Integration · Fundamental Theorem

The Fundamental Theorem of Calculus

The most important result in all of calculus: the bridge that connects differentiation and integration, turns area problems into antiderivative lookups, and unlocks the entire machinery of integral calculus. This page covers both parts, their proofs, the chain rule extension, and 12 fully worked examples.

Share this page

§ 01Why This Theorem Is Fundamental

Before the Fundamental Theorem, computing the area under a curve meant summing thousands of rectangles. Differentiation and integration appeared to be completely unrelated operations. The Fundamental Theorem revealed they are perfect inverses of one another.

The theorem comes in two parts. Part 1 tells you that differentiation undoes integration: if you integrate a function and then differentiate the result, you recover the original function. Part 2 tells you the practical payoff: instead of computing a limit of Riemann sums to find ∫ₐᵇ f(x) dx, you only need to find any antiderivative F and evaluate F(b) − F(a).

This is why calculus is powerful. A calculation that would take hours of summing rectangles collapses to a single subtraction. Newton and Leibniz's greatest insight was recognising this connection, and it transformed mathematics, physics, and engineering forever.

Prerequisites You should know: limits and continuity, the definition of the definite integral as a limit of Riemann sums, and basic differentiation. Familiarity with antiderivatives is assumed.

§ 02The Two Parts, Side by Side

Here are both parts stated precisely. Read them carefully, they say subtly different things.

FTC · Part 1
The Derivative of an Integral
d/dx ∫ₐˣ f(t) dt = f(x)

Conditions: f is continuous on an open interval containing a.

Meaning: Differentiation and integration are inverse operations. The function g(x) = ∫ₐˣ f(t) dt is an antiderivative of f(x).

FTC · Part 2
Evaluating Definite Integrals
∫ₐᵇ f(x) dx = F(b) − F(a)

Conditions: f is continuous on [a, b] and F is any antiderivative of f (F′ = f).

Meaning: To evaluate a definite integral, find an antiderivative and subtract its values at the endpoints.

The Notation F(b) − F(a) This is so commonly written that mathematicians have a shorthand: [F(x)]ₐᵇ or F(x)|ₐᵇ, both meaning "evaluate F at b then subtract F at a". You will see both notations in textbooks and examinations.

§ 03Proof Sketch — FTC Part 1

Let g(x) = ∫ₐˣ f(t) dt. We want to show g′(x) = f(x). By the definition of the derivative:

Derivative from First Principles
g′(x) = limh→0 [g(x+h) − g(x)] / h

  • Write g(x+h) − g(x) as a single integral using interval additivity:

    g(x+h) − g(x) = ∫ₐˣ⁺ʰ f(t) dt − ∫ₐˣ f(t) dt = ∫ₓˣ⁺ʰ f(t) dt

  • By the Mean Value Theorem for Integrals, there exists a number c between x and x+h such that:

    ∫ₓˣ⁺ʰ f(t) dt = f(c) · h

  • Divide by h:

    [g(x+h) − g(x)] / h = f(c)

  • As h → 0, c is squeezed between x and x+h, so c → x. Since f is continuous, f(c) → f(x).

    g′(x) = limh→0 f(c) = f(x) ✓
x y a x x+h g(x) = ∫ₐˣ f(t)dt ≈ f(x)·h f(x)
The accumulator function g(x) = ∫ₐˣ f(t) dt. Adding a thin strip of width h adds area ≈ f(x)·h, so dg/dx = f(x).

§ 04Proof Sketch — FTC Part 2

We want to show that ∫ₐᵇ f(x) dx = F(b) − F(a) for any antiderivative F of f.

  • From FTC Part 1, the function g(x) = ∫ₐˣ f(t) dt is an antiderivative of f. That is, g′(x) = f(x).

  • Any two antiderivatives of the same function differ by a constant. So if F is another antiderivative, then F(x) = g(x) + C for some constant C.

