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NotesMath AATopic 5.10
Unit 5 · Calculus · Topic 5.10

IB Math AA — Integration by substitution

Topic 5.10 of IB Mathematics: Analysis and Approaches covers Integration by substitution, which is part of Unit 5: Calculus. Students explore key concepts including Reverse chain rule, Substitution. A strong understanding of integration by substitution is essential for IB Math AA exams and builds the foundation for connected topics across the syllabus.

Exam technique guidePractice questions

Key concepts in Integration by substitution

Key Idea: This topic is about reading the steepness of a curve: the derivative is the gradient at a single point and the instantaneous rate of change. It's the conceptual foundation for all of calculus — mostly tested on Paper 1 (interpret and reason, no calculator).

📈 What the derivative is

A straight line has one gradient everywhere; a curve keeps changing steepness, so we use the gradient at a point — the gradient of the tangent there (never a chord). The same number is also a rate of change: how fast y changes per unit of x at an instant (if x is time, the instantaneous rate). Always carry units — (y-units) per (x-unit).

✍️ Notation: f′(x) and dy/dx

f′(x)=dydxf'(x) = \frac{dy}{dx}f′(x)=dxdy​
f′(x)f'(x)f′(x)
"f prime of x" — the gradient function
dydx\frac{dy}{dx}dxdy​
rate of change of y with respect to x (same thing)

± Sign of the gradient = shape of the curve

Sign of f′(x)What the curve is doing
f′(x) > 0Increasing — going up as x increases
f′(x) < 0Decreasing — going down as x increases
f′(x) = 0Stationary — momentarily flat (peak, trough or flat spot)

✏️ IB-style worked examples

IB-style question — interpret a rate of change

The height h (cm) of a plant after t weeks satisfies dh/dt = 3.5 when t = 6. Interpret this value, with units.

Step by step:

  1. dh/dt is the rate the height changes per week.

    dhdt=3.5 cm per week\frac{dh}{dt} = 3.5 \text{ cm per week}dtdh​=3.5 cm per week
  2. The value is given at t = 6 specifically.

    at t=6\text{at } t = 6at t=6
Final answer:

At t = 6 weeks the plant's height is increasing at 3.5 cm per week.

IB-style question — gradient at a point from f′(x)

The gradient function of a curve is f′(x) = 4x − 5. Find the gradient of the curve at x = 2.

Step by step:

  1. Substitute x = 2 into the gradient function.

    f′(2)=4(2)−5f'(2) = 4(2) - 5f′(2)=4(2)−5
  2. Evaluate.

    =3= 3=3
Final answer:

The gradient at x = 2 is 3.

IB-style question — increasing or decreasing?

For a function g, g′(−1) = 7 and g′(4) = −2. State whether g is increasing or decreasing at each point.

Step by step:

  1. At x = −1 the derivative is positive.

    g′(−1)=7>0⇒increasingg'(-1) = 7 > 0 \Rightarrow \text{increasing}g′(−1)=7>0⇒increasing
  2. At x = 4 the derivative is negative.

    g′(4)=−2<0⇒decreasingg'(4) = -2 < 0 \Rightarrow \text{decreasing}g′(4)=−2<0⇒decreasing
Final answer:

Increasing at x = −1; decreasing at x = 4.

Important: The gradient at a point is the gradient of the tangent there — not the average gradient of a chord between two points, and not a single fixed number for the whole curve. To get a value, substitute the x into f′(x).

Tap each card to reveal the answer.

What does dy/dx represent? The gradient of the curve at a point / the rate of change of y with respect to x.

f′(x) < 0 at a point — what's the curve doing? Decreasing (going downhill as x increases).

f′(x) = 0 at a point — what kind of point is it? A stationary point — momentarily flat (peak, trough or flat spot).

dV/dt = −8 L/min for a draining tank — meaning? The volume is decreasing at 8 litres per minute.

Exam Tips

  • Gradient at a point = gradient of the tangent there (not a chord, not one number for the whole curve).
  • f′(x) and dy/dx mean the same thing — the gradient function.
  • To get a gradient value, substitute the x-value into f′(x).
  • f′(x) > 0 increasing, f′(x) < 0 decreasing, f′(x) = 0 stationary.
  • A rate of change carries units: (y-units) per (x-unit), e.g. cm per week.

What you'll learn in Topic 5.10

  • 5.10.1 Reverse chain rule
  • 5.10.2 Substitution
Suggested study order: Read the notes for each sub-topic below → test yourself with flashcards → attempt practice questions → review exam technique.

Study resources — 5.10 Integration by substitution

5.10.1

Reverse chain rule

Notes
5.10.2

Substitution

Notes

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Topic 5.10 Integration by substitution forms a core part of Unit 5: Calculus in IB Math AA. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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