SACE Board SACE 2023-style practice: Calculator-assumed. Find the area enclosed between the curves y = 6x - x^2 and y = x.

Intersections: set 6x - x^2 = x, so 5x - x^2 = 0, giving x(5 - x) = 0 and x = 0 or x = 5. Decide which is on top: at x = 1, 6x - x^2 = 5 and y = x = 1, so the parabola 6x - x^2 is the upper curve. Integrate (top minus bottom): $A = _0^5 ((6x - x^2) - x)\,dx = _0^5 (5x - x^2)\,dx = [5x^22 - x^33]_0^5.$ At x = 5: 1252 - 1253 = 375 - 2506 = 1256. So A = 1256 ≈ 20.83 square units. Marks: one for the intersections, one for identifying the upper curve, one for the integral set-up, one for the value.

SACE Board SACE 2021-style practice: Calculator-free. Find the area enclosed between y = x^2 and y = 2x.

Intersections: x^2 = 2x gives x^2 - 2x = 0, so x(x - 2) = 0 and x = 0 or x = 2. Between these, test x = 1: y = 2x = 2 is above y = x^2 = 1, so the line is on top. $A = _0^2 (2x - x^2)\,dx = [x^2 - x^33]_0^2 = 4 - 83 = 43.$ So A = 43 square units. Marks: one for intersections, one for the integrand (top minus bottom), one for the value.

§-Syllabus dot point

SAMath MethodsSyllabus dot point

How do we find the area enclosed between two curves?

The area between two curves is the integral of the upper function minus the lower function over the interval where they enclose a region.

How to find the area enclosed between two curves: locate the intersection points, integrate (upper minus lower), and why this method avoids any sign problems.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The method
When the curves swap over
Common errors
Integrating with respect to y
A region bounded on three sides
Why it matters

What this dot point is asking

To find the area trapped between two curves you integrate the difference between them. Because you always subtract the lower curve from the upper curve, the integrand stays non-negative and you avoid the below-the-axis sign trap of the previous dot point.

The method

The three steps:

Find the intersection points by solving $f(x)=g(x)$ . These give the limits.
Decide which curve is on top between those limits (test a point or sketch).
Integrate (top − bottom) between the limits.

Area between a line and a parabola

Worked example

Find the area enclosed between $y=x+2$ and $y=x^2$ .

Step 1 - intersections: Set $x^2=x+2$ :
$x^2-x-2=0 \quad\Rightarrow\quad (x-2)(x+1)=0 \quad\Rightarrow\quad x=-1,\ 2.$
Step 2 - which is on top: Test $x=0$ : the line gives $2$ , the parabola gives $0$ , so the line $y=x+2$ is the upper curve between $-1$ and $2$ .
Step 3 - integrate (top − bottom): $A=\int_{-1}^{2}\big((x+2)-x^2\big)\,dx=\Big[\tfrac{x^2}{2}+2x-\tfrac{x^3}{3}\Big]_{-1}^{2}.$

At $x=2$ : $2+4-\tfrac{8}{3}=6-\tfrac{8}{3}=\tfrac{10}{3}$ .
At $x=-1$ : $\tfrac12-2+\tfrac13=-\tfrac{7}{6}$ .

A=\tfrac{10}{3}-\left(-\tfrac{7}{6}\right)=\tfrac{20}{6}+\tfrac{7}{6}=\frac{27}{6}=\frac{9}{2}=4.5 \text{ square units}.

When the curves swap over

If the curves cross between the outer limits, the "top" curve changes. Split the integral at the crossover and use (top − bottom) appropriate to each piece, taking each as positive.

Curves that swap which is on top

Find the total area between $y = x^3$ and $y = x$ from $x = -1$ to $x = 1$ .

Step 1: Find all intersection points inside the interval

To know where to split the region, solve $x^3 = x$ .

x^3 - x = 0 \quad\Rightarrow\quad x(x^2 - 1) = 0 \quad\Rightarrow\quad x = -1,\ 0,\ 1.

The curves meet at $x = -1$ , $x = 0$ , and $x = 1$ . The point $x = 0$ lies strictly inside the interval, so the curves cross there and swap which is on top. We must split the integral at $x = 0$ .

Step 2: Determine which curve is on top on each sub-interval

Test a point inside each sub-interval to decide the order.

