Fundamental Theorem of Calculus

April 22, 2026

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 bounded1 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 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 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 \(f\) 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 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 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 This theorem is fundamental because it unifies differentiation and integration, the two central operations of calculus.

—David Álvarez Rosa