Fundamental Theorem of Calculus

# Fundamental Theorem of Calculus | David Álvarez Rosa | Personal Website Source: [https://david.alvarezrosa.com/posts/fundamental-theorem-of-calculus/](https://david.alvarezrosa.com/posts/fundamental-theorem-of-calculus/) April 22, 2026Although 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[§](https://david.alvarezrosa.com/posts/fundamental-theorem-of-calculus/#riemann-integral) Given a bounded11Note 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 integrable22Every 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[§](https://david.alvarezrosa.com/posts/fundamental-theorem-of-calculus/#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 maximum33The 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,44Topological 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\.**55Geometrically: 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[§](https://david.alvarezrosa.com/posts/fundamental-theorem-of-calculus/#fundamental-theorem-of-calculus) We now have everything needed to prove the main result\. **Fundamental Theorem of Calculus\.**66A 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\.77For 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 - [Home](https://david.alvarezrosa.com/) - [About](https://david.alvarezrosa.com/about/) - [Posts](https://david.alvarezrosa.com/posts/) - [RSS](https://david.alvarezrosa.com/index.xml)