Integration — Worked Solutions (HSC Maths Advanced)
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Worked examples for HSC Maths Advanced integration. 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 Advanced — Integration. 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 — Indefinite integral of a polynomial
Question
Find $\displaystyle\int \left(6x^2 - 4x + 5\right)\,dx$.
Solution
Integrate term by term: raise each power by one and divide by the new power.
$\displaystyle\int 6x^2\,dx = 2x^3$, $\displaystyle\int -4x\,dx = -2x^2$, $\displaystyle\int 5\,dx = 5x$.
So the answer is $2x^3 - 2x^2 + 5x + C$.
Don't drop the $+C$ — an indefinite integral without it loses a mark.
Integration reverses differentiation, so for each power of $x$ we add one to the index and divide by that new index: $\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1}$.
Let's do each term. For $6x^2$: the index becomes $3$, so $\frac{6x^3}{3} = 2x^3$. For $-4x$: $\frac{-4x^2}{2} = -2x^2$. The constant $5$ integrates to $5x$.
Because there are infinitely many functions with the same derivative (they differ by a constant), we add an arbitrary constant $C$.
Putting it together: $2x^3 - 2x^2 + 5x + C$.
Rule: $\int x^n\,dx = \frac{x^{n+1}}{n+1}$.
- $\int 6x^2\,dx = 2x^3$
- $\int -4x\,dx = -2x^2$
- $\int 5\,dx = 5x$
Answer: $2x^3 - 2x^2 + 5x + C$.
Where the marks go
- 1 mark: Correctly integrates the $x^2$ and $x$ terms
- 1 mark: Correct full primitive including the constant of integration $C$
Key idea
Integrate a polynomial term by term using $\int x^n\,dx = \frac{x^{n+1}}{n+1}$, and always include $+C$ for an indefinite integral.
Example 2 — Integrating an exponential
Question
Find $\displaystyle\int \left(e^{2x} + \frac{3}{x}\right)\,dx$, for $x > 0$.
Solution
Two standard primitives: $\int e^{kx}\,dx = \frac{1}{k}e^{kx}$ and $\int \frac{1}{x}\,dx = \ln x$.
$\displaystyle\int e^{2x}\,dx = \tfrac{1}{2}e^{2x}$ and $\displaystyle\int \frac{3}{x}\,dx = 3\ln x$.
So the answer is $\tfrac{1}{2}e^{2x} + 3\ln x + C$.
Watch the $\frac{1}{2}$ factor on the exponential — that's the mark people lose.
Let's take the two pieces separately. For $e^{2x}$, integrating an exponential of the form $e^{kx}$ gives $\frac{1}{k}e^{kx}$, because differentiating $\frac{1}{2}e^{2x}$ brings down a $2$ that cancels the $\frac{1}{2}$. So $\int e^{2x}\,dx = \tfrac{1}{2}e^{2x}$.
For $\frac{3}{x}$, recall that $\frac{d}{dx}\ln x = \frac{1}{x}$, so $\int \frac{1}{x}\,dx = \ln x$ (for $x>0$), and the $3$ comes along for the ride: $3\ln x$.
Adding the constant of integration, the answer is $\tfrac{1}{2}e^{2x} + 3\ln x + C$.
Standard primitives.
- $\int e^{2x}\,dx = \tfrac{1}{2}e^{2x}$
- $\int \frac{3}{x}\,dx = 3\ln x$ (for $x>0$)
Answer: $\tfrac{1}{2}e^{2x} + 3\ln x + C$.
Where the marks go
- 1 mark: Correct $\tfrac{1}{2}e^{2x}$ (correct $\tfrac{1}{k}$ factor)
- 1 mark: Correct $3\ln x$ and constant of integration
Key idea
Use $\int e^{kx}\,dx = \frac{1}{k}e^{kx}$ and $\int \frac{1}{x}\,dx = \ln x$; keep track of the $\frac{1}{k}$ factor.
