What the Theorem Says

The Fundamental Theorem of Calculus (FTC) is the central result of the subject. It tells us that differentiation and integration are inverse processes of each other. It comes in two parts.

Part 1 — The Derivative of an Integral

If f is continuous on [a, b] and we define:

F(x) = ∫ax f(t) dt

Then F is differentiable and F'(x) = f(x). The derivative of an area-accumulation function returns the original function.

Intuition

Think of filling a tank with water. F(x) is the total volume at time x. The rate at which volume increases is exactly f(x) — the flow rate at that moment. Differentiating accumulated total gives back instantaneous rate.

Part 2 — Evaluating Definite Integrals

If F is any antiderivative of f, then:

∫ab f(x) dx = F(b) − F(a)
📋 Worked Example

Find: ∫₀³ x² dx

Step 1Antiderivative: F(x) = x³/3
Step 2F(3) − F(0) = 27/3 − 0 = 9

Why It Matters

Before the FTC, computing areas required exhausting Riemann sum calculations. The FTC turns that into simple algebra: find an antiderivative, plug in two numbers, subtract. This is why calculus became so powerful — a hard geometric problem became an algebraic one.

Frequently Asked Questions

What is the difference between Part 1 and Part 2?▾
Part 1 proves integration and differentiation undo each other. Part 2 gives the practical formula for evaluating definite integrals. Part 2 is what you use in calculations; Part 1 is the deeper theoretical result that justifies it.
Does the FTC work for all functions?▾
Part 1 requires f to be continuous. Part 2 requires f to be integrable and F to be a valid antiderivative. For discontinuous functions, more care is needed.

Why "Fundamental"?

Before the FTC, integration and differentiation appeared to be two completely unrelated problems. Integration was about area — an ancient geometric question. Differentiation was about tangent lines and rates — a modern algebraic question. Newton and Leibniz's great discovery was that these are not two problems but one: they are inverse operations, like multiplication and division.

This connection is so profound that it transforms calculus from a collection of techniques into a unified subject. Every time you evaluate a definite integral by finding an antiderivative, you are applying the FTC. It is not one tool among many — it is the tool that makes integration computable.

Part 1 — Why It's True

Define F(x) = ∫ₐˣ f(t) dt. We want to show F'(x) = f(x). By the definition of the derivative:

F'(x) = limh→0 [F(x+h) − F(x)] / h = limh→0 [∫ₐ^(x+h) f(t)dt − ∫ₐˣ f(t)dt] / h = limh→0 (1/h) ∫ₓ^(x+h) f(t) dt

For small h, f(t) ≈ f(x) on [x, x+h], so the integral ≈ f(x)·h. Dividing by h gives f(x). The formal proof uses the Mean Value Theorem for integrals to make this rigorous.

More Worked Examples

📋 Using FTC Part 2
∫₁⁴ √x dxF(x) = (2/3)x^(3/2). Answer: (2/3)(8) − (2/3)(1) = 16/3 − 2/3 = 14/3
∫₀^(π/2) cos x dxF(x) = sin x. Answer: sin(π/2) − sin(0) = 1 − 0 = 1
∫₁ᵉ (1/x) dxF(x) = ln x. Answer: ln(e) − ln(1) = 1 − 0 = 1
∫₋₁¹ x³ dxF(x) = x⁴/4. Answer: 1/4 − 1/4 = 0 (odd function, symmetric interval)

Using FTC Part 1 — Differentiating Integrals

📋 d/dx of an integral with variable upper limit
d/dx ∫₀ˣ t³ dt= x³ (direct application of Part 1)
d/dx ∫₀^(x²) sin t dt= sin(x²) · 2x (chain rule: inner function is x²)
d/dx ∫ₓ⁵ eᵗ dt= −eˣ (flip limits, apply Part 1)

The Most Common Misconception

⚠ Watch Out

The FTC says ∫ₐᵇ f(x)dx = F(b) − F(a) where F' = f. Students often forget that F must be differentiable on the entire interval [a, b]. If f has a discontinuity in [a, b], the FTC does not directly apply — you need to split the integral or use improper integral techniques.

AM
Dr. Aisha Malik, PhD Mathematics
Senior Lecturer in Applied Mathematics · 12 years teaching calculus
Dr. Malik holds a PhD in Applied Mathematics from the University of Edinburgh and has taught calculus to over 4,000 students. She has reviewed this article for mathematical accuracy and pedagogical clarity.
Technically reviewed by: Prof. James Chen, Stanford Mathematics Department · April 2026