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Module 6 · L15 of 15 ~35 min ⚡ +95 XP available

Module Synthesis — Further Calculus

Fourteen lessons. One toolkit. Antiderivatives, the Fundamental Theorem, areas and volumes, three integration techniques, differential equations, motion. This lesson draws the threads together and shows how each piece connects to the others — then prepares you for the HSC.

Today's hook — Newton invented calculus in two weeks. Leibniz formalised it over decades. What took the greatest minds in history years to build, you now hold in your toolkit. This lesson is about seeing the whole — how antiderivatives, the FTC, techniques, and differential equations form a single coherent theory of change.

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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

Module formula reference

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Every formula from Module 6, in one place. This is your HSC reference card.

Core integrals

$\int x^n dx = \frac{x^{n+1}}{n+1}+C$ · $\int e^x dx = e^x+C$ · $\int \frac{1}{x}dx = \ln|x|+C$

FTC & areas

$\int_a^b f\,dx = F(b)-F(a)$ · Area $= \int_a^b [\text{top}-\text{bottom}]\,dx$ · Volume $= \pi\int_a^b [f(x)]^2\,dx$

Techniques

Substitution: $u = g(x)$ · By parts: $\int u\,dv = uv - \int v\,du$ · LIATE for choosing $u$

$$\frac{dN}{dt} = kN \implies N = N_0 e^{kt} \qquad v = \frac{dx}{dt},\; a = \frac{dv}{dt},\; x = \int v\,dt,\; v = \int a\,dt$$

The structure of Further Calculus

Foundation (L01–L03)

Antiderivatives

Reverse of differentiation
Indefinite integrals + $C$
Power, exponential, log rules

Core theorem (L04–L05)

FTC + Definite integrals

$\int_a^b f = F(b) - F(a)$
Unifies slopes and areas
Signed area interpretation

Applications (L06–L07)

Areas & Volumes

Area between curves
Volume of revolution (disk)
Sketch first — always

Techniques (L08–L10)

Three methods

Substitution: composite functions
By parts: products (LIATE)
Partial fractions: rational functions

Dynamic systems (L11–L13)

Differential equations

Separable DEs
Exponential growth/decay
Real-world modelling

Physical applications (L14)

Motion

$a \to v \to x$ via integration
Distance vs displacement
Initial conditions

Key connections. FTC connects differentiation (slopes) to integration (areas). Areas and volumes are definite integral applications. Integration techniques expand what functions we can integrate. Differential equations use integration to solve for unknown functions from rates. Motion chains two integrations together. Every piece is the same operation — accumulation — in a different context.

HSC checklist

Power rule$\int x^n dx = \frac{x^{n+1}}{n+1}+C$, $n \neq -1$. Exception: $\int \frac{1}{x}dx = \ln|x|+C$.

FTC Part 2$\int_a^b f(x)\,dx = F(b) - F(a)$. Check: definite integral has no +C.

Area between curves$A = \int_a^b [\text{top} - \text{bottom}]\,dx$. Sketch first to identify which is on top.

Volume of revolution$V = \pi \int_a^b [f(x)]^2\,dx$. Washer: $V = \pi \int (R^2 - r^2)\,dx$ — square first, then subtract.

LIATEFor integration by parts: choose $u$ as the earlier letter in LIATE (Log, Inverse, Algebraic, Trig, Exp).

Separable DE$\frac{dy}{dx} = f(x)g(y)$: separate as $\frac{dy}{g(y)} = f(x)\,dx$, integrate both sides.

Pitfalls to avoid in the HSC

core concept

Seven errors that appear repeatedly in HSC marking. Each costs marks.

Error 01

Forgetting +C

Indefinite integrals always need +C. Check: is the integral definite (with limits)? If not, write +C. Markers deduct marks every time it's missing.

Error 02

Wrong power rule at n = −1

$\int \frac{1}{x}\,dx \neq \frac{x^0}{0}$ — that's division by zero. The answer is $\ln|x| + C$. This exponent is the one exception to the power rule.

