Fundamental Theorem of Calculus

Part 1 — derivatives of integrals

If f is continuous on [a, b], then the function F defined by:

F(x) = ∫_a^x f(t) dt

is differentiable on (a, b), and F'(x) = f(x). In words — the derivative of an accumulation function is the original function being accumulated. Differentiating an integral with respect to its upper limit recovers the integrand.

This means F is an antiderivative of f. Every continuous function HAS an antiderivative — namely, its accumulation function from any starting point. We can't always express that antiderivative in closed form (∫e^(−x²) dx has no elementary antiderivative), but it exists.

Part 2 — definite integrals from antiderivatives

If F is any antiderivative of a continuous function f on [a, b], then:

∫_a^b f(x) dx = F(b) − F(a)

Pick any antiderivative — F or F + C, both work. Evaluate at the endpoints. Subtract. That's the integral.

Without this, computing ∫_0^1 x² dx would mean computing lim_{n→∞} Σ (i/n)² · 1/n — a Riemann sum that requires careful algebra to evaluate. With the FTC, we know x³/3 is an antiderivative, so the integral is just (1/3) − 0 = 1/3. Same answer in one line vs. half a page.

A geometric proof of Part 1

Define F(x) = ∫_a^x f(t) dt. By definition, F(x + h) − F(x) is the area under f from x to x + h.

For small h, the strip from x to x + h is approximately a rectangle of width h and height f(x) — area f(x) · h. So:

F(x + h) − F(x) ≈ f(x) · h
(F(x + h) − F(x)) / h ≈ f(x)

The left side is the difference quotient — its limit as h → 0 is F'(x). The right side is f(x). So F'(x) = f(x), QED.

This is heuristic; the rigorous proof uses the Mean Value Theorem for integrals to make "approximately a rectangle" precise. But the intuition is exactly right — accumulating a function f and then taking the rate of change of that accumulation gives back f.

Worked examples

Example 1 — Compute ∫_0^1 x² dx

Antiderivative — F(x) = x³/3 (since (x³/3)' = x²).

∫_0^1 x² dx = F(1) − F(0) = 1/3 − 0 = 1/3

Example 2 — Compute ∫_0^π sin(x) dx

Antiderivative — F(x) = −cos(x).

∫_0^π sin(x) dx = −cos(π) − (−cos(0)) = −(−1) − (−1) = 1 + 1 = 2

Example 3 — Compute the derivative of an integral

What's d/dx ∫_2^x cos(t²) dt? By Part 1, this is cos(x²). The antiderivative isn't expressible in elementary functions, but its derivative IS — directly, without computing the integral first.

Example 4 — variable upper limit with a function

What's d/dx ∫_0^(x²) e^(−t) dt? Apply Part 1 with the chain rule. Let u = x². Then:

d/dx [∫_0^u e^(−t) dt] = e^(−u) · du/dx = e^(−x²) · 2x

Why this is the most important theorem

Without the FTC:

• Computing any definite integral requires evaluating a Riemann sum limit.
• Closed-form integration would be impossible — you'd compute every integral numerically.
• Most of physics and engineering would be intractable.

With the FTC:

• To integrate, find an antiderivative — a much simpler task.
• You can verify integrals by differentiating — a robust check.
• Differential equations become tractable — solving f'(x) = g(x) is "find F such that F' = g."
• Modern calculus practice (substitution, integration by parts, partial fractions) all rely on the FTC linkage.

Generalizations

Theorem	Dimension	Statement
Fundamental Theorem of Calculus	1D	∫_a^b F'(x) dx = F(b) − F(a)
Green's theorem	2D	Line integral around curve = double integral of curl over enclosed region
Stokes' theorem	2D surface in 3D	Line integral on boundary = surface integral of curl
Divergence theorem (Gauss)	3D	Triple integral of divergence = surface integral of flux on boundary
Generalized Stokes' theorem	Any manifold	∫_∂M ω = ∫_M dω — the integral of a form on a manifold equals integral of its differential on the boundary

The pattern is the same — local accumulation equals boundary difference. Each theorem is FTC at higher dimension. The most general form (Stokes' on manifolds) reduces to all the others as special cases.

JavaScript — verifying the FTC numerically

// Numerical Riemann sum vs antiderivative — should agree
function riemannSum(f, a, b, n = 100000) {
  const h = (b - a) / n;
  let sum = 0;
  for (let i = 0; i < n; i++) sum += f(a + i * h);
  return sum * h;
}

// f(x) = x², antiderivative F(x) = x³/3
function f(x) { return x * x; }
function F(x) { return x ** 3 / 3; }

const a = 0, b = 1;

const numerical = riemannSum(f, a, b);
const ftc = F(b) - F(a);

console.log('Riemann sum:', numerical);  // ≈ 0.3333...
console.log('FTC:        ', ftc);        // exactly 1/3 = 0.333...

// They agree — that's the FTC working in practice. The Riemann sum took
// 100,000 evaluations; the FTC took 2.

Common mistakes

• Forgetting that f must be continuous. Both parts assume continuity. Discontinuities (especially infinite ones) require splitting the integral or using improper-integral techniques.
• Confusing definite and indefinite integrals. Definite — ∫_a^b f(x) dx — is a number. Indefinite — ∫f(x) dx — is a family of functions (with + C). FTC bridges them — definite integrals computed via indefinite ones.
• Wrong order of evaluation. F(b) − F(a), upper minus lower. Reversing gives the negative — and corresponds to integrating "backwards," which has its own meaning (∫_a^b f = −∫_b^a f).
• Forgetting the chain rule with variable bounds. d/dx ∫_a^(g(x)) f(t) dt = f(g(x)) · g'(x). The "inner derivative" g'(x) is required when the upper bound is a function of x, not just x itself.
• Treating the integral as just a number when it's a function. ∫_0^x f(t) dt is a function of x, with derivative f(x). Don't compute it as a number, then try to differentiate the number — it depends on x.
• Adding constants to a definite integral. Definite integrals are numbers; no + C. The constant matters only in indefinite integrals.