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Module 9 · L20 of 20 · Final ~45 min ⚡ +95 XP available

Module 9 Synthesis & Exam Technique

You've worked through 19 lessons across three pillars — areas between curves, volumes of revolution, and differential equations. This final synthesis lesson stitches them together with one unified workflow you can run on any HSC question. The focus shifts from "how do I do it?" to "how do I get the marks?" — exam triage, working-out structure, common pitfalls, and the decision tree markers use.

Today's hook — Write down — without looking back — the three formulas at the heart of Module 9: (1) area between $f$ and $g$, (2) washer volume about the $x$-axis, (3) separation of variables. If you can't reproduce all three from memory now, you cannot retrieve them under exam pressure. Try it, then check against card 04.

0/5QUESTS

Recall — write the three core formulas

+5 XP warm-up

Without looking back: write the three pillar formulas of Module 9 — (a) area between $y=f(x)$ and $y=g(x)$, (b) washer volume about the $x$-axis, (c) the separation-of-variables form. Then write one sentence on the typical mistake students make in each.

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How HSC markers actually award marks

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HSC markers follow a marking rubric that splits each question into discrete steps. They award method marks even when the final answer is wrong, but only if your working is clearly structured. Three habits unlock those method marks: sketch first, write the integral in full, evaluate cleanly.

For a typical 4-mark calculus application question, the breakdown is usually: 1 mark for the correct setup (limits, integrand), 1 mark for the integration step, 1 mark for substituting limits, 1 mark for the final exact answer. Skipping the setup line costs 2 marks even if your final number is right.

Sketch · Setup · Integrate · Evaluate · Verify

$5 \text{ steps} \to 5 \text{ marks earned}$

Show every step

Don't skip the line where you write the integral with limits. That single line is usually worth one mark on its own — the cheapest mark in the paper.

Exact form by default

Unless the question says "to 2 decimal places," leave $\pi$, $\sqrt{2}$, $\ln 3$ in your answer. Decimal approximation when not asked costs the final mark.

Sketch even if not asked

Even when no sketch is required, drawing the region in margin space anchors your thinking and prevents sign errors. The sketch itself is rarely marked, but the answer it produces is.

What you'll master

Know

Key facts

The three pillar formulas: area, washer volume, separation of variables
The HSC marking allocation: setup, integrate, substitute, evaluate, verify
The most common mark-losing mistakes in each pillar

Understand

Concepts

Why each step on the workflow is its own mark
How to triage time across a multi-part question
When to leave a part blank vs. when to attempt for method marks

Can do

Skills

Solve any Module 9 question using a single five-step workflow
Maximise method marks even when stuck on the final algebra
Identify and avoid the three pillar-specific traps

The three pillar formulas (memorise & cross-check)

Area between curves$A = \displaystyle\int_a^b [f(x) - g(x)]\,dx$ with $f \geq g$. Always sketch and identify the top curve first.

Area w.r.t. $y$-axis$A = \displaystyle\int_c^d [F(y) - G(y)]\,dy$ when curves are naturally expressed as $x = F(y)$.

Disk volume$V = \pi\displaystyle\int_a^b [f(x)]^2\,dx$ — region touches the axis of revolution (no hole).

Washer volume$V = \pi\displaystyle\int_a^b [R(x)^2 - r(x)^2]\,dx$ — region has a hole. $R$ outer, $r$ inner. Never $(R-r)^2$.

Separation of variablesFor $\dfrac{dy}{dx} = f(x)g(y)$: rearrange to $\dfrac{1}{g(y)}\,dy = f(x)\,dx$, integrate both sides, apply IC for $C$.

ME12-4NESA outcome: uses calculus in the solution of applied problems, including differential equations and volumes of solids of revolution.

The unified five-step workflow

core concept

One workflow handles every Module 9 question, whether it's area, volume, DE, or a combination:

Sketch. Draw all curves, mark intersections, shade the region.
Classify. Is this an area, volume, or DE problem? Note which formula applies.
Setup. Write the integral or DE in full with explicit limits.
Evaluate. Integrate, substitute the limits, simplify to exact form.
Verify. Area > 0, volume > 0, units sensible, sign correct.

Decision branch by question type:

"Find the area" / "the region enclosed" $\to$ area formula, $f-g$ or $F-G$.
"Rotated about the $x$-axis" / "solid of revolution" $\to$ disk or washer.
"$\dfrac{dy}{dx} = \ldots$, find $y$" $\to$ separation of variables, apply IC.
"Show the rate of growth" / "is proportional to" $\to$ build a DE first, then solve.

Exam tip. If you cannot finish a question, still write the setup line. Setup is usually worth $\tfrac{1}{4}$ of the total marks for that part — a free mark for showing you knew the strategy.