Example 3 — Definite integral
Question
Evaluate $\displaystyle\int_{1}^{3} \left(3x^2 - 2x\right)\,dx$.
Solution
Find the primitive, then apply the limits.
$\displaystyle\int (3x^2 - 2x)\,dx = x^3 - x^2$.
Evaluate: $\big[x^3 - x^2\big]_1^3 = (27 - 9) - (1 - 1) = 18 - 0 = 18$.
No $+C$ on a definite integral — it cancels. Answer: $18$.
For a definite integral we first find a primitive, then use the fundamental theorem of calculus: subtract the primitive at the lower limit from its value at the upper limit.
The primitive of $3x^2 - 2x$ is $x^3 - x^2$ (no constant needed, since it cancels in the subtraction).
Now substitute the limits. At $x = 3$: $27 - 9 = 18$. At $x = 1$: $1 - 1 = 0$. Subtracting, $18 - 0 = 18$.
So $\displaystyle\int_{1}^{3}(3x^2 - 2x)\,dx = 18$.
Primitive: $x^3 - x^2$.
- Upper limit $x=3$: $27 - 9 = 18$
- Lower limit $x=1$: $1 - 1 = 0$
- $18 - 0 = 18$
Answer: $18$.
Where the marks go
- 1 mark: Correct primitive $x^3 - x^2$
- 1 mark: Correctly substitutes both limits
- 1 mark: Correct final value $18$
Key idea
A definite integral is the primitive evaluated at the upper limit minus its value at the lower limit; no constant of integration is needed.
Example 4 — Area under a curve
Question
Find the area of the region bounded by the curve $y = 4x - x^2$ and the $x$-axis.
Solution
First find where the curve meets the $x$-axis: $4x - x^2 = x(4 - x) = 0$, so $x = 0$ and $x = 4$. The parabola opens downward, so it's above the axis between them — no need to split for sign.
$\displaystyle\text{Area} = \int_0^4 (4x - x^2)\,dx = \Big[2x^2 - \tfrac{1}{3}x^3\Big]_0^4$.
$= \big(32 - \tfrac{64}{3}\big) - 0 = \tfrac{96 - 64}{3} = \tfrac{32}{3}$.
Area $= \dfrac{32}{3}$ square units. Check the curve sits above the axis before integrating, or you risk a sign error.
To find an enclosed area we need the boundaries, so let's find where the curve crosses the $x$-axis: set $y=0$, giving $x(4-x)=0$, so $x=0$ and $x=4$.
Between these, is the curve above or below the axis? Since the coefficient of $x^2$ is negative the parabola opens downward, and at, say, $x=2$ we get $y = 8 - 4 = 4 > 0$. So it's above the axis throughout and the area is simply the integral.
$\displaystyle\text{Area} = \int_0^4 (4x - x^2)\,dx = \Big[2x^2 - \tfrac{1}{3}x^3\Big]_0^4 = \Big(32 - \tfrac{64}{3}\Big) - 0$.
Writing $32 = \frac{96}{3}$ gives $\frac{96 - 64}{3} = \frac{32}{3}$.
So the area is $\dfrac{32}{3}$ square units.
Roots: $x(4-x)=0 \Rightarrow x=0,\,4$. Curve above axis on $[0,4]$ (downward parabola).
- $\int_0^4 (4x - x^2)\,dx = \big[2x^2 - \tfrac{1}{3}x^3\big]_0^4$
- $= 32 - \tfrac{64}{3} = \tfrac{96 - 64}{3} = \tfrac{32}{3}$
Area $= \dfrac{32}{3}$ square units.
Where the marks go
- 1 mark: Finds the limits $x = 0$ and $x = 4$ (and notes the curve is above the axis)
- 1 mark: Correct primitive and substitution of limits
- 1 mark: Correct area $\tfrac{32}{3}$ square units
Key idea
Area under a curve above the $x$-axis is the definite integral between the points where it cuts the axis; confirm the curve's sign first.
