Worked Solutions
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.
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.