# Fundamental Theorem of Calculus

> From the Riemann integral to the keystone of calculus.

By David Álvarez Rosa, 22 April 2026. Canonical: https://david.alvarezrosa.com/posts/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 {#riemann-integral}

Given a bounded[^fn:1] 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[^fn:2] 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 {#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[^fn:3] 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,[^fn: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.**[^fn: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 {#fundamental-theorem-of-calculus}

We now have everything needed to prove the main result.

**Fundamental Theorem of Calculus.**[^fn:6]  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.[^fn:7]  This theorem is fundamental
because it unifies differentiation and integration, the two central
operations of calculus.

[^fn:1]: Note that continuity is not required here;
    boundedness alone ensures the subinterval infima and suprema are
    finite.
[^fn: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.
[^fn:3]: The local minimum case is
    identical, with all inequalities reversed.
[^fn: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.
[^fn:5]: Geometrically: there is always a point where
    the tangent line is parallel to the secant through the endpoints.
[^fn: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.
[^fn:7]: For example, $\int_0^1 x^2\,dx = F(1)-F(0) = 1/3$ with
    $F(x)=x^3/3$.  No partitions needed.