Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus: The Bridge

The Fundamental Theorem of Calculus (FTC) reveals the deep connection between differentiation (finding slopes/rates of change) and integration (finding areas/accumulated change): the two operations are, in fact, inverses of each other. It has two parts — the first deals with the derivative of an integral function, the second gives a practical method to evaluate definite integrals without computing limits of sums.

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Part 1: The Derivative of an Integral (The Area Function)

For a continuous function $f$ on $[a, b]$ and any $x$ in $[a, b]$, define the area function giving the area under the curve $y = f(t)$ from the fixed starting point $a$ up to the variable point $x$:

What this means: the rate at which the accumulated area changes at a point $x$ is exactly the height of the original function $f$ at that same point. Differentiating an integral (with a variable upper limit) gives back the original integrand — integration and differentiation are inverse processes.

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Part 2: Evaluating Definite Integrals (The Evaluation Theorem)

This is the part of the FTC you will use most often for calculations.

To find the definite integral, find an antiderivative $F$ of $f$, evaluate it at the upper limit $b$ and the lower limit $a$, and subtract. This is often written with a special notation:

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Proof of the Second Fundamental Theorem of Calculus

›Proof

Step 1: Define the area function.

Let $A(x) = \int_a^x f(t) \, dt$. By the First FTC, $A'(x) = f(x)$, so $A(x)$ is an antiderivative of $f(x)$.

Step 2: Relate any antiderivative to the area function.

Let $F(x)$ be any antiderivative of $f(x)$. Since $A(x)$ is also an antiderivative, they differ only by a constant: $F(x) = A(x) + C$.

Step 3: Find the constant $C$.

At $x = a$: $F(a) = A(a) + C$. But $A(a) = \int_a^a f(t) \, dt = 0$, so $C = F(a)$.

Step 4: Write the general antiderivative. …