta x_{k}\] \[\,{\rm{<}}\,\frac{\epsilon}{b-a}\cdot\sum_{k=1}^{n}\Delta x _{k}\] \[=\frac{\epsilon}{b-a}\cdot (b-a)=\epsilon.\] As \(\epsilon\) is arbitrary, \[F(b)-F(a)=\lim_{n\to\infty}\sum_{k=1}^{n}f(x_{k-1})\Delta x_{k}.\] The desired result follows. It is clear that any other value \(f(z_{k-1})\) would work the same as \(f(x_{k-1})\) if we chose \(z_{k-1}\in [x_{k-1},x_{k}]\) \((k=1,\dots ,n\)), and this completes our proof.

Note that given any subdivision \(\{a=x_0\,{\rm{<}}\,x_1\,{\rm{<}}\,\cdots\,{\rm{<}}\,x_n =b\}\) (\(n\geq 1\)) of \([a,b]\), uniform or not, we have, by virtue of condition (B) of Definition 1, \[\sum_{k=1}^{n}\min_{u\in [x_{k-1},x_{k}]}f(u)\cdot\Delta x_{k}\leq I_{a}^{b}(f)\leq\sum_{k=1}^{n}\max_{u\in [x_{k-1},x_{k}]}f(u)\cdot\Delta x_{k}.\quad\quad\quad\quad(8)\] In the case of a continuous function \(y=f(x)\), the sums in inequalities (8) are in fact Riemann sums, and the inequalities are equivalent to  \[\sum_{k=1}^{n}\inf_{u\in [x_{k-1},x_{k}]}f(u)\cdot\Delta x_{k}\leq I_{a}^{b}(f)\leq\sum_{k=1}^{n}\sup_{u\in [x_{k-1},x_{k}]}f(u)\cdot\Delta x_{k}.\quad\quad\quad\quad(9)\] Of course, inequalities (9) make sense for bounded functions \(y=f(x)\), and this is clearly the starting point of Darboux integration.