What is not splitting at a crossing?

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

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 4.76 min answerUpdated 2026-05-29

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
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.

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

Area enclosed by $y=6-x^{2}$ and $y=x$

Find the intersections. Set $6-x^{2}=x$ , so $x^{2}+x-6=0$ , giving $(x+3)(x-2)=0$ and $x=-3$ or $x=2$ .

Between $x=-3$ and $x=2$ the parabola $y=6-x^{2}$ lies above the line $y=x$ .

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

Antidifferentiate.

= \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}.

The enclosed area is $\dfrac{125}{6}$ square units. Markers reward the intersection points, the upper-minus-lower integrand, and accurate evaluation.

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.