Unified workflow: (1) read; (2) classify (area/volume/DE/mixed); (3) write formula; (4) compute; (5) verify sign/units/scale.

Pause — copy the unified five-step workflow (read, classify, set up, compute, verify) and identify which step prevents each common error into your book.

Quick check: A 4-mark HSC question asks you to find the volume of revolution. You sketch the region and write the washer integral with limits, but cannot evaluate it. How many marks should you typically expect?

Pillar-by-pillar trap audit

core concept

We just saw the unified five-step workflow: read, classify (area/volume/DE/mixed), set up the formula, compute, and verify by checking sign/units/scale. That raises a question: each of the Module 9 pillars has a signature error students make under time pressure — what is the specific trap for area, volume, and DE problems? This card answers it → area: forgetting to split at a crossing; volume: squaring before subtracting not after; DEs: dropping the constant of integration.

Each pillar has a signature error students make under time pressure. Knowing them in advance is half the defence.

Area trap. Forgetting that area is always top minus bottom. If your integrand is negative on the interval, you wrote $g - f$ instead of $f - g$. Take the absolute value, or swap the order.
Washer trap. Writing $(R - r)^2$ instead of $R^2 - r^2$. The difference of squares formula doesn't apply here — each radius must be squared before subtraction.
DE trap. Forgetting the constant of integration $+ C$ on the right-hand side after separating. Without applying the initial condition, your particular solution is wrong by an unknown constant.

$$\text{Area: } \int(\text{top}-\text{bot}) \quad \text{Washer: } \pi\!\int\!(R^2-r^2) \quad \text{DE: } \int\!\tfrac{1}{g(y)}dy = \int f(x)dx + C$$

Memory hook. Three pillars, three common traps. Write them on the inside of your reference sheet and reread them before any Module 9 question.

Each pillar has a signature error students make under time pressure. Knowing them in advance is half the defence.

Pause — copy the three pillar traps: (1) area — split regions at crossings; (2) volume — $[R]^2-[r]^2$ not $[R-r]^2$; (3) DEs — always include $+C$ into your book.

True or false: When solving $\dfrac{dy}{dx} = \dfrac{x}{y}$ by separation of variables, the constant of integration $C$ can be safely left out and added at the very end.

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

PROBLEM 1 · FULL WORKFLOW · AREA

Find the area enclosed by $y = x^2 - 1$ and $y = 3 - x^2$. Use the five-step workflow.

Sketch & intersect: $x^2 - 1 = 3 - x^2 \Rightarrow 2x^2 = 4 \Rightarrow x = \pm\sqrt{2}$. Test $x=0$: top curve $y = 3-x^2 = 3$, bottom $y = x^2 - 1 = -1$. So $3 - x^2$ is on top.

Always identify the top curve by testing an interior point. This step alone earns 1 mark.

PROBLEM 2 · FULL WORKFLOW · VOLUME

The region enclosed by $y = 2x$ and $y = x^2$ is rotated about the $x$-axis. Find the volume.

Sketch & intersect: $2x = x^2 \Rightarrow x(x-2)=0 \Rightarrow x = 0, 2$. Test $x=1$: $2(1) = 2 > 1 = 1^2$, so $y = 2x$ is outer, $y = x^2$ is inner.

Outer = farther from axis; inner = closer to axis. Sketch is not strictly required but anchors your reasoning.

PROBLEM 3 · FULL WORKFLOW · DIFFERENTIAL EQUATION

A population $P$ satisfies $\dfrac{dP}{dt} = 0.2P$ with $P(0) = 500$. Find $P$ as a function of $t$, then find the time for the population to double.

Separate & integrate: $\dfrac{dP}{P} = 0.2\,dt \Rightarrow \int\dfrac{dP}{P} = \int 0.2\,dt \Rightarrow \ln P = 0.2t + C$.

Separation of variables with $+C$ on the right. Exponentiate next.

Fill the gap: The population $P = 500 e^{0.2t}$ doubles when $t = \dfrac{\ln }{0.2}$ time units.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Skipping the setup line

Writing only the final number costs the method mark for setup, even if your answer is correct. Markers cannot award method marks they cannot see. Always write the integral with limits as its own line.

Trap 02

Decimalising when exact form is expected

Writing $\dfrac{64\pi}{15} = 13.40$ on the last line throws away the final mark. Unless the question says "to N decimal places," leave $\pi$, $\sqrt{}$, and $\ln$ in your answer.

Trap 03

Forgetting the constant of integration in DEs

Many students integrate $\ln P = 0.2t$ (missing $+C$), then claim $P = e^{0.2t}$ — a particular solution that doesn't satisfy any non-trivial initial condition. Always write $+C$, then exponentiate to $Ae^{kt}$, then apply the IC.