On $(-1, 0)$ , test $x = -0.5$ : $x^3 = (-0.5)^3 = -0.125$ and $x = -0.5$ . Since $-0.125 > -0.5$ , the curve $y = x^3$ is above $y = x$ here.

On $(0, 1)$ , test $x = 0.5$ : $x^3 = 0.125$ and $x = 0.5$ . Since $0.5 > 0.125$ , the curve $y = x$ is above $y = x^3$ here.

Step 3: Notice the symmetry

The two sub-intervals $[-1, 0]$ and $[0, 1]$ are mirror images of each other (both functions are odd), so both pieces have the same area. Computing just the right piece is enough.

Step 4: Integrate the right piece (top minus bottom)

On $[0, 1]$ the upper curve is $y = x$ and the lower is $y = x^3$ , so integrate their difference.

\int_0^1 (x - x^3)\,dx = \Big[\tfrac{x^2}{2} - \tfrac{x^4}{4}\Big]_0^1 = \tfrac{1}{2} - \tfrac{1}{4} = \tfrac{1}{4}.

Step 5: Double the result to get the total area

A = 2 \times \tfrac{1}{4} = \frac{1}{2} \text{ square unit.}

Final answer: The total area is $\dfrac{1}{2}$ square unit.

Common errors

Integrating with respect to y

Sometimes the region is more naturally described by horizontal strips, especially when the curves are written as $x$ in terms of $y$ . The area is then $\displaystyle\int_c^d \big(x_{\text{right}}-x_{\text{left}}\big)\,dy$ , where $c$ and $d$ are the $y$ -coordinates of the intersection points. This is the mirror image of the usual method: rightmost curve minus leftmost curve, integrated over the $y$ -interval. Choosing the orientation that makes the curves single-valued over the interval often turns a two-piece problem into a one-piece one.

A region bounded on three sides

Not every region is bounded by exactly two curves over its whole width. A region might be capped by one curve on the left and a different curve on the right of some changeover point, or bounded below by the $x$ -axis on part of the interval. In these cases break the region at every changeover and add the pieces, each computed as the appropriate top minus bottom. The discipline is always the same: identify every boundary, find every intersection, and integrate each sub-region with the correct upper and lower function.

Why it matters

Area between curves is a staple extended-response question and the most general area technique in Stage 2 - area under a curve is just the special case where the lower curve is the x-axis ( $g(x)=0$ ). The intersection-then-integrate structure is exactly the modelling workflow rewarded in the Mathematical Investigation.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20234 marksCalculator-assumed. Find the area enclosed between the curves

y = 6x - x^2

and

y = x

Show worked answer →

Intersections: set $6x - x^2 = x$ , so $5x - x^2 = 0$ , giving $x(5 - x) = 0$ and $x = 0$ or $x = 5$ .

Decide which is on top: at $x = 1$ , $6x - x^2 = 5$ and $y = x = 1$ , so the parabola $6x - x^2$ is the upper curve.

Integrate (top minus bottom):

A = \int_0^5 \big((6x - x^2) - x\big)\,dx = \int_0^5 (5x - x^2)\,dx = \Big[\tfrac{5x^2}{2} - \tfrac{x^3}{3}\Big]_0^5.

At $x = 5$ : $\tfrac{125}{2} - \tfrac{125}{3} = \tfrac{375 - 250}{6} = \tfrac{125}{6}$ . So $A = \dfrac{125}{6} \approx 20.83$ square units.

Marks: one for the intersections, one for identifying the upper curve, one for the integral set-up, one for the value.

SACE 20213 marksCalculator-free. Find the area enclosed between

y = x^2

and

y = 2x

Show worked answer →

Intersections: $x^2 = 2x$ gives $x^2 - 2x = 0$ , so $x(x - 2) = 0$ and $x = 0$ or $x = 2$ .

Between these, test $x = 1$ : $y = 2x = 2$ is above $y = x^2 = 1$ , so the line is on top.

A = \int_0^2 (2x - x^2)\,dx = \Big[x^2 - \tfrac{x^3}{3}\Big]_0^2 = 4 - \tfrac{8}{3} = \tfrac{4}{3}.

So $A = \dfrac{4}{3}$ square units. Marks: one for intersections, one for the integrand (top minus bottom), one for the value.