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Module 3 · L15 of 15 ~40 min +95 XP available

Module 3 Review

Calculus connects rates of change, curves, and areas. This review consolidates differentiation rules, stationary points, optimisation, integration, and approximation methods into a coherent toolkit for analysing functions.

Today's hook — You can now differentiate, find stationary points, optimise, integrate, and approximate. How do all these tools connect? This lesson ties the threads together: differentiation finds rates and shapes, integration recovers functions and measures quantities, and together they solve real problems.

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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall — your gut answer first

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Before we start, what are the most important things you have learned across Module 3?

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The two moves

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Module 3 has many techniques. The two moves that unify them are:

Move 1: Identify the tool. Read the problem and decide whether you need differentiation, integration, or both. Look for key words: “rate of change”, “stationary”, “maximum/minimum”, “area”, “volume”, “approximate”.

Move 2: Execute carefully. Write down what you are finding before calculating. Check your answer: differentiate integrals, use the second derivative test, verify units.

$$\frac{d}{dx}(x^n) = nx^{n-1} \quad \int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$

Differentiation tools

Power rule, chain rule (composite), product rule (product), quotient rule (fraction). Identify structure first.

Integration tools

Power rule, definite integrals, area between curves, volumes of revolution, trapezoidal rule.

Check endpoints

In optimisation on a closed interval: always evaluate at stationary points AND both endpoints.

What you will master

Know

Key facts

All differentiation rules from Module 3
Integration formulas: power rule, definite integral, disc method, trapezoidal rule
That a stationary point has $f'(x) = 0$

Understand

Concepts

How differentiation and integration are inverses
Why optimisation requires checking all critical points and endpoints
The relationship between signed area, total area, and definite integrals

Can do

Skills

Differentiate using all Module 3 rules
Find and classify stationary points using first and second derivatives
Solve optimisation problems and evaluate definite integrals

Key terms

DifferentiationThe process of finding the derivative; measures the instantaneous rate of change of a function.

IntegrationThe reverse of differentiation; also used to find areas, volumes, and accumulated quantities.

Stationary pointA point where $f'(x) = 0$; can be a local maximum, local minimum, or horizontal inflection.

OptimisationUsing calculus to find maximum or minimum values in practical contexts.

Signed areaThe value of a definite integral; regions below the $x$-axis contribute negatively.

Second derivative testIf $f'(x) = 0$ and $f''(x) < 0$, local max; if $f''(x) > 0$, local min; if $f''(x) = 0$, inconclusive.

Concept — the big picture of calculus

core concept

Module 3 covers the core calculus tools:

Differentiation rules: power, chain, product, quotient
Applications of differentiation: stationary points, increasing/decreasing intervals, optimisation, curve sketching
Integration: antiderivatives, definite integrals, areas between curves, volumes of revolution
Numerical methods: trapezoidal rule for approximating integrals

$$\begin{aligned} &\text{Power: } \frac{d}{dx}(x^n) = nx^{n-1} \quad & &\text{Chain: } \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \\ &\text{Product: } (uv)' = u'v + uv' \quad & &\text{Quotient: } \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \\ &\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad & &V = \pi \int_a^b [f(x)]^2 \, dx \end{aligned}$$

Trapezoidal rule: $\frac{h}{2}[y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n]$ where $h = \frac{b-a}{n}$

Differentiation and integration are inverses. If $F'(x) = f(x)$, then $\int f(x) \, dx = F(x) + C$ and $\int_a^b f(x) \, dx = F(b) - F(a)$. This is the Fundamental Theorem of Calculus — the deepest result in Module 3. Always check integration answers by differentiating.

To differentiate: identify structure (composite / product / quotient), choose the right rule; Stationary points: solve $f'(x) = 0$; classify using $f''(x)$ or first derivative sign table

Pause — copy the rule-selection guide (identify composite/product/quotient structure first, then choose chain/product/quotient rule) and the stationary point classification decision tree into your book.

Quick check: True or false — when finding the maximum value of a function on a closed interval, it is sufficient to only check stationary points (you do not need to check the endpoints).

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · DIFFERENTIATION WITH MULTIPLE RULES

Differentiate $y = x^2 \sin(3x)$ using appropriate rules.

Product rule needed: $u = x^2$, $v = \sin(3x)$

Two functions are multiplied together — product rule applies.

