Calculus — Worked Solutions (HSC Maths Extension 1)
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Worked examples for HSC Maths Extension 1 further calculus. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Maths Extension 1 — Calculus. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Integration by substitution
Question
Use the substitution $u = x^2 + 1$ to evaluate $\displaystyle\int_0^2 x(x^2+1)^3\,dx$.
Solution
Let $u = x^2 + 1$, so $du = 2x\,dx$, i.e. $x\,dx = \tfrac12\,du$.
Change the limits: $x = 0 \Rightarrow u = 1$; $x = 2 \Rightarrow u = 5$.
$$\int_0^2 x(x^2+1)^3\,dx = \frac12\int_1^5 u^3\,du = \frac12\left[\frac{u^4}{4}\right]_1^5 = \frac18(625 - 1) = 78.$$
Change the limits to $u$ and you never have to back-substitute. Forgetting the $\tfrac12$ from $du$ is the classic dropped mark.
Substitution works because $x\,dx$ is sitting right next to a function of $x^2+1$ — that's the signal to let $u = x^2+1$.
Differentiate: $\dfrac{du}{dx} = 2x$, so $du = 2x\,dx$, which rearranges to $x\,dx = \tfrac12\,du$. That replaces the whole $x\,dx$ chunk neatly.
Because it's a definite integral, convert the limits too: when $x = 0$, $u = 0^2+1 = 1$; when $x = 2$, $u = 2^2+1 = 5$. Now everything is in $u$:
$$\int_0^2 x(x^2+1)^3\,dx = \frac12\int_1^5 u^3\,du = \frac12\cdot\frac{u^4}{4}\Big|_1^5 = \frac18(5^4 - 1^4) = \frac{624}{8} = 78.$$
Converting the limits is the time-saver — it means you finish in $u$ and never undo the substitution.
- $u = x^2+1$, $du = 2x\,dx$ ⇒ $x\,dx = \frac12\,du$
- Limits: $x=0 \to u=1$, $x=2 \to u=5$
- $\frac12\int_1^5 u^3\,du = \frac12\cdot\frac{u^4}{4}\Big|_1^5 = \frac18(625-1) = 78$
Where the marks go
- 1 mark: Correct substitution $du = 2x\,dx$ and $x\,dx = \frac12 du$
- 1 mark: Changes the limits to $u = 1$ and $u = 5$
- 1 mark: Evaluates to $78$
Key idea
With a substitution in a definite integral, convert the limits to the new variable so you can evaluate directly without back-substituting.
Example 2 — Volume of revolution
Question
The region bounded by $y = \sqrt{x}$, the $x$-axis and the line $x = 4$ is rotated about the $x$-axis. Find the exact volume of the solid generated.
Solution
Rotating about the $x$-axis: $V = \pi\displaystyle\int_a^b y^2\,dx$.
Here $y^2 = (\sqrt x)^2 = x$, with $a = 0$, $b = 4$:
$$V = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi\cdot\frac{16}{2} = 8\pi.$$
The region starts at $x = 0$ where the curve meets the axis — get the lower limit right or the whole integral is off.
For a rotation about the $x$-axis we add up thin disks; each disk has radius $y$, so its area is $\pi y^2$ and the volume is $V = \pi\displaystyle\int_a^b y^2\,dx$.
Square the curve first: $y = \sqrt x$ gives $y^2 = x$, which is lovely and simple. The limits come from the region — it runs from where $y = \sqrt x$ meets the $x$-axis, at $x = 0$, out to the line $x = 4$.
$$V = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi\left(\frac{16}{2} - 0\right) = 8\pi.$$
We use $y^2$, not $y$, because the disk's area depends on the square of its radius — that's the heart of the disk method.
- $V = \pi\int_a^b y^2\,dx$, axis $= x$-axis
- $y^2 = (\sqrt x)^2 = x$; limits $0$ to $4$
- $V = \pi\int_0^4 x\,dx = \pi\cdot\frac{x^2}{2}\Big|_0^4 = \pi\cdot 8 = 8\pi$
Where the marks go
- 1 mark: States $V = \pi\int y^2\,dx$
- 1 mark: Correct $y^2 = x$ and limits $0$ to $4$
- 1 mark: Integrates to $\pi\frac{x^2}{2}$
- 1 mark: Exact volume $8\pi$
Key idea
Volume about the $x$-axis is $\pi\int y^2\,dx$; square the function first, and read the limits from where the region begins and ends.
Frequently asked questions
Step-by-step solutions to exam-style questions on Calculus in HSC Maths Extension 1, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Maths Extension 1 syllabus for Calculus, using the methods and notation expected in the exam.