Fundamental Theorem of Calculus

⚡ Key Takeaways

One-line statement: Differentiation and integration are inverse operations — the derivative of an integral gives back the original function, and a definite integral can be evaluated using an antiderivative.

Key result in words: To find the area under a curve from $a$ to $b$ , find any function whose derivative is the integrand and subtract its values at the endpoints.

Where it applies: Continuous functions on closed intervals; the cornerstone of all single-variable calculus, physics, engineering, and economics.

Who/when: Synthesized by Isaac Newton (1660s) and Gottfried Wilhelm Leibniz (1670s), building on Barrow, Fermat, and others; rigorized by Cauchy and Riemann in the 1800s.

💡 The Core Insight

Integration and differentiation are inverse processes. Accumulating a quantity (integration) and measuring its instantaneous rate of change (differentiation) are reverse operations, exactly like multiplication and division. This single insight transforms area computation from an infinite limiting process (summing infinitely many rectangles) into a finite, algebraic procedure: find an antiderivative, plug in two numbers, subtract.

DefinitionAntiderivative

A function $F$ is an antiderivative of $f$ on an interval if $F'(x) = f(x)$ for every $x$ in that interval. Antiderivatives are not unique: if $F$ is one antiderivative, then $F + C$ is also an antiderivative for any constant $C$ , since the derivative of a constant is zero. The collection of all antiderivatives is the indefinite integral, written $\int f(x)\,dx = F(x) + C$ .

DefinitionDefinite Integral

For a function $f$ continuous on $[a,b]$ , the definite integral is the signed area between the graph of $f$ and the $x$ -axis, defined as a limit of Riemann sums:

1

\quad \int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x, \qquad \Delta x = \frac{b-a}{n}

Derivation hint: Partition $[a,b]$ into $n$ subintervals of width $\Delta x$ , choose a sample point $x_i^*$ in each, build rectangles of height $f(x_i^*)$ , and let $n \to \infty$ .

TheoremFundamental Theorem of Calculus, Part 1

Let $f$ be continuous on $[a,b]$ . Define the accumulation function

2

\quad g(x) = \int_a^x f(t)\,dt, \qquad a \le x \le b.

Then $g$ is differentiable on $(a,b)$ and

3

\quad g'(x) = \frac{d}{dx}\int_a^x f(t)\,dt = f(x).

Derivation hint: Form the difference quotient $\frac{g(x+h)-g(x)}{h}$ , recognize the numerator as an integral over $[x, x+h]$ , and apply the squeeze of continuity as $h\to 0$ .

Symbols: $f$ is the integrand (continuous), $t$ is a dummy variable of integration, $x$ is the upper limit (the genuine variable), $g$ is the area accumulated from $a$ up to $x$ .

TheoremFundamental Theorem of Calculus, Part 2 (Evaluation Theorem)

Let $f$ be continuous on $[a,b]$ and let $F$ be any antiderivative of $f$ (so $F' = f$ ). Then

4

\quad \int_a^b f(x)\,dx = F(b) - F(a) \equiv \big[F(x)\big]_a^b.

Derivation hint: Apply Part 1 to show the accumulation function and $F$ differ by a constant, then evaluate at $a$ and $b$ to cancel that constant.

Symbols: $F(b)-F(a)$ is the net change of $F$ across $[a,b]$ ; the choice of antiderivative does not matter because the constant $C$ cancels in the subtraction.

ProofProof of Part 1

Goal: Show $g'(x) = f(x)$ where $g(x) = \int_a^x f(t)\,dt$ .

Step 1. Write the difference quotient for $g$ at $x$ :

5

\quad \frac{g(x+h) - g(x)}{h} = \frac{1}{h}\left[\int_a^{x+h} f(t)\,dt - \int_a^x f(t)\,dt\right].

Derivation hint: This is the definition of the derivative before taking the limit.

Step 2. Use the additivity of integrals, $\int_a^{x+h} = \int_a^x + \int_x^{x+h}$ , to collapse the bracket:

6

\quad \frac{g(x+h) - g(x)}{h} = \frac{1}{h}\int_x^{x+h} f(t)\,dt.

Derivation hint: The two integrals from $a$ to $x$ cancel, leaving only the sliver from $x$ to $x+h$ .

Step 3. By the Extreme Value Theorem, on the small interval $[x, x+h]$ the continuous $f$ attains a minimum $m_h = f(u)$ and maximum $M_h = f(v)$ . Then the area of the sliver is squeezed:

7

\quad m_h \cdot h \le \int_x^{x+h} f(t)\,dt \le M_h \cdot h.

Derivation hint: A region of width $h$ trapped between heights $m_h$ and $M_h$ has area between $m_h h$ and $M_h h$ .

Step 4. Divide by $h$ and take the limit. As $h \to 0$ , both $u$ and $v$ are forced toward $x$ , so by continuity $m_h \to f(x)$ and $M_h \to f(x)$ :

8

\quad g'(x) = \lim_{h\to 0}\frac{1}{h}\int_x^{x+h} f(t)\,dt = f(x).

