The Fundamental Theorem of Calculus

⚡ Key Takeaways

The Fundamental Theorem of Calculus (FTC) links the two central operations of calculus — differentiation and integration — proving they are inverse processes. In words: the rate at which the area under a curve accumulates equals the height of the curve, and the total accumulated change equals the difference of an antiderivative at the endpoints. Applies to continuous functions on a closed interval. Discovered/formalized independently by Isaac Newton (c. 1666) and Gottfried Wilhelm Leibniz (c. 1675), with rigorous proof refined through the 19th century.

💡 The Core Insight

Integration and differentiation undo each other. Accumulating a quantity and then measuring its instantaneous rate of growth returns the original quantity; conversely, the total accumulated change of a quantity over an interval is just the net difference of its antiderivative. This single bridge converts hard area problems into easy antiderivative lookups, and converts hard rate problems into accumulation problems.

DefinitionThe Accumulation (Area) Function

Given a function $f$ continuous on $[a,b]$ , define the accumulation function $A$ by

1

\quad A(x) = \int_a^x f(t)\,dt

Derivation hint: This is the signed area between the graph of $f$ and the horizontal axis, from the fixed lower limit $a$ to the variable upper limit $x$ . Here:

$f(t)$ — the integrand, a continuous function of the dummy variable $t$ .
$t$ — the dummy variable of integration (any letter works; it disappears after integrating).
$a$ — the fixed lower limit.
$x$ — the variable upper limit; $A$ is a function of $x$ .
$A(x)$ — the signed area accumulated from $a$ to $x$ (regions below the axis count negative).

TheoremFTC Part 1 — Differentiation of the Integral

If $f$ is continuous on $[a,b]$ , then $A(x)=\int_a^x f(t)\,dt$ is differentiable on $(a,b)$ and

2

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

Derivation hint: The instantaneous rate at which area accumulates equals the current height of the curve.

TheoremFTC Part 2 — Evaluation Theorem

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ (that is, $F'=f$ ), then

3

\quad \int_a^b f(x)\,dx = F(b) - F(a)

Derivation hint: Part 1 makes $A$ one such antiderivative; any two antiderivatives differ by a constant, which cancels under subtraction. The right-hand side is often written $\big[F(x)\big]_a^b$ or $F(x)\Big|_a^b$ .

ProofProof of Part 1

We compute $A'(x)$ directly from the definition of the derivative.

Step 1. Form the difference quotient. For a small increment $h>0$ ,

4

\quad A(x+h) - A(x) = \int_a^{x+h} f(t)\,dt - \int_a^{x} f(t)\,dt = \int_x^{x+h} f(t)\,dt

Derivation hint: Use the additivity of the integral over adjacent intervals, $\int_a^{x+h}=\int_a^x+\int_x^{x+h}$ .

Step 2. Apply the Mean Value Theorem for Integrals. Since $f$ is continuous on $[x,x+h]$ , there exists a point $c_h \in [x, x+h]$ with

5

\quad \int_x^{x+h} f(t)\,dt = f(c_h)\cdot h

Derivation hint: The integral over a small interval equals the interval width times the function's value at some intermediate height $c_h$ .

Step 3. Form the difference quotient and take the limit.

6

\quad \frac{A(x+h)-A(x)}{h} = f(c_h)

Derivation hint: Divide equation [5] by $h$ ; the area increment over $h$ is exactly $f(c_h)$ .

Step 4. As $h\to 0$ , the point $c_h$ is squeezed toward $x$ (since $x\le c_h\le x+h$ ). By continuity of $f$ ,

7

\quad A'(x) = \lim_{h\to 0} f(c_h) = f(x)

Derivation hint: Continuity guarantees $f(c_h)\to f(x)$ as $c_h\to x$ ; the one-sided argument is symmetric for $h<0$ . This proves $A'(x)=f(x)$ . $\blacksquare$

∎

ProofProof of Part 2

Step 1. Let $F$ be any antiderivative of $f$ , and let $A(x)=\int_a^x f(t)\,dt$ . By Part 1, $A'(x)=f(x)=F'(x)$ , so $A$ and $F$ have equal derivatives on $(a,b)$ .

Step 2. Two functions with the same derivative differ by a constant (a corollary of the Mean Value Theorem). Hence

8

\quad A(x) = F(x) + C \quad\text{for some constant } C

Derivation hint: $(A-F)'=0$ everywhere implies $A-F$ is constant.

