Fundamental theorems of calculus (FTC1 and FTC2)

Intuition [Beginner]

A derivative tells you how fast something is changing right now. An integral tells you how much has accumulated over a stretch of time. The fundamental theorems of calculus say these two operations are inverses: each one undoes the other.

Imagine driving at a steadily increasing speed. Your speedometer reads the rate of change (derivative). The odometer tracks the total distance covered (integral). If you know the speed at every instant, you can compute the total distance. If you know the total distance at every moment, you can read off the speed. That two-way relationship is the entire content of the two fundamental theorems.

The first theorem says: build a running-total function by accumulating area under a curve, then take the derivative of that running total, and you get back the original curve. The second says: the total area under a curve between two points equals the change in any antiderivative (a function whose derivative is the original curve).

This concept exists because without it, every area computation requires a separate geometric argument. The FTC reduces all of them to a single algebraic step: find an antiderivative, plug in the endpoints, subtract.

Visual [Beginner]

The picture shows a coordinate plane with a rising curve $y=f(x)$ drawn above the horizontal axis from $x=0$ to $x=5$. A shaded region fills the area between the curve and the axis, growing wider from left to right. Below the shaded region, a second graph shows the accumulated-area function $A(x)$, which rises steeply where $f$ is tall and rises gently where $f$ is small. A tangent line on the lower graph at $x=3$ has slope equal to $f(3)$ on the upper graph.

The slope of the lower graph at any point equals the height of the upper graph directly above it. This is the first fundamental theorem in visual form.

Worked example [Beginner]

A car moves along a straight road. Its speed at time $t$ minutes is $v=2t$ kilometres per minute. How far does the car travel from $t=0$ to $t=3$?

Step 1. The total distance equals the accumulated area under the speed curve. The speed curve $v=2t$ is a straight line through the origin with slope $2$. The area from $t=0$ to $t=3$ is a triangle with base $3$ and height $2\times 3=6$.

Step 2. Triangle area is one-half times base times height: $1/2\times 3\times 6=9$.

Step 3. Verify via the second theorem: an antiderivative of $v=2t$ is the distance function $d=t^2$. The total change is $3^2-0^2=9$.

What this tells us: the area method (Step 2) and the antiderivative method (Step 3) always agree. That agreement is the second fundamental theorem.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Let $f:[a,b]\to \mathbb{R}$ be a function. Define the accumulated-area function (also called the integral function) by

$$F(x)={\int }_a^{x}f(t)dt,x\in [a,b].$$

Theorem (FTC1). If $f$ is continuous on $[a,b]$, then $F$ is differentiable on $(a,b)$ and

$$F^{\prime }(x)=f(x)\text{for all }x\in (a,b).$$

In words: the derivative of the integral function is the original integrand.

Theorem (FTC2). If $f$ is Riemann-integrable on $[a,b]$ and $G$ is any antiderivative of $f$ on $[a,b]$ (meaning $G^{\prime }(x)=f(x)$ for all $x\in (a,b)$ and $G$ is continuous on $[a,b]$), then

$${\int }_a^{b}f(x)dx=G(b)-G(a).$$

In words: the integral of $f$ over $[a,b]$ equals the net change in any antiderivative.

Counterexamples to common slips

• FTC1 requires continuity. The function $f(x)=\mathrm{sgn}(x)$ on $[-1,1]$ is Riemann-integrable (its integral is $0$), but the integral function $F(x)=|x|-1$ is not differentiable at $x=0$. Discontinuity of $f$ at a single point can break differentiability of $F$.

• FTC2 requires an antiderivative that exists everywhere on the closed interval. The function $f(x)=1/x^2$ on $[-1,1]$ has no antiderivative valid on all of $[-1,1]$ because $f$ blows up at $0$. Blindly applying $G(x)=-1/x$ and writing $G(1)-G(-1)=-1-1=-2$ gives a negative result for a non-negative function.

• Two antiderivatives can differ by more than a constant on a disconnected domain. On $\mathbb{R}\setminus \{0\}$, the function $G_1(x)=\ln |x|$ and $G_2(x)=\ln |x|+1_{x>0}$ both satisfy $G^{\prime }=1/x$, yet $G_2-G_1$ is not constant on the full domain. FTC2 applies on each connected component separately.

