The Fundamental Theorem of Calculus

Integration and differentiation were defined by unrelated limits — one a supremum of areas, the other a limit of slopes. The fundamental theorem of calculus proves they are inverse operations. It comes in two forms: one evaluates a definite integral through an antiderivative, the other differentiates an integral with a moving upper limit. Almost all of integral calculus rests on the pair.¹

First form: evaluation

If a function is the derivative of something, its integral is the net change of that something.

Evaluation form via the mean value theorem. On each subinterval a point

c_{i}

makes the rectangle of height

f (c_{i})

have area equal to the change in

F

across that piece; the changes telescope to

F (b) - F (a)

A finite set of points where $F$ fails to be differentiable is harmless, provided $F$ stays continuous and an integrable $f$ agrees with $F^{'}$ elsewhere.¹

Second form: differentiation of the area function

Running the integral as a function of its upper limit produces an antiderivative of the integrand.

Geometrically, $F^{'} (c)$ is the height of the thin strip of area added as $x$ passes $c$ .

The area function

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

grows by a thin strip when

x

advances by

h

. That strip has area about

f (x) h

, so the difference quotient tends to

f (x)

and

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

When $f$ is continuous on all of $[a, b]$ , $F$ is differentiable everywhere with $F^{'} = f$ : every continuous function has an antiderivative, namely its own area function. The two forms then compose into the pair of identities

a \int b F^{'} = F (b) - F (a), \frac{d}{d x} a \int x f = f (x) .

The two forms as inverse operations: differentiating the area function returns the integrand, and integrating a derivative returns the net change of the function.

Differentiating an integral with a variable limit

Composing the differentiation form with the chain rule differentiates an integral whose limit is itself a function.

Continuity is necessary, and the converse fails

Both forms hinge on continuity, and two examples mark the limits.

Where $f$ jumps, $F$ need not be differentiable. Let $f (x) = - 1$ for $x < 0$ and $f (x) = 1$ for $x \geq 0$ . Then $F (x) = \int_{0}^{x} f = ∣ x ∣$ , which has no derivative at $0$ — exactly where $f$ jumps.¹
The converse of the differentiation form is false. Let $g (x) = 0$ for $x \neq = 0$ and $g (0) = 1$ . Then $G (x) = \int_{0}^{x} g = 0$ for all $x$ , so $G^{'} (0) = 0$ exists even though $g$ is discontinuous at $0$ . Differentiability of the area function does not require continuity of the integrand.¹

Many antiderivatives have no elementary formula. The logarithm is defined as $ln x = \int_{1}^{x} \frac{1}{s} d s$ , and the error function $erf (x) = \frac{2}{π} \int_{0}^{x} e^{- s^{2}} d s$ has no closed form; the differentiation form still guarantees each is an antiderivative, and numerical approximation of the integral is how their values are found.¹

Integration by parts

The product rule, run through the evaluation form, becomes a rule for integrals.

The geometry is a rectangle in the $(f, g)$ plane: as $t$ runs from $a$ to $b$ the point $(f (t), g (t))$ traces a curve, and the two integrals are the areas on the two sides of that curve, adding to the change in the corner rectangle's area.

Integration by parts as an area split. As

t

runs over

[a, b]

the point

(f, g)

traces a curve; the area under it (giving

\int g df

) plus the area left of it (giving

\int f d g

) equal the change in the corner rectangle

f (b) g (b) - f (a) g (a)

Change of variables

Substitution is the chain rule read backwards through the integral.

The hypotheses carry weight: $\int_{- 1}^{1} \frac{ln ∣ x ∣}{x} d x$ cannot be attacked by $u = ln ∣ x ∣$ , because the integrand is unbounded and not even defined at $0$ , so the substitution manipulates symbols that carry no meaning.¹

Form	Statement	Reads
Evaluation	$\int_{a}^{b} F^{'} = F (b) - F (a)$	integral of a derivative is net change
Differentiation	$\frac{d}{d x} \int_{a}^{x} f = f (x)$	derivative of area is the integrand
By parts	$\int f g^{'} = f g_{a}^{b} - \int f^{'} g$	product rule, integrated
Substitution	$\int f (g) g^{'} = \int_{g (a)}^{g (b)} f$	chain rule, integrated

Lebl, Basic Analysis I, §5.3 — the evaluation form (Thm. 5.3.1), the differentiation form with Lipschitz continuity of the area function (Thm. 5.3.3), the $∣ x ∣$ and constant-zero counterexamples, change of variables (Thm. 5.3.5), and the $ln$ / $erf$ remarks. ↩ ↩² ↩³ ↩⁴ ↩⁵ ↩⁶ ↩⁷ ↩⁸
Shkoller, MAT125B Lecture Notes, §1.11 — the fundamental theorem of calculus (Thm. 1.34), the non-differentiable area-function example, the antiderivative computation of $\int_{0}^{1} x^{2}$ , and integration by parts (Thm. 1.39). ↩ ↩² ↩³

First form: evaluation

Second form: differentiation of the area function

Differentiating an integral with a variable limit

Continuity is necessary, and the converse fails

Integration by parts

Change of variables

Footnotes