What is not splitting at a crossing?

If the curves swap order inside the interval, a single integral cancels and underestimates the area.

§-Syllabus dot point

WAMath MethodsSyllabus dot point

How do we find the area enclosed between two curves, including where they cross?

Calculate the area enclosed between two curves by integrating the difference of the upper and lower functions over the correct interval

WACE Year 12 Mathematics Methods Unit 4 area between curves: finding intersection points, integrating upper minus lower, splitting where curves cross, with worked SCSA-style examples and the common sign mistake.

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 area formula
Finding the limits
Integrating with respect to $y$
Area between a curve and the axis
When the curves cross

What this dot point is asking

SCSA Unit 4 extends area calculation to the region enclosed between two curves. This dot point asks you to find the intersection points, integrate the difference of the functions, and handle cases where the upper and lower curves change. It is examined in both sections.

The area formula

When one curve lies above the other on $[a,b]$ , the enclosed area is the integral of the gap between them.

Subtracting lower from upper guarantees a non-negative integrand, so the result is a genuine area. This works even where both curves are below the $x$ -axis, because the vertical gap is what matters. The constants from each antiderivative cancel in the subtraction, so the area between curves never carries a constant of integration; only the limits and the order of the curves matter.

Finding the limits

The limits are the $x$ -values where the curves meet, found by solving $f(x)=g(x)$ . Between consecutive intersection points the order of the curves is fixed, so identify which is on top there before integrating.

Area between a line and a parabola

Find the area enclosed by $y=6-x^{2}$ and $y=x$ .

Step 1: Find the intersection points

The limits of integration come from where the curves meet, so solve $6-x^{2}=x$ to find those $x$ -values. Rearranging gives $x^{2}+x-6=0$ , which factors as $(x+3)(x-2)=0$ , so $x=-3$ or $x=2$ .

Step 2: Identify the upper curve and set up the integral

Between $x=-3$ and $x=2$ the parabola $y=6-x^{2}$ lies above the line $y=x$ , so subtract the line from the parabola to obtain a non-negative integrand throughout the interval.

A = \int_{-3}^{2}\big((6-x^{2})-x\big)\,dx = \int_{-3}^{2}\big(6-x-x^{2}\big)\,dx.

Step 3: Antidifferentiate and evaluate

Apply the power rule term by term, then substitute the limits and subtract to obtain the area.

A = \left[6x-\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{-3}^{2}.

At $x=2$ : $12-2-\tfrac{8}{3}=10-\tfrac{8}{3}=\tfrac{22}{3}$ . At $x=-3$ : $-18-\tfrac{9}{2}+9=-9-\tfrac{9}{2}=-\tfrac{27}{2}$ .

A = \frac{22}{3}-\left(-\frac{27}{2}\right) = \frac{44}{6}+\frac{81}{6} = \frac{125}{6}.

Final answer: The enclosed area is $\dfrac{125}{6}$ square units.

Markers reward the intersection points, the upper-minus-lower integrand, and accurate evaluation.

Integrating with respect to $y$

Some regions are described more naturally as a width in the $x$ -direction at each height $y$ , especially when the boundaries are given as $x=f(y)$ . In that case the area is $\int_c^d (x_{\text{right}}-x_{\text{left}})\,dy$ , integrating the horizontal gap over the range of $y$ . The principle is identical, right minus left in place of upper minus lower, and the limits are the $y$ -coordinates of the intersection points. Choosing the orientation that makes the integrand simplest can turn a two-piece split into a single clean integral.

Area between a curve and the axis

A region bounded by a curve and the $x$ -axis is the special case where the lower (or upper) curve is the line $y=0$ . The method is identical: find where the curve meets the axis, then integrate the curve over each interval, taking the absolute value of any part below the axis. Confusing this signed integral with total area is the most common slip, so always check whether the region dips below the axis before reporting an area.

When the curves cross

If the curves intersect inside the interval, the upper and lower functions swap there. Split the integral at the crossing and use upper minus lower on each piece, so the integrand stays non-negative throughout.

Splitting where curves cross

Find the area between $y=x^{3}$ and $y=x$ on $[-1,1]$ .

Step 1: Find where the curves cross and identify which is on top in each piece

Solving $x^{3}=x$ gives $x(x^{2}-1)=0$ , so the curves meet at $x=-1,0,1$ . On $[-1,0]$ the cubic $y=x^{3}$ is above $y=x$ ; on $[0,1]$ the line $y=x$ is above. Because the upper and lower functions swap at $x=0$ , the interval must be split there to keep each integrand non-negative.

Step 2: Integrate each piece separately and add

Using upper minus lower on each sub-interval ensures both integrands are non-negative, so their sum gives the true total area rather than a cancellation.

A = \int_{-1}^{0}(x^{3}-x)\,dx + \int_{0}^{1}(x-x^{3})\,dx.

Each integral evaluates to $\tfrac{1}{4}$ , so the total area is $\tfrac{1}{4}+\tfrac{1}{4}=\tfrac{1}{2}$ square units. A single unsplit integral would wrongly give zero by cancellation.

Final answer: $\dfrac{1}{2}$ square units.

Exam-style practice questions

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

WACE 20226 marksCalculator-assumed. The curves

y=x^{2}

and

y=2x

enclose a region. (a) Find the

x

-coordinates of the points of intersection. (b) Find the area of the enclosed region.

Show worked answer →

A line-and-parabola enclosed area.

(a) Set $x^{2}=2x$ , so $x^{2}-2x=0$ , $x(x-2)=0$ , giving $x=0$ and $x=2$ .

(b) Between $0$ and $2$ the line $y=2x$ is above the parabola $y=x^{2}$ . So $A=\displaystyle\int_{0}^{2}(2x-x^{2})\,dx=\left[x^{2}-\dfrac{x^{3}}{3}\right]_{0}^{2}=4-\dfrac{8}{3}=\dfrac{4}{3}$ square units.

Markers reward the intersection points, identifying the line as the upper curve, and the correct definite integral.

WACE 20247 marksCalculator-assumed. Find the total area enclosed between the curve

y=x^{3}-4x

and the

x

-axis.

Show worked answer →

A total-area question requiring splitting at intercepts.

Intercepts: $x^{3}-4x=x(x^{2}-4)=x(x-2)(x+2)=0$ , so $x=-2,0,2$ . The curve is above the axis on $[-2,0]$ and below on $[0,2]$ (by symmetry the magnitudes match).

On $[-2,0]$ : $\displaystyle\int_{-2}^{0}(x^{3}-4x)\,dx=\left[\dfrac{x^{4}}{4}-2x^{2}\right]_{-2}^{0}=0-(4-8)=4$ .

On $[0,2]$ the signed integral is $-4$ , so the area is $4$ . Total area $=4+4=8$ square units.

Markers reward finding all three intercepts, splitting at $x=0$ , and adding the absolute values rather than the signed integrals.

What this dot point is asking

The area formula

Finding the limits

Integrating with respect to yyy

Area between a curve and the axis

When the curves cross

Exam-style practice questions

Integrating with respect to $y$