Key theorem with proof [Intermediate+]

Theorem (FTC1 — derivative of the integral). Let $f:[a,b]\to \mathbb{R}$ be continuous. Define $F(x)={\int }_a^{x}f(t)dt$. Then $F$ is differentiable on $(a,b)$ and $F^{\prime }(x)=f(x)$ for every $x\in (a,b)$.

Proof. Fix $x\in (a,b)$. For $h\ne 0$ with $x+h\in [a,b]$, the difference quotient is

$$\frac{F(x+h)-F(x)}{h}=\frac{1}{h}{\int }_x^{x+h}f(t)dt.$$

By the mean value theorem for integrals (a consequence of the intermediate value theorem applied to continuous functions), there exists a point $c_h$ between $x$ and $x+h$ such that

$${\int }_x^{x+h}f(t)dt=f(c_h)\cdot h.$$

Therefore the difference quotient equals $f(c_h)$. As $h\to 0$, the point $c_h$ is squeezed between $x$ and $x+h$, so $c_h\to x$. By continuity of $f$,

$$F^{\prime }(x)=\underset{h\to 0}{\lim }f(c_h)=f(x).$$

$\square$

Theorem (FTC2 — integral via antiderivative). Let $f$ be Riemann-integrable on $[a,b]$ and let $G$ be continuous on $[a,b]$ with $G^{\prime }(x)=f(x)$ for all $x\in (a,b)$. Then

$${\int }_a^{b}f(x)dx=G(b)-G(a).$$

Proof. Let $P=\{x_0,x_1,\dots ,x_n\}$ be any partition of $[a,b]$ with $a=x_0<x_1<\cdots <x_n=b$. On each subinterval $[x_{i-1},x_i]$, the mean value theorem for derivatives gives a point $c_i\in (x_{i-1},x_i)$ with

$$G(x_i)-G(x_{i-1})=G^{\prime }(c_i)(x_i-x_{i-1})=f(c_i)\Delta x_i.$$

Summing from $i=1$ to $n$:

$$G(b)-G(a)=\sum _{i=1}^{n}f(c_i)\Delta x_i.$$

The right side is a Riemann sum for $f$ over $P$. As $\Vert P\Vert \to 0$, every Riemann sum for a Riemann-integrable function converges to the integral. Therefore

$$G(b)-G(a)=\underset{\Vert P\Vert \to 0}{\lim }\sum _{i=1}^{n}f(c_i)\Delta x_i={\int }_a^{b}f(x)dx.$$

$\square$

Bridge. The FTC is the foundational reason that integration and differentiation are inverse operations, and this is exactly the bridge between the geometric theory of area and the algebraic theory of antiderivatives. FTC1 builds toward the Lebesgue differentiation theorem 02.11.04 where the derivative-of-the-integral result holds almost everywhere for locally integrable functions, and appears again in 02.05.04 as the multivariable chain rule that underpins change of variables in higher dimensions. The central insight is that continuity of $f$ forces the integral function $F$ to be differentiable, and putting these together with the mean value theorem closes the loop: every continuous function has an antiderivative (by FTC1), and the bridge is that evaluating that antiderivative at the endpoints yields the integral (by FTC2).

Exercises [Intermediate+]

Advanced results [Master]

Theorem 1 (Lebesgue differentiation theorem). If $f\in L_{\mathrm{loc}}^1(\mathbb{R})$, then for almost every $x$,

$$\underset{h\to 0}{\lim }\frac{1}{2h}{\int }_{x-h}^{x+h}f(t)dt=f(x).$$

This is the measure-theoretic strengthening of FTC1: the derivative of the indefinite Lebesgue integral recovers $f$ at almost every point, even when $f$ is only integrable, not continuous. The proof goes through the Hardy-Littlewood maximal inequality.