  • Note that g(a) = ∫ₐᵃ f(t) dt = 0 (integral over a zero-width interval).

  • Now evaluate F(b) − F(a):

    F(b) − F(a) = [g(b) + C] − [g(a) + C] = g(b) − g(a) = g(b) − 0 = g(b)

  • But g(b) = ∫ₐᵇ f(t) dt by definition. Therefore:

    ∫ₐᵇ f(x) dx = F(b) − F(a) ✓
Why the Constant C Disappears This is the beautiful reason you do not need to include +C when evaluating definite integrals. The constant cancels: [F(x) + C]ₐᵇ = F(b) + C − F(a) − C = F(b) − F(a). Always write the antiderivative without +C for definite integrals.

§ 05Why They Are the Same Theorem

Parts 1 and 2 are sometimes presented as separate results, but they are two faces of the same coin. Part 1 gives the theoretical engine: differentiation and integration invert each other. Part 2 gives the practical tool that flows from Part 1.

Aspect FTC Part 1 FTC Part 2
What it says d/dx ∫ₐˣ f(t) dt = f(x) ∫ₐᵇ f(x) dx = F(b) − F(a)
Direction Integration then differentiation Find antiderivative, then subtract
Primary use Differentiating integral expressions Evaluating definite integrals
Key condition f continuous near a f continuous on [a, b]
Output A function (the derivative) A number (the area)

§ 06FTC Part 1 with the Chain Rule

A very common exam question combines FTC Part 1 with the chain rule. What if the upper limit is a function of x rather than just x itself?

General Form — Variable Upper Limit
d/dx ∫ₐ^(g(x)) f(t) dt = f(g(x)) · g′(x)

This follows directly from FTC Part 1 plus the chain rule. Think of it as: evaluate f at the upper limit, then multiply by the derivative of the upper limit.

The rule extends further: if both limits are functions of x, split the integral at a constant and apply the rule twice:

Both Limits Are Functions
d/dx ∫_(h(x))^(g(x)) f(t) dt = f(g(x))·g′(x) − f(h(x))·h′(x)
The Most Common Mistake with FTC Part 1 Students often forget to multiply by the derivative of the upper limit. If the upper limit is x², the answer is f(x²) · 2x, not just f(x²). Always apply the chain rule.

§ 07Worked Examples

Example 1 Basic FTC Part 2 — polynomial

Evaluate ∫₁⁴ (3x² − 2x + 1) dx.

  1. Find the antiderivative: F(x) = x³ − x² + x.
  2. Apply FTC Part 2: F(4) − F(1).
  3. F(4) = 64 − 16 + 4 = 52.
  4. F(1) = 1 − 1 + 1 = 1.
∫₁⁴ (3x²−2x+1) dx = 52 − 1 = 51
Example 2 FTC Part 2 — trig function

Evaluate ∫₀^(π/2) cos x dx.

  1. Antiderivative of cos x is sin x.
  2. [sin x]₀^(π/2) = sin(π/2) − sin(0) = 1 − 0 = 1.
∫₀^(π/2) cos x dx = 1
Example 3 FTC Part 2 — exponential

Evaluate ∫₀² eˣ dx.

  1. Antiderivative of eˣ is eˣ (the magic of e!).
  2. [eˣ]₀² = e² − e⁰ = e² − 1.
∫₀² eˣ dx = e² − 1 ≈ 6.389
Example 4 FTC Part 2 — rational function

Evaluate ∫₁^e (1/x) dx.

  1. Antiderivative of 1/x is ln|x|.
  2. [ln x]₁^e = ln(e) − ln(1) = 1 − 0 = 1.
∫₁^e (1/x) dx = 1
Example 5 FTC Part 1 — basic application

Find g′(x) where g(x) = ∫₂ˣ (t³ + sin t) dt.

  1. By FTC Part 1, g′(x) = f(x) where f(t) = t³ + sin t.
  2. Simply replace t with x.
g′(x) = x³ + sin x
Example 6 FTC Part 1 — variable lower limit

Find d/dx ∫ₓ⁵ (t² + 1) dt.

