April 22, 2026

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 bounded1 1 Note that continuity is not required here; boundedness alone ensures the subinterval infima and suprema are finite.  function \(f:[a,b]\to\mathbb{R}\), 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 integrable2 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 \(f:I\to\mathbb{R}\) differentiable at \(a\in I\). If \(f\) has a local extremum at \(a\), then \(f^{\prime}(a)=0\).

Proof. Assume \(f\) has a local maximum3 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 \(f:[a,b]\to\mathbb{R}\) 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 \(f:[a,b]\to\mathbb{R}\) be Riemann integrable, and let \(F:[a,b]\to\mathbb{R}\) 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.

—David Álvarez Rosa

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