Theorem 2 (Absolute continuity and the Lebesgue FTC). A function $F:[a,b]\to \mathbb{R}$ is absolutely continuous if and only if there exists $f\in L^1([a,b])$ such that $F(x)=F(a)+{\int }_a^{x}f(t)dt$ for all $x\in [a,b]$. In that case $F^{\prime }(x)=f(x)$ almost everywhere.

Absolute continuity replaces the continuity hypothesis of the classical FTC1 with a stronger condition (uniform control of increments) and in return gains existence almost everywhere without requiring $f$ to be continuous.

Theorem 3 (Integration by parts for the Lebesgue integral). If $F$ and $G$ are absolutely continuous on $[a,b]$, then

$${\int }_a^b{F}^{\prime }(x)G^{\prime }(x)dx=F(b)G(b)-F(a)G(a)-{\int }_a^b{F}^{\prime }(x)G(x)dx.$$

This follows from the product rule $(FG{)}^{\prime }=F^{\prime }G+F{G}^{\prime }$ and the Lebesgue FTC applied to the absolutely continuous function $FG$.

Theorem 4 (FTC for the Riemann-Stieltjes integral). If $f$ is continuous on $[a,b]$ and $\alpha$ is of bounded variation on $[a,b]$, then

$${\int }_a^{b}fd\alpha =f(b)\alpha (b)-f(a)\alpha (a)-{\int }_a^b\alpha df$$

whenever either side exists. If $\alpha$ is continuously differentiable, the Riemann-Stieltjes integral reduces: ${\int }_a^{b}fd\alpha ={\int }_a^{b}f(x){\alpha }^{\prime }(x)dx$.

Theorem 5 (Fundamental theorem for line integrals). Let $U\subset {\mathbb{R}}^n$ be open and connected, and let $\mathbf{F}:U\to {\mathbb{R}}^n$ be a continuous vector field. Then $\mathbf{F}=\nabla \varphi$ for some $C^1$ function $\varphi :U\to \mathbb{R}$ if and only if ${\int }_{\gamma }\mathbf{F}\cdot d\mathbf{r}$ depends only on the endpoints of $\gamma$ for every piecewise-$C^1$ curve $\gamma$ in $U$. In that case,

$${\int }_{\gamma }\mathbf{F}\cdot d\mathbf{r}=\varphi (\mathbf{b})-\varphi (\mathbf{a})$$

where $\mathbf{a}$ and $\mathbf{b}$ are the start and end of $\gamma$. This is the multivariable FTC2: the line integral of a gradient field equals the change in the potential function.

Theorem 6 (FTC1 fails without absolute continuity). The Cantor function $\phi :[0,1]\to [0,1]$ is continuous, non-decreasing, and has ${\phi }^{\prime }=0$ almost everywhere. Yet $\phi (1)-\phi (0)=1\ne 0={\int }_0^1{\phi }^{\prime }(x)dx$. The Cantor function is not absolutely continuous; it is singular. This demonstrates that continuity of the integrand is not sufficient to replace absolute continuity in the Lebesgue FTC.

Theorem 7 (Change of variables via the FTC). If $\phi :[a,b]\to [\phi (a),\phi (b)]$ is a $C^1$ diffeomorphism and $f$ is continuous, then

$${\int }_{\phi (a)}^{\phi (b)}f(u)du={\int }_a^{b}f(\phi (t)){\phi }^{\prime }(t)dt.$$

The proof applies FTC2 to both sides: $G(u)={\int }_{\phi (a)}^{u}f(s)ds$ has $G^{\prime }=f$, so the left side is $G(\phi (b))-G(\phi (a))=(G\circ \phi )(b)-(G\circ \phi )(a)$, and by the chain rule $(G\circ \phi {)}^{\prime }=f(\phi (t)){\phi }^{\prime }(t)$, so the right side is also $(G\circ \phi )(b)-(G\circ \phi )(a)$.

