What are step 1 - intersections?

Set x^2=x+2: $x^2-x-2=0 (x-2)(x+1)=0 x=-1,\ 2.$

What is step 3 - integrate?

$A=_-1^2((x+2)-x^2)\,dx=[x^22+2x-x^33]_-1^2.$

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 4.78 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 method
When the curves swap over
Common errors
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

Worked example

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

They intersect where $x^3=x$ , i.e. $x(x^2-1)=0$ , so $x=-1,0,1$ .

On $(-1,0)$ test $x=-0.5$ : $x^3=-0.125$ , $x=-0.5$ , so $x^3$ is on top.
On $(0,1)$ test $x=0.5$ : $x^3=0.125$ , $x=0.5$ , so $x$ is on top.

By symmetry the two pieces are equal. Compute the right piece:

\int_0^1 (x-x^3)\,dx=\Big[\tfrac{x^2}{2}-\tfrac{x^4}{4}\Big]_0^1=\tfrac12-\tfrac14=\tfrac14.

Total area $=2\times\tfrac14=\dfrac12$ square unit.

Common errors

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.