Fundamental theorem of calculus

IB Mathematics: Analysis and Approaches HL· 5.6 Calculus· 15 min read

1. Part 1 of the Fundamental Theorem of Calculus★★☆☆☆HL only⏱ 5 min

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📘 Definition

FTC Part 1

If $f$ is continuous on $[a, b]$ , and $F (x) = \int_{a}^{x} f (t) d t$ for $x \in [a, b]$

$F$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and $F^{'} (x) = f (x)$ for all $x \in (a, b)$ . This proves differentiation and integration are inverse operations.

Example:

For $F (x) = \int_{2}^{x} t^{2} d t$ , $F^{'} (x) = x^{2}$ directly by FTC Part 1.

When the upper bound is a function of $x$ rather than $x$ itself, we apply the chain rule to find the derivative. For an upper bound $u (x)$ , the derivative becomes $\frac{d}{d x} \int_{a}^{u (x)} f (t) d t = f (u (x)) \cdot u^{'} (x)$ .

📐 Worked Example

Differentiate $G (x) = \int_{0}^{x^{2}} cos (t) d t$ with respect to $x$ .

1
Recognize $G (x)$ uses a composite upper bound $u (x) = x^{2}$ . By chain rule: $G^{'} (x) = \frac{d G}{d u} \cdot \frac{d u}{d x}$ .
2
By FTC Part 1, $\frac{d G}{d u} = cos (u)$ .
3
Calculate the derivative of the upper bound: $\frac{d u}{d x} = 2 x$ .
4
Substitute back $u = x^{2}$ to get the final result:
5
$G'(x) = \cos(x^2) \cdot 2x = 2x \cos(x^2)$

Exam tip:

Most IB exam questions on FTC Part 1 use variable composite bounds, never forget the chain rule derivative of the bound.

2. Part 2 of the Fundamental Theorem of Calculus★★☆☆☆HL only⏱ 6 min

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📘 Definition

FTC Part 2

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ ( $F^{'} = f$ )

The definite integral of $f$ over $[a, b]$ equals $F (b) - F (a)$ . This lets us calculate exact definite integral values without approximating Riemann sums.

Example:

$\int_{1}^{2} x d x = [\frac{1}{2} x^{2}]_{1}^{2} = 2 - \frac{1}{2} = \frac{3}{2}$

The constant of integration from indefinite integrals always cancels out in $F (b) - F (a)$ , so you never need to include it when evaluating definite integrals.

📐 Worked Example

Evaluate the definite integral $\int_{0}^{\frac{π}{2}} (2 sin x + 3 x^{2}) d x$ .

1
Find the antiderivative of each term: antiderivative of $2 sin x$ is $- 2 cos x$ , antiderivative of $3 x^{2}$ is $x^{3}$ .
2
Combine to get $F (x) = - 2 cos x + x^{3}$ (ignore the constant of integration).
3
Evaluate at the upper bound $\frac{π}{2}$ :
4
$F\left(\frac{\pi}{2}\right) = -2 \cos\left(\frac{\pi}{2}\right) + \left(\frac{\pi}{2}\right)^3 = 0 + \frac{\pi^3}{8} = \frac{\pi^3}{8}$
5
Evaluate at the lower bound $0$ :
6
$F(0) = -2 \cos(0) + 0^3 = -2(1) = -2$
7
Subtract lower bound from upper bound to get the final result:
8
$\int_0^{\frac{\pi}{2}} (2 \sin x + 3x^2) dx = F\left(\frac{\pi}{2}\right) - F(0) = \frac{\pi^3}{8} - (-2) = \frac{\pi^3}{8} + 2 \approx 5.87$

Exam tip:

Always show $F (b) - F (a)$ explicitly in your working, even if you use a calculator for the final value, to earn full method marks.

3. Core Proof and Connections★★★☆☆HL only⏱ 7 min

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The FTC is the unifying result of calculus, formalizing the inverse relationship between differentiation (finding rates of change) and integration (finding net area). This connection underpins all applications of calculus from kinematics to optimization.

