Calculus

The Fundamental Theorem of Calculus

This is the theorem that ties all of calculus together. It says something astonishing: differentiation and integration are inverse operations.

Finding areas and finding slopes — two seemingly unrelated problems — turn out to be two sides of the same coin.

1. The Area Function

Start with a function f(x). Now define a new function F(x) that measures the accumulated area under f from 0 to x:

F(x) = \int_0^x f(t)\,dt

As x moves to the right, F(x) collects more area. When f is positive, F increases. When f is negative, F decreases. Let’s watch this happen.

x (upper bound of integral)2

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f(t) = t, \quad F(x) = \int_0^{0} t \, dt = \frac{0^2}0

Try This

Try this: Move x from 0 to the right. The area under the blue line grows, and the red curve (the area function) rises. Move x into negative territory: now you are integrating “backwards,” and the red curve shows negative area becoming positive (since f is negative for negative x, the integral accumulates negative values first).

2. The Punchline: F’(x) = f(x)

Here is the miracle. If you take the derivative of the area function F(x), you get back the original function f(x):

\frac00\int_0^x f(t)\,dt = f(x)

Let’s verify this visually with f(t) = sin(t). The area function is F(x) = 1 - cos(x), and its derivative is sin(x) — we get f back.

x (examine this point)2

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f(x) = \sin(x), \quad F(x) = 1 - \cos(x), \quad F'(x) = \sin(x) = f(x) \checkmark

Connection

The blue curve and the green curve are identical — that is the whole point! F’(x) = f(x). The derivative of the area function gives back the original function. This means integration and differentiation undo each other.

3. Why It Works (Intuitively)

Think about what F(x + h) - F(x) means. It is the area under f from x to x + h — a thin strip. If h is tiny, that strip is approximately a rectangle with height f(x) and width h:

F(x+h) - F(x) \approx f(x) \cdot h

\frac{F(x+h) - F(x)}0 \approx f(x)

As h goes to zero, the approximation becomes exact, and the left side is the definition of F’(x). So F’(x) = f(x).

4. Part Two: Evaluating Definite Integrals

The second part of the fundamental theorem gives you the practical payoff. If F is any antiderivative of f, then:

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

No Riemann sums needed. Just find the antiderivative and subtract.

a (lower bound)0

b (upper bound)3

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\int_{{0}}^{0} x^2 \, dx = \frac{0^3}0 - \frac{{0}^3}0

Try This

Try this: Set a = 0 and b = 3 to get the integral of x^2 from 0 to 3 = 27/3 = 9. Now move a to 1: the area from 1 to 3 is 27/3 - 1/3 = 26/3. The antiderivative lets you compute any definite integral instantly.

5. The Big Picture: Two Operations, One Theorem

Starting from cos(x): integrate to get sin(x), then differentiate sin(x) to get back cos(x). Or differentiate cos(x) to get -sin(x), then integrate -sin(x) to get back cos(x). Round trips in both directions.

Challenge

Challenge: The function f(x) = 3x^2 has antiderivative F(x) = x^3. Use the fundamental theorem to compute the integral of 3x^2 from x = 1 to x = 4. Then check your answer by noting that 3x^2 is the derivative of x^3.

The Big Idea

The Fundamental Theorem of Calculus says that integration and differentiation are inverse operations. The derivative of the area function gives back the original function, and any definite integral can be computed using antiderivatives.

This one theorem unifies the two halves of calculus. Finding slopes (derivatives) and finding areas (integrals) are not separate subjects — they are mirror images of each other. That is why the course is called “calculus” (singular), not “two unrelated things about functions.”

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