PROBLEM 2 · OPTIMISATION ON A CLOSED INTERVAL

Find the maximum value of $f(x) = x^3 - 6x^2 + 9x + 2$ on $[0, 4]$.

$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$

Differentiate and factorise to find where the gradient is zero.

PROBLEM 3 · AREA BETWEEN CURVES

Find the area enclosed by $y = x^2$ and $y = 2x$.

Intersections: $x^2 = 2x$ gives $x = 0$ or $x = 2$

Find the bounds by solving where the two curves meet.

Quick check: Which rule is needed to differentiate $y = (3x^2 + 1)^5$?

Common errors · the 3 traps that cost marks

Trap 01

Mixing up differentiation rules

Chain rule for composites, product rule for products, quotient rule for quotients. Identify the structure before choosing the rule. When in doubt, simplify or expand first.

Trap 02

Forgetting to check endpoints in optimisation

On a closed interval, the maximum or minimum may occur at an endpoint, not at a stationary point. Always evaluate the function at all critical points AND both endpoints.

Trap 03

Confusing signed area with total area

The definite integral gives signed area. For total area, split at $x$-intercepts and take absolute values. This matters when parts of the curve are below the $x$-axis.

Odd one out: Three of the following are applications of differentiation. Which one is NOT?

Work mode · how are you completing this lesson?

Quick-fire practice · 5 problems

Differentiate $y = (2x + 1)^5$.

Find $f''(x)$ for $f(x) = x^4 - 3x^2 + x$.

Find the stationary points of $f(x) = x^3 - 3x^2$ and classify them.

Evaluate $\int_1^3 (2x + 1) \, dx$.

Use the trapezoidal rule with 4 strips to approximate $\int_0^2 x^2 \, dx$.

Fill the blanks: drag each token into the matching blank.

chain product endpoints integration

A composite function needs the ___ rule. Two functions multiplied together need the ___ rule. Finding areas under curves requires ___. In optimisation on a closed interval, always check the ___.

Teach it back: Explain in your own words how differentiation and integration are connected, and give one example where you use both in the same problem.

Revisit your thinking

Earlier you were asked to list the most important things from Module 3. The key threads are: differentiation finds rates and shapes; integration recovers functions and measures areas; together they solve optimisation and modelling problems. The Fundamental Theorem of Calculus — that differentiation and integration are inverses — unifies the whole module.

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Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

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Q1. Differentiate $y = \dfrac{x^2}{x + 1}$. Show all working. 3 MARKS

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Q2. A rectangle has perimeter 48 cm. Find the dimensions that maximise the area. Show all working including verification. 4 MARKS

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Q3. The region bounded by $y = \sqrt{x}$, $x = 1$, $x = 4$ and the $x$-axis is rotated about the $x$-axis. Find the volume of the solid. 4 MARKS

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📖 Comprehensive answers (click to reveal)

Q1 (3 marks): Quotient rule: $u = x^2$, $v = x+1$ [0.5]. $u' = 2x$, $v' = 1$ [0.5]. $y' = \frac{2x(x+1) - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2} = \frac{x(x+2)}{(x+1)^2}$ [2].

Q2 (4 marks): $2l + 2w = 48 \Rightarrow l = 24 - w$ [0.5]. $A = w(24-w) = 24w - w^2$ [0.5]. $\frac{dA}{dw} = 24 - 2w = 0 \Rightarrow w = 12$ [1]. $l = 12$, max area $= 144$ cm$^2$ [1]. $\frac{d^2A}{dw^2} = -2 < 0$ confirming maximum [1].

Q3 (4 marks): $y = \sqrt{x} \Rightarrow y^2 = x$ [0.5]. $V = \pi \int_1^4 x \, dx$ [1]. $= \pi \left[\frac{x^2}{2}\right]_1^4 = \pi(8 - 0.5) = \frac{15\pi}{2}$ [2]. Volume $= \frac{15\pi}{2}$ cubic units [0.5].

Drill answers: D1: $10(2x+1)^4$ · D2: $12x^2 - 6$ · D3: local max at $(0,0)$, local min at $(2,-4)$ · D4: $10$ · D5: $2.75$

Boss battle · Module 3 Final Boss

earn bronze · silver · gold

Five timed questions across all of Module 3. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering Module 3 review questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you have finished the practice and review. You have completed Module 3!

← Lesson 14 · The Trapezoidal Rule Module 4 · Lesson 1 →

Module overview · Maths Advanced