The variable is in the lower limit. Flip the integral using ∫ₓ⁵ = −∫₅ˣ.

  1. d/dx [−∫₅ˣ (t²+1) dt] = −(x²+1).
d/dx ∫ₓ⁵ (t²+1) dt = −(x²+1)
Example 7 FTC Part 1 with Chain Rule — upper limit is x²

Find d/dx ∫₁^(x²) √(1+t³) dt.

  1. Upper limit g(x) = x², so g′(x) = 2x.
  2. Integrand evaluated at upper limit: f(x²) = √(1+(x²)³) = √(1+x⁶).
  3. Apply general FTC: f(g(x)) · g′(x).
d/dx = √(1+x⁶) · 2x = 2x√(1+x⁶)
Example 8 FTC Part 1 with Chain Rule — upper limit is sin x

Find d/dx ∫₀^(sin x) e^(t²) dt.

  1. Upper limit g(x) = sin x, g′(x) = cos x.
  2. f(g(x)) = e^(sin²x).
  3. Answer: e^(sin²x) · cos x.
d/dx = e^(sin²x) · cos x
Example 9 Both limits variable — use split rule

Find d/dx ∫_(x)^(x²) cos(t²) dt.

  1. Split at any constant c: ∫_(x)^(x²) = ∫_(x)^(c) + ∫_(c)^(x²) = −∫_(c)^(x) + ∫_(c)^(x²).
  2. Differentiate each part separately using FTC+chain rule:
  3. From −∫_(c)^(x): −cos(x²) · 1 = −cos(x²).
  4. From ∫_(c)^(x²): cos((x²)²) · 2x = cos(x⁴) · 2x.
d/dx = 2x·cos(x⁴) − cos(x²)
Example 10 Net area — positive and negative regions

Evaluate ∫₀^(2π) sin x dx and explain the result geometrically.

  1. Antiderivative of sin x is −cos x.
  2. [−cos x]₀^(2π) = −cos(2π) − (−cos(0)) = −1 + 1 = 0.

The result is 0 because sin x is positive on (0, π) and negative on (π, 2π), and the two regions have equal area. The definite integral gives net signed area, not total area.

∫₀^(2π) sin x dx = 0 (net area is zero; total area is 4)
Example 11 Using FTC to find total area (absolute value)

Find the total area enclosed between y = sin x and the x-axis on [0, 2π].

  1. Split into regions where sin x ≥ 0 and sin x ≤ 0.
  2. On [0, π]: sin x ≥ 0. Area = ∫₀^π sin x dx = [−cos x]₀^π = 2.
  3. On [π, 2π]: sin x ≤ 0. Area = |∫_π^(2π) sin x dx| = |[−cos x]_π^(2π)| = |−2| = 2.
  4. Total area = 2 + 2 = 4.
Total area = 4 square units
Example 12 Using FTC to evaluate a limit of a Riemann sum

Evaluate: lim_(n→∞) Σ_(k=1)^n (k/n)² · (1/n).

Recognise this as a Riemann sum for ∫₀¹ x² dx with partition width 1/n and right endpoints k/n.

  1. The limit equals ∫₀¹ x² dx.
  2. Antiderivative: x³/3. Evaluate: [x³/3]₀¹ = 1/3 − 0 = 1/3.
lim_(n→∞) Σ (k/n)²(1/n) = 1/3