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

Step 4. Evaluate at $x=b$ :

9

\quad \int_a^b f(t)\,dt = A(b) = F(b)+C = F(b)-F(a)

Derivation hint: Substitute $C=-F(a)$ into $A(b)=F(b)+C$ . This proves the evaluation formula. $\blacksquare$

∎

mathematician · Isaac Newton & Gottfried Wilhelm Leibniz | 1643–1727 / 1646–1716

Newton developed his "method of fluxions" around 1666, treating quantities as flowing in time and recognizing that the area swept out has a rate of change equal to the curve's height. Leibniz, working independently in the 1670s, introduced the elegant notation we still use — the elongated-S integral sign $\int$ (from summa) and the differential $dx$ — and explicitly framed integration and differentiation as inverse operations. A bitter priority dispute followed, but Leibniz's superior notation and Newton's physical insight together founded modern calculus. Rigorous foundations awaited Cauchy and Riemann in the 19th century.

🧠 Intuition

Imagine water flowing into a tank, where $f(t)$ is the flow rate (litres per second) at time $t$ . The accumulation function $A(x)=\int_a^x f(t)\,dt$ is the total water added between time $a$ and time $x$ .

Part 1 says: the instantaneous rate at which the total volume grows is exactly the current flow rate — $A'(x)=f(x)$ . Of course! How fast the tank fills right now is just how fast water is coming in right now.
Part 2 says: to find the total water added from $a$ to $b$ , you don't need to track every drop. If $F$ is any function whose rate of change is $f$ (a "volume so far" function), the net change is simply $F(b)-F(a)$ .

The deep point is the bridge: areas (a "static" geometric pile-up) and slopes (a "dynamic" instantaneous rate) are inverses. Stacking up tiny slivers of area and then asking how fast the stack grows returns you to the height you started with.

Figure

FTC Part 1: the shaded area

A(x)

grows at rate

f(x)

. The thin strip of width

h

has area

\approx f(x)\,h

.

ExampleEvaluating a Definite Integral (Part 2)

Problem. A particle moves with velocity $v(t) = 3t^2 + 2$ metres per second. Find the displacement between $t = 1\text{ s}$ and $t = 4\text{ s}$ .

Given. $v(t)=3t^2+2$ , $a=1$ , $b=4$ . Displacement $= \int_1^4 v(t)\,dt$ .

Step 1 — Find an antiderivative. We need $F$ with $F'(t)=3t^2+2$ :

10

\quad F(t) = t^3 + 2t

Derivation hint: Reverse the power rule: $\frac{d}{dt}t^3=3t^2$ and $\frac{d}{dt}(2t)=2$ .

Step 2 — Apply FTC Part 2.

11

\quad \int_1^4 (3t^2+2)\,dt = \big[t^3+2t\big]_1^4 = (4^3 + 2\cdot 4) - (1^3 + 2\cdot 1)

Derivation hint: Substitute the upper and lower limits into $F$ and subtract.

Step 3 — Arithmetic.

12

\quad = (64 + 8) - (1 + 2) = 72 - 3 = 69

Derivation hint: $4^3=64$ , $2\cdot4=8$ ; $1^3=1$ , $2\cdot1=2$ .

\boxed{\text{Displacement} = 69 \text{ metres}}

ExampleDifferentiating an Integral with a Variable Limit (Part 1)

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

Given. Integrand $f(t)=\sin t$ (continuous everywhere); upper limit is $u(x)=x^2$ , not just $x$ .

Step 1 — Recognize the chain rule applies. Because the upper limit is $u(x)=x^2$ , FTC Part 1 combines with the chain rule:

13

\quad G'(x) = f\big(u(x)\big)\cdot u'(x)

Derivation hint: Write $G(x)=A(u(x))$ where $A(u)=\int_2^u \sin t\,dt$ ; then $G'=A'(u)\cdot u'$ and $A'(u)=\sin u$ .

Step 2 — Substitute. Here $f(u)=\sin(x^2)$ and $u'(x)=2x$ :

14

\quad G'(x) = \sin(x^2)\cdot 2x

Derivation hint: Evaluate $\sin$ at the upper limit $x^2$ , then multiply by the derivative of $x^2$ .

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

⚠️ Common mistake

" $\int_a^b f(x)\,dx$ gives the geometric area under the curve." Not quite — it gives the signed (net) area. Regions where $f<0$ contribute negatively. For example $\int_0^{2\pi}\sin x\,dx = 0$ , even though the actual geometric area enclosed is $4$ . To get true geometric area, integrate $|f(x)|$ , splitting at the zeros of $f$ .