Derivation hint: The squeeze theorem forces the middle quantity to the common limit $f(x)$ . $\blacksquare$

∎

ProofProof of Part 2

Goal: Show $\int_a^b f(x)\,dx = F(b) - F(a)$ for any antiderivative $F$ .

Step 1. Let $g(x) = \int_a^x f(t)\,dt$ . By Part 1, $g'(x) = f(x)$ , so $g$ is one antiderivative of $f$ .

Step 2. Since $F$ is also an antiderivative, $F$ and $g$ differ by a constant:

9

\quad F(x) = g(x) + C \quad\text{for all } x \in [a,b].

Derivation hint: Two functions with equal derivatives on an interval differ by a constant — a consequence of the Mean Value Theorem.

Step 3. Evaluate at $x=a$ . Since $g(a) = \int_a^a f\,dt = 0$ , we get $F(a) = 0 + C$ , so $C = F(a)$ .

Step 4. Evaluate at $x=b$ and substitute $C$ :

10

\quad F(b) = g(b) + F(a) \;\Longrightarrow\; g(b) = \int_a^b f(t)\,dt = F(b) - F(a).

Derivation hint: Rearrange to isolate the integral $g(b)$ . $\blacksquare$

∎

IN

Isaac Newton

1643–1727

Newton developed his "method of fluxions" in 1665–1666 during the plague years at Woolsthorpe. He conceived of quantities as flowing in time — a "fluent" $x$ with rate of flow (a "fluxion") $\dot{x}$ . For Newton, integration was finding the fluent from its fluxion, making the inverse relationship between the two operations physically natural: position and velocity are an antiderivative–derivative pair. His geometric and kinematic instincts let him use the theorem decades before publishing it in the Principia (1687) and De Analysi (circulated 1669, printed 1711). The notorious priority dispute with Leibniz over who "invented" calculus dominated his later years, though both arrived independently at the same fundamental connection.

🧠 Intuition

Think of $f(t)$ as the rate at which water flows into a tank at time $t$ , measured in liters per second. The accumulation function $g(x) = \int_0^x f(t)\,dt$ is then the total water in the tank at time $x$ .

Part 1 says: if you ask "how fast is the total water changing right now?" the answer is just the current flow rate $f(x)$ — obviously! The rate of change of the accumulated total is the thing being accumulated.

Part 2 says: to find how much water accumulated between time $a$ and time $b$ , you don't need to track the flow second by second. If you know a formula $F$ for the total water level, just subtract: $F(b) - F(a)$ . The net accumulation equals the net change in the total. Another image: $f$ is your car's speedometer, $F$ is your odometer. Part 2 says the distance traveled (the integral of speed) equals the odometer reading at the end minus at the start. You never had to add up infinitely many tiny distances — the odometer did it for you.

Figure

The accumulation function: shaded area under

f

from

a

to

x

equals the height

g(x)

.

ExampleEvaluating a definite integral via FTC Part 2

Given: Find $\displaystyle\int_1^4 \left(3x^2 - 2x + 1\right)\,dx$ . (Suppose this is a velocity in m/s and we want displacement in meters over $t\in[1,4]$ s.)

Step 1. Find an antiderivative $F$ by reversing the power rule term by term:

11

\quad F(x) = x^3 - x^2 + x.

Derivation hint: $\frac{d}{dx}x^3 = 3x^2$ , $\frac{d}{dx}x^2 = 2x$ , $\frac{d}{dx}x = 1$ .

Step 2. Evaluate at the upper limit:

F(4) = 4^3 - 4^2 + 4 = 64 - 16 + 4 = 52.

Step 3. Evaluate at the lower limit:

F(1) = 1^3 - 1^2 + 1 = 1 - 1 + 1 = 1.

Step 4. Subtract:

12

\quad \int_1^4 (3x^2 - 2x + 1)\,dx = F(4) - F(1) = 52 - 1 = 51.

Derivation hint: Apply equation [4] with $a=1$ , $b=4$ .

\boxed{\int_1^4 (3x^2 - 2x + 1)\,dx = 51 \text{ m}}

ExampleDifferentiating an accumulation function via FTC Part 1

Given: Let $\displaystyle g(x) = \int_2^{x^2} \sin(t)\,dt$ . Find $g'(x)$ .

Step 1. The upper limit is $x^2$ , not $x$ , so we need the chain rule combined with Part 1. Write $g(x) = G(u)$ where $G(u) = \int_2^u \sin t\,dt$ and $u = x^2$ .

Step 2. By Part 1, $\dfrac{dG}{du} = \sin(u)$ .

Step 3. By the chain rule:

13

\quad g'(x) = \frac{dG}{du}\cdot\frac{du}{dx} = \sin(u)\cdot 2x = \sin(x^2)\cdot 2x.

Derivation hint: Part 1 gives the derivative with respect to the upper limit; multiply by the derivative of that limit.

Verification at $x = 0$ : $g'(0) = \sin(0)\cdot 0 = 0$ , consistent with the area changing slowly near $x=0$ .