Synthesis. The foundational reason the FTC holds across all its versions is that differentiation and integration are inverse operations at the level of measures and densities. The central insight is that absolute continuity of $F$ is both necessary and sufficient for $F$ to be an indefinite Lebesgue integral, and this is exactly the condition that identifies the absolutely continuous functions with the Sobolev space $W^{1,1}$. Putting these together with the Lebesgue differentiation theorem, every locally integrable function $f$ has an integral function $F$ differentiable almost everywhere with $F^{\prime }=f$, and the bridge is that the Stieltjes integral generalises both the Riemann and Lebesgue settings by encoding the integrator's variation into a measure, while the fundamental theorem for line integrals 02.05.04 extends the same pattern to gradient fields on ${\mathbb{R}}^n$. The pattern generalises to differential forms via Stokes' theorem, where the integral of $d\omega$ over a region equals the integral of $\omega$ over the boundary — the FTC2 statement $G(b)-G(a)={\int }_a^{b}f$ is the $0$-dimensional case.

Full proof set [Master]

Proposition 1 (FTC1 via the mean value theorem). Restated from the Key theorem. Let $f:[a,b]\to \mathbb{R}$ be continuous and $F(x)={\int }_a^{x}f(t)dt$. Then $F^{\prime }(x)=f(x)$ for all $x\in (a,b)$.

Proof. For $h>0$ (the case $h<0$ is analogous), consider

$$F(x+h)-F(x)={\int }_x^{x+h}f(t)dt.$$

Since $f$ is continuous on the compact interval $[x,x+h]$, the extreme value theorem gives $m_h={\min }_{[x,x+h]}f$ and $M_h={\max }_{[x,x+h]}f$. Then

$$m_h\cdot h\le {\int }_x^{x+h}f(t)dt\le M_h\cdot h.$$

Dividing by $h>0$:

$$m_h\le \frac{F(x+h)-F(x)}{h}\le M_h.$$

As $h\to 0^{+}$, both $m_h$ and $M_h$ converge to $f(x)$ by continuity. By the squeeze theorem, $F_{+}^{\prime }(x)=f(x)$. The same argument with $h\to 0^{-}$ gives $F_{-}^{\prime }(x)=f(x)$. Hence $F^{\prime }(x)=f(x)$. $\square$

Proposition 2 (Absolute continuity characterises indefinite Lebesgue integrals). A function $F:[a,b]\to \mathbb{R}$ is absolutely continuous if and only if there exists $f\in L^1([a,b])$ with $F(x)=F(a)+{\int }_a^{x}fd\mu$ for all $x$, and in that case $F^{\prime }=f$ almost everywhere.

Proof (sketch). ($\Leftarrow$) If $F(x)=F(a)+{\int }_a^{x}fd\mu$, then for any finite collection of disjoint intervals $(a_i,b_i)$,

$$\sum _i|F(b_i)-F(a_i)|\le \sum _i{\int }_{a_i}^{b_i}|f|d\mu ={\int }_E|f|d\mu$$

where $E={\bigcup }_i(a_i,b_i)$. If ${\sum }_i(b_i-a_i)<\delta$, then $\mu (E)<\delta$, and by the absolute continuity of the Lebesgue integral, ${\int }_E|f|d\mu <ϵ$ for $\delta$ chosen appropriately. Hence $F$ is absolutely continuous.

($\Rightarrow$) If $F$ is absolutely continuous, then $F$ is of bounded variation, hence differentiable almost everywhere, and $F^{\prime }\in L^1$. Define $G(x)=F(a)+{\int }_a^x{F}^{\prime }(t)dt$. Then $G$ is absolutely continuous by the forward direction, and $G^{\prime }=F^{\prime }$ a.e. Thus $(F-G{)}^{\prime }=0$ a.e., and an absolutely continuous function with zero derivative a.e. is constant (this uses the Vitali covering lemma). Since $G(a)=F(a)$, we get $F=G$. $\square$

Proposition 3 (The Cantor function violates FTC2). The Cantor function $\phi$ is continuous and non-decreasing on $[0,1]$ with $\phi (0)=0$, $\phi (1)=1$, and ${\phi }^{\prime }=0$ almost everywhere.