Error 03

Negative area below x-axis

$\int_a^b f\,dx$ gives signed area. If asked for the geometric area, sketch first and use $\int |f|\,dx$, splitting at the x-intercepts.

Error 04

Washer formula: $(R-r)^2 \neq R^2-r^2$

For volume with a hole: $V = \pi\int (R^2 - r^2)\,dx$. Not $\pi\int (R-r)^2\,dx$. Square each radius separately, then subtract.

Error 05

Substitution: forgetting du

After choosing $u$, you must write $\frac{du}{dx}$ and solve for $dx$. Substituting $u$ but keeping $dx$ is wrong — the integral hasn't been fully converted.

Error 06

Distance vs displacement

Distance $= \int |v|\,dt$ (split where $v = 0$). Displacement $= \int v\,dt$ (can be negative). Using displacement when asked for distance will give a wrong answer even if the integration is perfect.

Technique selection: substitution = composite; by parts = product; partial fractions = rational; Area: sketch first; Volume: $\pi\int [f(x)]^2\,dx$, square before subtracting for washer

Pause — copy the technique-selection guide (substitution = composite, by-parts = product/LIATE, partial fractions = rational) and the washer-volume reminder ($R^2 - r^2$, not $(R-r)^2$) into your book.

Did you get this? True or false: when computing $\int_0^2 x^2\,dx$ using the FTC, you need to add a constant of integration.

HSC strategies · 6 for exam success

HSC examination strategies

We just saw the six pitfalls — missing $+C$, wrong rule at $n=-1$, unsigned area, washer formula, forgotten $du$, and distance vs displacement. That raises a question: beyond avoiding errors, what positive habits earn the most marks in the HSC? This card answers it → six active strategies: sketch first, check by differentiating, track constants $C_1/C_2$, show all working, choose the right technique, and interpret signed area correctly.

Sketch first

For area and volume problems, a sketch reveals intersections, which function is on top, and whether curves cross the x-axis.

Check by differentiating

After integrating, differentiate your answer to verify it matches the original integrand. Takes 10 seconds and can save 2 marks.

Track constants

In motion problems, label constants $C_1$, $C_2$ and use initial conditions systematically. Don't rush the constant-finding step.

Write working clearly

HSC markers award partial marks. Show substitution, show antiderivatives, show evaluation — every line of working is worth marks even if the final answer is wrong.

Know technique selection

Substitution for composite functions, by parts for products (LIATE), partial fractions for rational functions. Choosing the wrong technique first costs time.

Interpret negative integrals

Definite integrals give signed area. When asked for area, a negative result means the region is below the x-axis — take the absolute value.

6 HSC strategies: sketch · check by differentiating · track constants · show working · choose technique · interpret negatives; LIATE: Log > Inverse > Algebraic > Trig > Exponential

Pause — copy the six HSC strategies (sketch, check by differentiating, track constants, show working, choose technique, interpret negatives) and the LIATE priority order for by-parts into your book.

Quick check: Which technique is best for $\int x e^x\,dx$?

Mixed revision · worked examples from across the module

PROBLEM 1 · MIXED INTEGRATION

Evaluate $\displaystyle\int_0^2 x\sqrt{x^2+1}\,dx$ using substitution.

Let $u = x^2 + 1$, so $\dfrac{du}{dx} = 2x$, i.e. $x\,dx = \dfrac{du}{2}$

Spot the composite: $\sqrt{x^2+1}$ and its derivative $2x$ is present.

PROBLEM 2 · AREA BETWEEN CURVES

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

$x^2 = 2-x \implies x^2+x-2=0 \implies (x+2)(x-1)=0$

Find intersections: solve simultaneously. Limits: $x = -2$ and $x = 1$.

PROBLEM 3 · SEPARABLE DE

Solve $\dfrac{dy}{dx} = 3x^2 y$ with $y(0) = 2$.

$\dfrac{dy}{y} = 3x^2\,dx$

Separate variables: all $y$ terms on one side, all $x$ terms on the other.

Fill in the blank: To find the area between $y = x$ and $y = x^3$ from $x = 0$ to $x = 1$, since $x \geq x^3$ on $[0,1]$, the area is $\int_0^1 (x - x^3)\,dx = $ ___.