§ 08Common Mistakes

Mistake 1 — Including +C in a definite integral When applying FTC Part 2, never write +C. The constant cancels automatically. Writing ∫₁² x dx = [x²/2 + C]₁² is needlessly confusing and wrong in exams that require the simplified answer.
Mistake 2 — Forgetting the chain rule in FTC Part 1 d/dx ∫₀^(x³) f(t) dt = f(x³) · 3x², not f(x³). The 3x² factor is essential. Every time the upper limit is not simply x, chain rule applies.
Mistake 3 — Applying FTC to a discontinuous function If f has a discontinuity inside [a, b], FTC Part 2 does not apply directly. You must split the integral at the discontinuity and treat each piece as an improper integral. Applying F(b)−F(a) blindly gives a wrong answer.
Mistake 4 — Confusing net area with total area ∫ₐᵇ f(x) dx gives signed (net) area, not total area. If f dips below the x-axis, the negative contributions cancel the positive ones. For total area, integrate |f(x)| by splitting the interval at the zeros of f.
Mistake 5 — Wrong direction when the lower limit contains the variable d/dx ∫ₓᵇ f(t) dt = −f(x), NOT f(x). Flip the limits (reversing sign) before applying FTC Part 1.

§ 09Test Yourself — 10-Question Quiz

Questions alternate between fill-in-the-blank computations and multiple-choice conceptual questions. Exact or decimal answers accepted where specified.

Score: 0 / 0
Question 1 of 10

Evaluate ∫₀³ (2x + 1) dx. Enter your answer.

  1. F(x) = x² + x.
  2. F(3) = 9 + 3 = 12. F(0) = 0.
  3. Answer: 12 − 0 = 12.
Question 2 of 10

What does FTC Part 1 state?

FTC Part 1 says that the accumulator function g(x) = ∫ₐˣ f(t) dt is an antiderivative of f. Differentiating it recovers f(x). Option (b) is FTC Part 2.

Question 3 of 10

Evaluate ∫₀^(π/2) sin x dx. Enter your answer.

  1. Antiderivative of sin x is −cos x.
  2. [−cos x]₀^(π/2) = −cos(π/2) − (−cos 0) = 0 + 1 = 1.
Question 4 of 10

Find d/dx ∫₁^(x³) √(t⁴+1) dt.

  1. Upper limit g(x) = x³, g′(x) = 3x².
  2. Integrand at upper limit: √((x³)⁴+1) = √(x¹²+1).
  3. By FTC + chain rule: √(x¹²+1) · 3x².
Question 5 of 10

Evaluate ∫₁^e (3/x) dx. Enter your answer.

  1. Antiderivative: 3 ln x.
  2. [3 ln x]₁^e = 3 ln(e) − 3 ln(1) = 3·1 − 3·0 = 3.
Question 6 of 10

Why does +C disappear when evaluating a definite integral?

The constant C is added and subtracted: F(b)+C − (F(a)+C) = F(b)−F(a). The C terms cancel exactly, regardless of what value C takes. This is why any antiderivative works in FTC Part 2.

Question 7 of 10

Let g(x) = ∫₃ˣ (t² − 4) dt. Find g′(5).

  1. By FTC Part 1: g′(x) = x² − 4.
  2. g′(5) = 25 − 4 = 21.
Question 8 of 10

What is d/dx ∫ₓ^4 cos(t³) dt?

  1. Variable is in the lower limit, so flip: d/dx[−∫₄ˣ cos(t³)dt].
  2. = −cos(x³) · 1 = −cos(x³).
Question 9 of 10

Evaluate ∫₋₁^2 (x² − x) dx. Enter your answer.

  1. Antiderivative: F(x) = x³/3 − x²/2.
  2. F(2) = 8/3 − 4/2 = 8/3 − 2 = 2/3.
  3. F(−1) = −1/3 − 1/2 = −2/6 − 3/6 = −5/6.
  4. Answer: 2/3 − (−5/6) = 4/6 + 5/6 = 9/6 = 3/2 = 1.5.
Question 10 of 10

Which of the following is the correct interpretation of ∫₀^(2π) sin x dx = 0?

On [0, π], sin x is positive (area = 2). On [π, 2π], sin x is negative (area = −2). The definite integral gives net signed area: 2 + (−2) = 0. The total (unsigned) area is 4. This distinction is crucial in applications.

—

Quiz complete!

← Previous
Riemann Sums

Cookie Settings