Example 5 — Area between two curves
Question
Find the area of the region enclosed between the curve $y = x^2$ and the line $y = x + 2$.
Solution
Find the intersections: $x^2 = x + 2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1) = 0$, so $x = -1$ and $x = 2$.
On this interval the line is on top (test $x=0$: line $=2$, curve $=0$). Integrate top minus bottom:
$\displaystyle\text{Area} = \int_{-1}^{2}\big[(x+2) - x^2\big]\,dx = \Big[\tfrac{1}{2}x^2 + 2x - \tfrac{1}{3}x^3\Big]_{-1}^{2}$.
At $x=2$: $2 + 4 - \tfrac{8}{3} = 6 - \tfrac{8}{3} = \tfrac{10}{3}$. At $x=-1$: $\tfrac{1}{2} - 2 + \tfrac{1}{3} = -\tfrac{7}{6}$.
$\text{Area} = \tfrac{10}{3} - \big(-\tfrac{7}{6}\big) = \tfrac{20}{6} + \tfrac{7}{6} = \tfrac{27}{6} = \tfrac{9}{2}$.
Area $= \dfrac{9}{2}$ square units. Always integrate "upper minus lower" — get the order wrong and the sign flips.
Let's first find where the two graphs meet, because those points are the limits of our region. Setting them equal: $x^2 = x + 2$, so $x^2 - x - 2 = 0$, which factors as $(x-2)(x+1)=0$, giving $x=-1$ and $x=2$.
Next we decide which graph is on top between these values. Testing $x=0$, the line gives $2$ and the curve gives $0$, so the line $y = x+2$ is above the parabola throughout.
The area between two curves is the integral of (upper $-$ lower): $\displaystyle\text{Area} = \int_{-1}^{2}\big[(x+2) - x^2\big]\,dx = \Big[\tfrac{1}{2}x^2 + 2x - \tfrac{1}{3}x^3\Big]_{-1}^{2}$.
At the upper limit $x=2$: $\tfrac{1}{2}(4) + 4 - \tfrac{1}{3}(8) = 2 + 4 - \tfrac{8}{3} = \tfrac{10}{3}$. At the lower limit $x=-1$: $\tfrac{1}{2}(1) - 2 - \tfrac{1}{3}(-1) = \tfrac{1}{2} - 2 + \tfrac{1}{3} = -\tfrac{7}{6}$.
Subtracting: $\tfrac{10}{3} - \left(-\tfrac{7}{6}\right) = \tfrac{20}{6} + \tfrac{7}{6} = \tfrac{27}{6} = \tfrac{9}{2}$.
So the enclosed area is $\dfrac{9}{2}$ square units.
Intersections: $x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1)=0 \Rightarrow x = -1,\,2$. Line above curve on $[-1,2]$.
- $\int_{-1}^{2}\big[(x+2) - x^2\big]\,dx = \big[\tfrac{1}{2}x^2 + 2x - \tfrac{1}{3}x^3\big]_{-1}^{2}$
- $x=2$: $2 + 4 - \tfrac{8}{3} = \tfrac{10}{3}$
- $x=-1$: $\tfrac{1}{2} - 2 + \tfrac{1}{3} = -\tfrac{7}{6}$
- $\tfrac{10}{3} + \tfrac{7}{6} = \tfrac{27}{6} = \tfrac{9}{2}$
Area $= \dfrac{9}{2}$ square units.
Where the marks go
- 1 mark: Finds intersection points $x = -1$ and $x = 2$
- 1 mark: Sets up the integral of (upper $-$ lower), i.e. $(x+2) - x^2$
- 1 mark: Correct primitive and substitution of both limits
- 1 mark: Correct area $\tfrac{9}{2}$ square units
Key idea
Area between curves is $\int (\text{upper} - \text{lower})\,dx$ taken between their points of intersection.
Frequently asked questions
Step-by-step solutions to exam-style questions on Integration in HSC Maths Advanced, 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 Advanced syllabus for Integration, using the methods and notation expected in the exam.