Activities · mixed revision problems

Find $\displaystyle\int \left(3x^2 + \frac{1}{x} + e^{2x}\right)dx$.

Evaluate $\displaystyle\int_0^2 x\sqrt{x^2+1}\,dx$ using substitution.

Find the area between $y = x^2$ and $y = 2 - x$.

Explain how the FTC connects the two main ideas of calculus.

A particle has $a(t) = 2t + 1$, $v(0) = 2$, $x(0) = 1$. Find $x(3)$.

Odd one out: Three of these require integration by parts; one is best solved by substitution. Which is the odd one out?

The whole picture

Over 14 lessons we built a complete theory of change and accumulation. Antiderivatives reverse differentiation. The FTC unifies slopes and areas. Integration techniques unlock new classes of functions. Differential equations describe dynamic systems. Motion chains it all together. Every application — from rocket landings to radioactive dating to pandemic modelling — uses the same underlying idea: integration is accumulation.

The insight Newton had, that differentiation and integration are inverse operations, is among the most powerful mathematical discoveries in history. You now have it in your toolkit.

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Multiple choice

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Pick your answer, then rate your confidence.

Short answer

ApplyBand 53 marks

Q1. Evaluate $\displaystyle\int_0^{\pi/2} x \sin x\,dx$ using integration by parts. Show all working. (3 marks)

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ApplyBand 54 marks

Q2. Find the volume when $y = x^3$ from $x = 0$ to $x = 1$ is rotated about the x-axis. Then find the area enclosed between $y = x^3$ and $y = x$. Show all working. (4 marks)

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AnalyseBand 54 marks

Q3. Write an essay-style response (150–200 words) explaining how the Fundamental Theorem of Calculus, integration techniques, and differential equations connect to form a unified theory of change. Use at least two real-world examples. (4 marks)

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Comprehensive answers (click to reveal)

Activity 1:

1. $x^3 + \ln|x| + \frac{1}{2}e^{2x} + C$.

2. $u = x^2+1$, $du = 2x\,dx$. $\frac{1}{2}\int_1^5 u^{1/2}\,du = \frac{1}{3}(5^{3/2}-1) = \frac{5\sqrt{5}-1}{3}$.

3. $x^2 = 2-x \Rightarrow x = 1, -2$. $A = \int_{-2}^{1}(2-x-x^2)\,dx = \frac{9}{2}$.

4. FTC shows differentiation and integration are inverse operations — slopes connect to areas.

5. $v(t) = t^2+t+2$, $x(t) = \frac{t^3}{3}+\frac{t^2}{2}+2t+1$. $x(3) = 9+4.5+6+1 = 20.5$ m.

Q1 (3 marks): $u = x$, $dv = \sin x\,dx$, $v = -\cos x$ [1]. $[-x\cos x]_0^{\pi/2} + \int_0^{\pi/2}\cos x\,dx = 0 + [\sin x]_0^{\pi/2} = 1$ [2].

Q2 (4 marks): Volume: $V = \pi\int_0^1 x^6\,dx = \frac{\pi}{7}$ [2]. Area: $x^3 = x$ at $x = 0, 1$. $x \geq x^3$ on $[0,1]$. $A = \int_0^1(x-x^3)\,dx = \frac{1}{2}-\frac{1}{4} = \frac{1}{4}$ [2].

Q3 (4 marks): FTC connects differentiation and integration as inverse operations [1]. Techniques (substitution, by parts, partial fractions) extend range of solvable problems [1]. DEs use integration to find functions from rates of change [1]. At least two real-world examples with clear connection to theory [1].

Boss battle · The Module Synthesiser

earn bronze · silver · gold

Five timed questions from the entire Module 6. Mixed problems — the hardest challenge on the module. Beat the boss to prove mastery.

Enter the arena

Science Jump · platform challenge

Climb platforms with mixed problems from across the entire Further Calculus module.

Mark lesson as complete

Tick when you've finished the practice and review.

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Module overview · Maths Advanced · Checkpoint 3