Did you get this? True or false: HSC markers can award method marks for a wrong final answer as long as the integral was set up correctly and the working shows the method.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Find the area enclosed by $y = 4 - x^2$ and the $x$-axis. Use the full five-step workflow and label each step.

The region between $y = x$ and $y = x^3$ on $[0,1]$ is rotated about the $x$-axis. Find the volume using the washer method. Write each step explicitly.

A radioactive substance decays at a rate proportional to its mass: $\dfrac{dM}{dt} = -kM$ with $M(0) = 100$ grams and $M(10) = 80$ grams. Find $k$, then find the time for the mass to halve.

Without evaluating, set up the integral for the volume of the solid formed by rotating the region between $y = e^x$ and $y = e^{-x}$ on $[0, \ln 2]$ about the $x$-axis.

Reflect: write a personal one-page checklist of the steps and traps you will use on the HSC. Include the five workflow steps and the three traps you most often fall into.

Odd one out: Three of these are valid steps in the five-step Module 9 workflow. Which is NOT?

Revisit your thinking

Earlier you tried to write the three pillar formulas from memory.

The three pillars are: area $A = \int_a^b(f - g)\,dx$, washer volume $V = \pi\int_a^b(R^2 - r^2)\,dx$, and separation of variables $\int \tfrac{1}{g(y)}\,dy = \int f(x)\,dx + C$. If any of these were missing or wrong, copy them into your reference sheet now — they are the spine of Module 9.

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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

ApplyBand 33 marks

Q1. Find the area enclosed by $y = 4 - x^2$ and the $x$-axis. Show all five workflow steps. (3 marks)

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ApplyBand 44 marks

Q2. The region between $y = x$ and $y = x^3$ on $[0, 1]$ is rotated about the $x$-axis. Find the volume of the resulting solid. Give your answer in exact form. (4 marks)

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AnalyseBand 54 marks

Q3. A bacterial population grows according to $\dfrac{dN}{dt} = kN$. Initially $N(0) = 200$, and after 5 hours $N(5) = 500$. (a) Find $k$ in exact form. (b) Find the time for the population to reach 1000. (4 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $y = 4 - x^2 = 0 \Rightarrow x = \pm 2$. $A = \int_{-2}^{2}(4-x^2)\,dx = [4x - \tfrac{x^3}{3}]_{-2}^{2} = 2(8 - \tfrac{8}{3}) = \tfrac{32}{3}$ sq units.

2. $V = \pi\int_0^1(x^2 - x^6)\,dx = \pi[\tfrac{x^3}{3} - \tfrac{x^7}{7}]_0^1 = \pi(\tfrac{1}{3} - \tfrac{1}{7}) = \tfrac{4\pi}{21}$.

3. $M = 100 e^{-kt}$. From $M(10)=80$: $80 = 100 e^{-10k} \Rightarrow e^{-10k} = 0.8 \Rightarrow k = -\tfrac{\ln 0.8}{10} = \tfrac{\ln 1.25}{10}$. Half-life: $50 = 100 e^{-kt} \Rightarrow t = \tfrac{\ln 2}{k} = \tfrac{10\ln 2}{\ln 1.25}$.

4. $V = \pi\int_0^{\ln 2}(e^{2x} - e^{-2x})\,dx$ (setup only).

5. Personal answer — should list the five workflow steps and at least two pillar-specific traps from the lesson.

Q1 (3 marks): Intersections at $x = \pm 2$ [1]. $A = \int_{-2}^{2}(4-x^2)\,dx$ [1]. $= [4x-\tfrac{x^3}{3}]_{-2}^{2} = \tfrac{32}{3}$ sq units [1].

Q2 (4 marks): Identify outer $R=x$, inner $r=x^3$, limits $0$ to $1$ [1]. $V = \pi\int_0^1(x^2-x^6)\,dx$ [1]. $= \pi[\tfrac{x^3}{3}-\tfrac{x^7}{7}]_0^1$ [1]. $= \tfrac{4\pi}{21}$ cubic units [1].

Q3 (4 marks): $N = 200e^{kt}$ [1]. $500 = 200e^{5k} \Rightarrow e^{5k} = 2.5 \Rightarrow k = \tfrac{\ln 2.5}{5}$ [1]. For $N = 1000$: $1000 = 200e^{kt} \Rightarrow e^{kt} = 5 \Rightarrow t = \tfrac{\ln 5}{k}$ [1] $= \tfrac{5\ln 5}{\ln 2.5}$ hours [1].

Boss battle · The Synthesis Showdown

earn bronze · silver · gold

Five timed questions drawn from across Module 9 — areas, volumes, and differential equations. 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 synthesis questions from the whole module. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review. This completes Module 9.

← Lesson 19 · Mixed Application Problems Module 9 overview →

Module overview · Extension 1 · Checkpoint 4