"You must use the same antiderivative as everyone else." FTC Part 2 works for any antiderivative. Adding a constant $C$ changes $F(b)$ and $F(a)$ by the same amount, so $C$ cancels in the subtraction. That is why definite integrals need no "+C".

"The dummy variable matters." $\int_a^x f(t)\,dt$ and $\int_a^x f(s)\,ds$ are identical functions of $x$ . Never reuse $x$ as both the limit and the integration variable — write $\int_a^x f(t)\,dt$ , not $\int_a^x f(x)\,dx$ .

CounterexampleContinuity Cannot Be Dropped

FTC Part 2 requires $f$ to actually possess an antiderivative on the whole interval and the integral to make sense. Consider

15

\quad \int_{-1}^{1} \frac{1}{x^2}\,dx \stackrel{?}{=} \left[-\frac{1}{x}\right]_{-1}^{1} = (-1)-(1) = -2

Derivation hint: Blindly applying Part 2 with $F(x)=-1/x$ gives a finite negative number. This is nonsense: the integrand $1/x^2$ is positive everywhere it is defined, so any genuine area must be positive — yet we got $-2$ . The catch: $f(x)=1/x^2$ is discontinuous (it blows up) at $x=0$ , which lies inside $[-1,1]$ . FTC Part 2 does not apply, and the integral in fact diverges to $+\infty$ . Always check the integrand is continuous on the entire closed interval.

connection

FTC is the keystone joining the two halves of calculus. It depends on The Definite Integral (defined as a limit of Riemann Sums) and on The Derivative. The proof of Part 1 rests on the Mean Value Theorem (via its integral form), and Part 2 rests on the corollary that zero-derivative functions are constant. Once FTC is in hand, every differentiation rule becomes an integration rule run backward: the chain rule becomes Integration by Substitution, and the product rule becomes Integration by Parts. The theorem also underwrites the Net Change Theorem in physics and generalizes, at higher levels, to The Generalized Stokes Theorem, which unifies it with Green's, the divergence, and the classical Stokes theorems.

🔬 Where this appears

⏳ History

- c. 287–212 BCE — Archimedes. Computes areas (e.g. the parabola segment) by the method of exhaustion, a precursor to integration, without any general theorem linking it to slopes.

- 1635 — Bonaventura Cavalieri. Develops the "method of indivisibles," treating areas as sums of infinitely thin lines, anticipating the integral.

- 1660s — Isaac Barrow. Newton's teacher at Cambridge; states a geometric theorem essentially equivalent to FTC, recognizing the inverse relation between tangent and area problems.

- c. 1666 — Isaac Newton. Develops the method of fluxions, treating the area under a curve as a flowing quantity whose rate equals the curve's ordinate.

- c. 1675 — Gottfried Wilhelm Leibniz. Independently formulates calculus with the modern

\int

and

dx

notation and explicitly frames differentiation and integration as inverse operations.

- 1684 & 1693 — Leibniz publishes. His differential and integral calculus appear in Acta Eruditorum, predating Newton's published account.

- 1687 — Newton's Principia. Uses calculus-based reasoning (in geometric guise) to derive the laws of motion and gravitation.

- 1823 — Augustin-Louis Cauchy. Provides the first rigorous definition of the definite integral as a limit and a careful proof of FTC for continuous functions.

- 1854 — Bernhard Riemann. Generalizes the integral (the Riemann integral), clarifying exactly which functions are integrable and sharpening the hypotheses of FTC.

open-problem

The "definitive" FTC for continuous functions hides deep refinements explored in advanced analysis:

How discontinuous can $f$ be and still satisfy Part 2? For the Riemann integral, every continuous function has an antiderivative, but a function can have an antiderivative without being Riemann-integrable, and vice versa — the relationship between having an antiderivative and being integrable is subtle.
The Lebesgue version. The correct generalization replaces "antiderivative" with absolutely continuous functions: $\int_a^b F'\,dx = F(b)-F(a)$ holds iff $F$ is absolutely continuous. A function can be continuous and increasing yet have $F'=0$ almost everywhere (the Cantor function, or "Devil's staircase"), for which FTC fails despite continuity — a striking counterexample to naive intuition.
Henstock–Kurzweil integral. This gauge integral restores the cleanest possible FTC: every derivative is integrable and recovers its primitive exactly, repairing pathologies the Lebesgue integral cannot.

These questions show that the elegant high-school statement is the visible tip of a rich theory about exactly when accumulation and rate-of-change remain perfect inverses.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also