Proof (construction). Let $C$ be the standard Cantor set. At stage $n$ of the construction, remove $2^{n-1}$ open middle thirds. Define $\phi$ on the complement of $C$ by assigning values: on the removed interval at stage $1$ (the middle third $(1/3,2/3)$), set $\phi =1/2$. At stage $2$, set $\phi =1/4$ on $(1/9,2/9)$ and $\phi =3/4$ on $(7/9,8/9)$. Continue: at each stage, $\phi$ takes the average of the values at the adjacent intervals already assigned. This defines $\phi$ on $[0,1]\setminus C$, and $\phi$ extends continuously to $C$ because the jumps shrink to zero.

The function $\phi$ is constant on each removed interval, so ${\phi }^{\prime }=0$ there. The Cantor set has Lebesgue measure zero, so ${\phi }^{\prime }=0$ a.e. Yet $\phi (1)-\phi (0)=1\ne 0={\int }_0^{1}0dx$. The function is not absolutely continuous: for the $2^n$ intervals of $C$'s complement at stage $n$, the total length is $(2/3{)}^n\to 0$ but $\phi$ increases by $1$ across them. $\square$

Connections [Master]

• Multivariable chain rule and change of variables 02.05.03. The change-of-variables formula for multivariable integrals generalises the one-dimensional substitution rule (Theorem 7 above). The chain rule 02.05.03 provides the derivative computation that makes the substitution work, and the FTC supplies the proof mechanism — evaluate antiderivatives at the transformed endpoints.

• Implicit and inverse function theorems 02.05.04. The inverse function theorem's proof integrates the derivative to construct a local inverse, relying on FTC1 to guarantee that the constructed integral function has the correct derivative. The implicit function theorem is a corollary, and both appear in the proof of Stokes' theorem where the FTC for line integrals handles the one-dimensional base case.

• Banach spaces and bounded linear operators 02.11.01. The Lebesgue FTC (Theorem 2 above) identifies the absolutely continuous functions with the Sobolev space $W^{1,1}([a,b])$, a subspace of $L^1$. The bounded linear operators framework 02.11.01 provides the functional-analytic setting in which the Lebesgue differentiation theorem and the Hardy-Littlewood maximal inequality are formulated, and the operator $f\mapsto {\int }_a^{x}f(t)dt$ is bounded from $L^1$ to $W^{1,1}$.

• Lyapunov stability and conserved quantities 02.12.08. The direct method of Lyapunov 02.12.08 constructs Lyapunov functions whose time derivative is negative definite. Proving that a Lyapunov function decreases along trajectories requires FTC2 to relate the integral of the derivative to the function's net change. First integrals 02.12.12 — conserved quantities whose derivative along the flow vanishes — are characterised through FTC1 applied to the flow map.

Historical & philosophical context [Master]

Newton 1665–1666, in his method of fluxions ^[Newton1665], discovered the inverse relationship between what he called fluxions (rates of change) and fluents (accumulated quantities). His geometric reasoning treated the area under a curve as a quantity generated by the motion of an ordinate sweeping along the axis, and recognised that the rate of generation at each instant equals the height of the ordinate. Leibniz 1675, independently, introduced the elongated-S notation for the integral and articulated the inverse relationship in the notation itself: the differential $\mathrm{d}$ and the integral $\int$ are inverse operations ^{[Leibniz1675]}. Cauchy 1823, in his Resume des lecons at the Ecole Polytechnique ^[Cauchy1823], gave the first rigorous proof of the FTC for continuous functions using the mean value theorem, replacing the geometric intuitions of Newton and Leibniz with an $ϵ$-$\delta$ argument.

The Lebesgue strengthening came with Lebesgue 1904 in his Lecons sur l'integration ^{[Lebesgue1904]}, which identified absolute continuity as the precise condition under which a function is an indefinite integral. This result closed a gap exposed by the Cantor function (a continuous, monotone function with zero derivative almost everywhere yet non-constant), demonstrating that continuity alone is insufficient. The Stieltjes integral, introduced by Stieltjes 1894 in connection with moment problems, generalises the FTC to integrators of bounded variation. The fundamental theorem for line integrals extends the FTC to gradient fields on ${\mathbb{R}}^n$ and is the one-dimensional precursor to Stokes' theorem on manifolds.

Bibliography [Master]

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}

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}

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}

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}

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}

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