Worked Solutions
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.
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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.
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