Mathematics · Ch 16 — Integration

First Fundamental Theorem of Integral Calculus

16.8.2

First Fundamental Theorem of Integral Calculus

The First Fundamental Theorem of Integral Calculus

This theorem establishes the crucial link between the two branches of calculus: differentiation and integration. It tells us that the rate at which the area under a curve changes is exactly the value of the function itself.

The Area Function and Its Derivative

For a continuous function $f$ on $[a, b]$ and any $x$ in $[a, b]$ , define the area function — the area under $y = f(t)$ from the fixed left endpoint $a$ to the variable right endpoint $x$ :

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

(The variable of integration is $t$ to avoid confusion with the upper limit $x$ .) The First Fundamental Theorem states that the derivative of this area function is the original function evaluated at $x$ :

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

This holds for all $x$ in the closed interval $[a, b]$ .

Why This Makes Sense (A Geometric Proof)

›Proof

Geometric Derivation of $A'(x) = f(x)$

For a small increment $h > 0$ , the change in area is the thin vertical strip under the curve between $x$ and $x+h$ :

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

For a continuous function and small $h$ , $f(t)$ on $[x, x+h]$ is approximately the constant $f(x)$ , so this strip is approximately a rectangle of height $f(x)$ and width $h$ :

$\Delta A \approx f(x) \cdot h \quad\Longrightarrow\quad \frac{\Delta A}{h} \approx f(x)$

Taking the limit as $h \to 0$ makes the approximation exact:

$A'(x) = \lim_{h \to 0} \frac{A(x+h) - A(x)}{h} = \lim_{h \to 0} \frac{1}{h} \int_x^{x+h} f(t) \, dt = f(x)$

The instantaneous rate of change of the accumulated area is precisely the height of the curve at that point.

The Theorem in Context …

Theorem 1

Theorem 1 (First Fundamental Theorem of Integral Calculus)

Let $f$ be a continuous function on the closed interval $[a, b]$ , and define the area function

A(x) = \int_{a}^{x} f(t) \, dt, \quad x \in [a, b].

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

A'(x) = f(x) \quad \text{for all } x \in [a, b].

Important

The theorem says: the derivative of the area under the curve from a fixed point $a$ to a variable point $x$ is exactly the original function $f$ at $x$ .

This is the first link between integration and differentiation — it tells us that integration (building area) and differentiation (finding slope) are inverse processes.

Complete Proof

›Proof

Step 1 – Set up the difference quotient.

For a small increment $h \neq 0$ such that $x+h$ still lies in $[a, b]$ , consider

$A(x+h) - A(x) = \int_{a}^{x+h} f(t) \, dt - \int_{a}^{x} f(t) \, dt.$

By the property of definite integrals (additivity of limits),

$A(x+h) - A(x) = \int_{x}^{x+h} f(t) \, dt.$

Step 2 – Apply the Mean Value Theorem for integrals.

Since $f$ is continuous on $[x, x+h]$ , by the Mean Value Theorem for integrals there exists some $c$ in $[x, x+h]$ such that

$\int_{x}^{x+h} f(t) \, dt = f(c) \cdot h.$

Therefore,

$A(x+h) - A(x) = f(c) \cdot h.$

Step 3 – Form the difference quotient.

Divide both sides by $h$ (remember $h \neq 0$ ):

$\frac{A(x+h) - A(x)}{h} = f(c).$

Step 4 – Take the limit as $h \to 0$ .

As $h \to 0$ , the interval $[x, x+h]$ shrinks to the single point $x$ . The point $c$ lies in that interval, so $c \to x$ . Because $f$ is continuous at $x$ ,

$\lim_{h \to 0} f(c) = f(x).$

Hence,

$\lim_{h \to 0} \frac{A(x+h) - A(x)}{h} = f(x).$

Step 5 – Conclude.

The limit on the left is precisely the derivative $A'(x)$ . Thus

$A'(x) = f(x) \quad \text{for all } x \in [a, b].$

This completes the proof.

When Is This Theorem Used? …

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