\boxed{g'(x) = 2x\,\sin(x^2)}

⚠️ Common mistake

Pitfall 1 — The continuity hypothesis is not optional. FTC Part 2 requires $f$ to be continuous (or at least integrable with an antiderivative) on the entire interval $[a,b]$ . Blindly applying $F(b)-F(a)$ across a discontinuity gives nonsense.

Pitfall 2 — Forgetting the chain rule. $\frac{d}{dx}\int_a^{x^2} f(t)\,dt$ is not $f(x)$ ; it is $f(x^2)\cdot 2x$ . Part 1 in its basic form applies only when the upper limit is exactly $x$ .

Pitfall 3 — Confusing the dummy variable. In $\int_a^x f(t)\,dt$ , the variable $t$ is internal and disappears after integration; only $x$ survives. Writing $\int_a^x f(x)\,dx$ is sloppy and invites errors.

Pitfall 4 — Sign and orientation. $\int_b^a f = -\int_a^b f$ . Reversing limits flips the sign. A definite integral is signed area: regions below the $x$ -axis count as negative.

Counterexample

Consider $f(x) = \dfrac{1}{x^2}$ on $[-1, 1]$ . A naive student picks the antiderivative $F(x) = -\dfrac{1}{x}$ and computes

\int_{-1}^{1}\frac{1}{x^2}\,dx \stackrel{?}{=} F(1) - F(-1) = (-1) - (1) = -2.

This is absurd: $\frac{1}{x^2}$ is always positive, so its integral cannot be negative. The error is that $f$ has an infinite discontinuity at $x=0 \in [-1,1]$ , violating the continuity hypothesis. The integral is in fact a divergent improper integral, not the finite value $-2$ . Lesson: always check that the integrand is continuous (and finite) on the whole interval before invoking FTC Part 2.

connection

The FTC sits at the center of a web of calculus ideas. It depends on the Definite Integral (defined via Riemann sums) and the Derivative (defined via limits), welding them together. The proof of Part 2 leans crucially on the Mean Value Theorem (to conclude two functions with equal derivatives differ by a constant) and on the Squeeze Theorem and Extreme Value Theorem within Part 1. It generalizes upward into multivariable calculus, where Green's Theorem, the Divergence Theorem, and Stokes' Theorem are all higher-dimensional incarnations of "integrate a derivative over a region = evaluate the function on its boundary." It also underwrites Techniques of Integration such as Integration by Substitution and Integration by Parts, which are simply the chain rule and product rule read backwards.

🔬 Where this appears

⏳ History

- c. 250 BCE — Archimedes uses the method of exhaustion to compute the area of a parabolic segment and the volume of a sphere, anticipating integration by summing infinitely refined slices, but without a general theorem.

- 1630s — Pierre de Fermat and Bonaventura Cavalieri develop "indivisibles" and quadrature methods, computing areas under power curves

\int x^n\,dx

for many

n

.

- 1660s — Isaac Barrow (Newton's teacher at Cambridge) proves a geometric form of the inverse relationship between tangent (derivative) and area (integral) problems — arguably the first statement of the FTC, though framed geometrically.

- 1665–1666 — Isaac Newton develops the method of fluxions during the plague years, treating integration as the inverse of differentiation in a kinematic framework.

- 1675–1677 — Gottfried Wilhelm Leibniz independently invents calculus, introducing the integral sign

\int

(an elongated S for summa) and the

dx

notation, making the inverse relationship notationally transparent.

- 1684–1687 — Leibniz publishes his calculus (Nova Methodus, 1684); Newton publishes the Principia (1687). The bitter priority dispute, fanned by the Royal Society, lasts decades.

- 1823 — Augustin-Louis Cauchy gives the first rigorous definition of the definite integral as a limit of sums and proves the FTC with modern standards of rigor.

- 1854 — Bernhard Riemann generalizes the integral (the Riemann integral), clarifying exactly which functions are integrable and solidifying the theorem's hypotheses.

- 1902 — Henri Lebesgue introduces measure-theoretic integration, extending the FTC to far wider classes of functions and identifying its precise limits (e.g., for absolutely continuous functions).

open-problem

The classical FTC assumes a continuous integrand, but mathematicians have long probed how far it can be pushed. Subtlety 1: For the Riemann integral, Part 2 can fail if $f$ has an antiderivative that is not Riemann-integrable — there exist functions $F$ that are differentiable everywhere with derivative $F'$ that is not Riemann-integrable (Volterra's function, 1881). This motivated stronger integration theories. Subtlety 2: Under the Lebesgue integral, the correct condition for $\int_a^x f'(t)\,dt = f(x)-f(a)$ to hold is absolute continuity of $f$ — merely being continuous and differentiable almost everywhere is not enough (the Cantor function is a famous counterexample whose derivative is zero almost everywhere yet rises from 0 to 1). Subtlety 3: The Henstock–Kurzweil (gauge) integral, developed in the 1950s–60s, achieves the "perfect" FTC: every derivative is integrable and Part 2 holds without extra hypotheses — a still-active corner of analysis research. These refinements show that the deceptively simple statement "differentiation and integration are inverse" conceals deep questions about the very meaning of integration.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also