Fundamental Theorem of Calculus

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

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 Fundamental Theorem of Calculus (FTC) is the bridge between the two great operations of calculus: differentiation (finding rates of change) and integration (finding accumulated quantities or areas). It states, in two complementary parts, that these operations undo each other. The first part says that if you accumulate the area under a function and then take the derivative of that accumulation, you recover the original function. The second part — the workhorse for computation — says that a definite integral can be evaluated by finding an antiderivative and subtracting its endpoint values. The surprising fact is that the seemingly unrelated geometric problem of computing areas and the analytic problem of computing slopes are, in fact, two sides of the same coin — a connection that took nearly two thousand years to fully grasp.

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.

Formal Statement

Antiderivative

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 $\int f (x) d x = F (x) + C$ .

Definite 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:

\int_{a}^{b} f (x) d x = n \to \infty lim i = 1 \sum n f (x_{i}^{*}) Δ x, Δ x = \frac{b - a}{n}

⟳Partition

[a, b]

into

n

subintervals of width

Δ x

, choose a sample point

x_{i}^{*}

in each, build rectangles of height

f (x_{i}^{*})

, and let

n \to \infty

Fundamental Theorem of Calculus, Part 1

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

g (x) = \int_{a}^{x} f (t) d t, a \leq x \leq b .

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

g^{'} (x) = \frac{d}{d x} \int_{a}^{x} f (t) d t = f (x) .

⟳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

Fundamental 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

\int_{a}^{b} f (x) d x = F (b) - F (a) \equiv [F (x)]_{a}^{b} .

⟳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.

Why It Holds / Derivation

Proof of Part 1

Goal: Show

g^{'} (x) = f (x)

where

g (x) = \int_{a}^{x} f (t) d t

. Step 1. Write the difference quotient for

g

x

\frac{g ( x + h ) - g ( x )}{h} = \frac{1}{h} [\int_{a}^{x + h} f (t) d t - \int_{a}^{x} f (t) d t] .

⟳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:

\frac{g ( x + h ) - g ( x )}{h} = \frac{1}{h} \int_{x}^{x + h} f (t) d t .

⟳The two integrals from

a

x

cancel, leaving only the sliver from

x

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:

m_{h} \cdot h \leq \int_{x}^{x + h} f (t) d t \leq M_{h} \cdot h .

⟳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)

g^{'} (x) = h \to 0 lim \frac{1}{h} \int_{x}^{x + h} f (t) d t = f (x) .

⟳The squeeze theorem forces the middle quantity to the common limit

f (x)

■

Proof of Part 2

Goal: Show

\int_{a}^{b} f (x) d x = F (b) - F (a)

for any antiderivative

F

. Step 1. Let

g (x) = \int_{a}^{x} f (t) d t

. 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:

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

⟳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 d t = 0

, we get

F (a) = 0 + C

, so

C = F (a)

. Step 4. Evaluate at

x = b

and substitute

C

F (b) = g (b) + F (a) ⟹ g (b) = \int_{a}^{b} f (t) d t = F (b) - F (a) .

⟳Rearrange to isolate the integral

g (b)

■

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") $\overset{x}{˙}$ . 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.

Building 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 $g (x) = \int_{0}^{x} f (t) d t$ 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.

The accumulation function: shaded area under $f$ from $a$ to $x$ equals the height $g(x)$.

Worked Examples

Evaluating a definite integral via FTC Part 2

Given: Find

\int_{1}^{4} (3 x^{2} - 2 x + 1) d x

. (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:

F (x) = x^{3} - x^{2} + x .

⟳

\frac{d}{d x} x^{3} = 3 x^{2}

\frac{d}{d x} x^{2} = 2 x

\frac{d}{d x} 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:

\int_{1}^{4} (3 x^{2} - 2 x + 1) d x = F (4) - F (1) = 52 - 1 = 51.

⟳Apply equation [4] with

a = 1

b = 4

\int_{1}^{4} (3 x^{2} - 2 x + 1) d x = 51 m

Differentiating an accumulation function via FTC Part 1

Given: Let

g (x) = \int_{2}^{x^{2}} sin (t) d t

. 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 d t

and

u = x^{2}

. Step 2. By Part 1,

\frac{d G}{d u} = sin (u)

. Step 3. By the chain rule:

g^{'} (x) = \frac{d G}{d u} \cdot \frac{d u}{d x} = sin (u) \cdot 2 x = sin (x^{2}) \cdot 2 x .

⟳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

g^{'} (x) = 2 x sin (x^{2})

Common Misconceptions

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}{d x} \int_{a}^{x^{2}} f (t) d t

is not

f (x)

; it is

f (x^{2}) \cdot 2 x

. 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) d t

, the variable

t

is internal and disappears after integration; only

x

survives. Writing

\int_{a}^{x} f (x) d x

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.

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

\int_{- 1}^{1} \frac{1}{x ^{2}} d x = ? 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.

Connections

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.

Real-World Applications

Physics — kinematics. Displacement is the integral of velocity, and velocity the integral of acceleration. Given an acceleration profile $a (t)$ , NASA computes spacecraft velocity and position by $\int a d t$ and $\int v d t$ , evaluated via FTC.

Electrical engineering — charge and current. Current $i (t)$ is the rate of charge flow; total charge delivered to a capacitor over $[t_{1}, t_{2}]$ is $Q = \int_{t_{1}}^{t_{2}} i (t) d t = F (t_{2}) - F (t_{1})$ where $F$ is accumulated charge.

Economics — total cost from marginal cost. If marginal cost $M C (q)$ is the derivative of total cost, then total cost of producing units $a$ to $b$ is $\int_{a}^{b} M C (q) d q$ , recovering total cost from rates.

Probability and statistics. The cumulative distribution function $F (x) = \int_{- \infty}^{x} p (t) d t$ accumulates a probability density $p$ ; by Part 1, $F^{'} (x) = p (x)$ , linking densities and cumulative probabilities.

Medicine — drug dosage (pharmacokinetics). The "area under the curve" (AUC) of a plasma drug-concentration graph quantifies total exposure and is computed as a definite integral over the dosing interval.

Engineering — work and energy. Work done by a variable force is $W = \int_{a}^{b} F (x) d x$ ; e.g., the energy to stretch a spring or pump fluid from a tank is evaluated by antiderivatives.

Hydrology and climate. Total rainfall or river discharge over a period is the integral of an instantaneous rate, computed from accumulated-volume functions.

Historical Development

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 $\int x^{n} d x$ 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 $d x$ 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).

Subtleties & Limits

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 $\int_{a}^{x} f^{'} (t) d t = f (x) - f (a)$ 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.