Fundamental Theorem of Calculus

Although the notion of area is intuitive, its mathematical treatment requires a rigorous definition. This post introduces the Riemann integral, and proves the fundamental theorem of calculus—a beautiful result that connects integrals and derivatives.

Riemann integral §

Given a bounded^{1 1} Note that continuity is not required here; boundedness alone ensures the subinterval infima and suprema are finite.  function $f\colon[a,b]\to\mathbb{R}$, we can approximate the area under its graph by rectangles. Choose a partition of its domain

$$\mathcal{P}=\{x_0,x_1,\ldots,x_n\mid a=x_0<x_1<\cdots<x_n=b\}.$$

For each subinterval $[x_{k-1},x_k]$, define the width $\Delta x_k=x_k-x_{k-1}$, and let $m_k$ and $M_k$ denote the infimum and supremum of $f$ on that subinterval. The lower and upper sums are

$$L(f,\mathcal{P})=\sum_{k=1}^{n}m_k\Delta x_k, \qquad U(f,\mathcal{P})=\sum_{k=1}^{n}M_k\Delta x_k.$$

We define $f$ to be Riemann integrable^{2 2} Every continuous function on $[a,b]$ is Riemann integrable; so is every monotone function. The exact characterization is Lebesgue’s criterion: $f$ is Riemann integrable iff it is bounded and continuous almost everywhere.  on $[a,b]$ iff for every $\varepsilon>0$ there exists a partition $\mathcal{P}$ such that $U(f,\mathcal{P})-L(f,\mathcal{P})<\varepsilon$, in which case

$$\int_a^b f =\sup_{\mathcal{P}}L(f,\mathcal{P}) =\inf_{\mathcal{P}}U(f,\mathcal{P}).$$

Calculus machinery §

The proof requires the mean value theorem, which in turn rests on Rolle’s theorem and Fermat’s proposition.

Fermat’s Proposition. Let $I\subset\mathbb{R}$ be open and $f\colon I\to\mathbb{R}$ differentiable at $a\in I$. If $f$ has a local extremum at $a$, then $f^{\prime}(a)=0$.

Proof. Assume $f$ has a local maximum^{3 3} The local minimum case is identical, with all inequalities reversed.  at $a$. Then there exists $\delta>0$ such that $f(x)-f(a)\le 0$ for all $x\in(a-\delta,a+\delta)$. Therefore

$$\frac{f(x)-f(a)}{x-a}\ge 0 \quad (x<a), \qquad \frac{f(x)-f(a)}{x-a}\le 0 \quad (x>a).$$

Taking limits, $f^{\prime}_-(a)\ge 0$ and $f^{\prime}_+(a)\le 0$. Since $f$ is differentiable at $a$, $f^{\prime}_-(a)=f^{\prime}_+(a)=f^{\prime}(a)$, hence $f^{\prime}(a)=0$. $\square$

Rolle’s Theorem. If $f\colon[a,b]\to\mathbb{R}$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there exists $\xi\in(a,b)$ such that $f^{\prime}(\xi)=0$.

Proof. By the extreme value theorem,^{4 4} Topological result: $[a,b]$ is compact (Heine-Borel), the continuous image of a compact set is compact, and compact subsets of $\mathbb{R}$ are closed and bounded, so they contain their $\inf$ and $\sup$, which are finite.  $f$ attains its minimum $m$ and maximum $M$ on $[a,b]$. If $m=M$, then $f$ is constant and any $\xi\in(a,b)$ works. Otherwise, since $f(a)=f(b)$, at least one extremum is attained at some $\xi\in(a,b)$; by Fermat, $f^{\prime}(\xi)=0$. $\square$

Mean Value Theorem.^{5 5} Geometrically: there is always a point where the tangent line is parallel to the secant through the endpoints.  If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $\xi\in(a,b)$ such that

$$f^{\prime}(\xi)=\frac{f(b)-f(a)}{b-a}.$$

Proof. Define

$$g(x)=f(a)+\frac{f(b)-f(a)}{b-a}(x-a), \qquad h(x)=f(x)-g(x).$$

Then $h$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $h(a)=h(b)=0$. By Rolle’s theorem, there exists $\xi\in(a,b)$ with $h^{\prime}(\xi)=0$, which gives

$$f^{\prime}(\xi)-\frac{f(b)-f(a)}{b-a}=0.\,\square$$

Fundamental theorem of calculus §

We now have everything needed to prove the main result.

Fundamental Theorem of Calculus.^{6 6} A broader formulation also includes the statement that $x\mapsto\int_a^x f(t)\,dt$ is an antiderivative of $f$ under suitable regularity assumptions.  Let $f\colon[a,b]\to\mathbb{R}$ be Riemann integrable, and let $F\colon[a,b]\to\mathbb{R}$ be continuous on $[a,b]$, differentiable on $(a,b)$, and satisfy $F^{\prime}(x)=f(x)$ for all $x\in(a,b)$. Then

$$\int_a^b f = F(b)-F(a).$$

Proof. Fix a partition $\mathcal{P}=\{x_0,\ldots,x_n\}$. For each $[x_{k-1},x_k]$, the mean value theorem applied to $F$ gives $z_k\in(x_{k-1},x_k)$ such that

$$F(x_k)-F(x_{k-1})=f(z_k)\,\Delta x_k.$$

Since $m_k\le f(z_k)\le M_k$, we obtain

$$L(f,\mathcal{P}) \le \sum_{k=1}^{n}\left(F(x_k)-F(x_{k-1})\right) = F(b)-F(a) \le U(f,\mathcal{P}).$$

Taking supremum and infimum over all partitions and using integrability, we get

$$\int_a^b f=F(b)-F(a).\,\square$$

Thus computing an area reduces to evaluating an antiderivative at two points.^{7 7} For example, $\int_0^1 x^2\,dx = F(1)-F(0) = 1/3$ with $F(x)=x^3/3$. No partitions needed.  This theorem is fundamental because it unifies differentiation and integration, the two central operations of calculus.