🔬 Derivation

Goal:

Prove FTC Part 1: $\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$

Starting from:

Definition of the derivative: $F^{'} (x) = lim_{h \to 0} \frac{F ( x + h ) - F ( x )}{h}$

1
Substitute $F (x) = \int_{a}^{x} f (t) d t$ : $\frac{F ( x + h ) - F ( x )}{h} = \frac{1}{h} (\int_{a}^{x + h} f (t) d t - \int_{a}^{x} f (t) d t) = \frac{1}{h} \int_{x}^{x + h} f (t) d t$
2
By the Mean Value Theorem for integrals, there exists some $c$ between $x$ and $x + h$ such that $\int_{x}^{x + h} f (t) d t = f (c) h$ .
3
Substitute back to get $\frac{1}{h} \cdot f (c) h = f (c)$ .
4
As $h \to 0$ , $c \to x$ , and since $f$ is continuous, $lim_{h \to 0} f (c) = f (x)$ .

Result:

Therefore, $F^{'} (x) = f (x)$ , which completes the proof.

✓ Quick check

Test your understanding

If $F (x) = \int_{5}^{x} \frac{1}{1 + t ^{3}} d t$ , what is $F^{'} (2)$ ?
- A) $\frac{1}{9}$
- B) $\frac{1}{1 + 8}$
- C) $\frac{1}{2}$
- D) $0$
Reveal answer
A —
By FTC Part 1, $F^{'} (x) = \frac{1}{1 + x ^{3}}$ , so $F^{'} (2) = \frac{1}{1 + 8} = \frac{1}{9}$ .

4. Common Pitfalls

Wrong move:

Forgetting the chain rule when differentiating integrals with variable bounds

Why:

FTC Part 1 is stated for an upper bound of $x$ , not a function of $x$ , so the chain rule is required for composite bounds

Correct move:

Always apply: $\frac{d}{d x} \int_{a}^{u (x)} f (t) d t = f (u (x)) \cdot u^{'} (x)$

Wrong move:

Calculating $F (a) - F (b)$ instead of $F (b) - F (a)$ for definite integrals

Why:

Mixing up the order of lower and upper bounds, leading to a negative result for positive area

Correct move:

Always subtract the value at the lower bound from the value at the upper bound

Wrong move:

Retaining the dummy variable of integration after evaluation

Why:

Confusing the bound variable with the integration variable inside the integral

Correct move:

The dummy variable is eliminated after substitution of the bound values, so the final result will never contain the integration variable

Wrong move:

Applying FTC to discontinuous functions over intervals containing discontinuities

Why:

The theorem requires continuity of $f$ on the entire closed interval $[a, b]$

Correct move:

Split the integral at the discontinuity and evaluate each part separately, checking for convergence

5. Quick Reference Cheatsheet

Result Type	Formula	Use Case
FTC Part 1	$\frac{d}{d x} \int_{a}^{u (x)} f (t) d t = f (u (x)) u^{'} (x)$	Differentiate variable-bound integrals
FTC Part 2	$\int_{a}^{b} f (x) d x = F (b) - F (a), F^{'} = f$	Evaluate exact definite integrals
Two variable bounds	$\frac{d}{d x} \int_{v (x)}^{u (x)} f (t) d t = f (u) u^{'} - f (v) v^{'}$	Differentiate integrals with two variable bounds
Constant of integration	$C$ cancels in $F (b) - F (a)$	Definite integral evaluation

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Differentiate variable bound integral
2023 · 2
Evaluate definite integral
2021 · 1
Proof of FTC result

Going deeper

syllabusIB AA HL Official SyllabusPage 42, Calculus topic

What's Next

Now that you've mastered the Fundamental Theorem of Calculus, you hold the core tool for all advanced work in integral calculus. The FTC enables you to move beyond approximations to solve applied problems from kinematics to volumes of revolution, and to tackle more complex integration techniques. The inverse relationship between differentiation and integration it establishes is also foundational for solving differential equations, a major exam-weighted topic in IB AA HL that appears across both papers. This topic opens the door to all further calculus work you will do for the exam and in future university study.