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Areas between curves (WACE Year 12 Mathematics Specialist Unit 4)

Syllabus dot point

“Find the area between curves using definite integration, accounting for intersection points and which curve is on top”

7 min read

Quick answer

The area between curves y=f(x)y = f(x) (upper) and y=g(x)y = g(x) (lower) from x=ax = a to x=bx = b is ∫ab[f(x)−g(x)] dx\int_a^b [f(x) - g(x)]\,dx. Find the limits from the intersection points, always integrate top minus bottom, and split the region wherever the curves swap order.

Jump to a section
  1. What this dot point is asking
  2. Top minus bottom
  3. Finding the limits
  4. Why top minus bottom always works
  5. When the curves cross inside
  6. Integrating with respect to y
  7. Choosing the right strip orientation
  8. A note on signed area versus enclosed area
  9. Exam-style questions

What this dot point is asking

SCSA wants you to set up and evaluate the area between two curves, locate the boundaries by solving for intersections, and handle regions where the upper and lower curves change.

Top minus bottom

Area between two curves

A=∫ab[f(x)−g(x)] dx,f(x)≥g(x) on [a,b].A = \int_a^b \big[f(x) - g(x)\big]\,dx, \qquad f(x) \ge g(x) \text{ on } [a, b].

The integrand is the vertical gap between the curves at each xx. Because we subtract the lower curve, the result is positive and there is no need to worry about whether either curve dips below the xx-axis: the difference handles that automatically.

Finding the limits

The limits aa and bb are usually the xx-coordinates of the intersection points, found by solving f(x)=g(x)f(x) = g(x). Sketch the region to confirm which curve is on top, since that determines the order of subtraction.

Why top minus bottom always works

The reason the formula needs no absolute-value signs is geometric. At a fixed xx, the height of the thin strip between the curves is the upper yy-value minus the lower yy-value, which is f(x)−g(x)f(x) - g(x) whenever ff is on top. Summing these strip areas as the width dx→0dx \to 0 is exactly the definite integral. Because f(x)−g(x)≥0f(x) - g(x) \ge 0 throughout the interval, the integral is positive automatically, even where both curves are below the xx-axis, because shifting both curves up by the same amount leaves their difference unchanged. This is why you should never integrate ff and gg separately and subtract the signed areas: subtract the functions first, then integrate once.

When the curves cross inside

Split at every crossing

If the curves intersect between the outer limits, the upper curve changes. Split the integral at each crossing and use top minus bottom separately on each piece, so every contribution stays positive.

Integrating with respect to y

When the region is bounded more naturally on the left and right, it can be cleaner to write xx as a function of yy and integrate ∫cd[xright−xleft] dy\int_c^d [x_{\text{right}} - x_{\text{left}}]\,dy. Choose the variable that makes the bounds simplest. A region trapped between a sideways parabola x=y2x = y^2 and a line, for instance, has a single clean description in terms of yy but would need splitting in terms of xx, so the choice of variable can turn two integrals into one.

Choosing the right strip orientation

Deciding whether to integrate in xx or in yy is a judgement worth making before any algebra. Use vertical strips (integrate in xx) when the region has a clear top curve and bottom curve over a single xx-interval. Use horizontal strips (integrate in yy) when the region has a clear right boundary and left boundary over a single yy-interval, which is typical when the curves are given as xx in terms of yy or open sideways. If neither orientation avoids a split, pick whichever produces the simpler integrand. A quick sketch with one sample strip drawn in shows immediately which curve is the upper or right boundary and which is the lower or left boundary.

A note on signed area versus enclosed area

The phrase area between curves always means a positive, enclosed area, so the top-minus-bottom rule with a split at every crossing is the safe method. This is different from a single definite integral ∫abf(x) dx\int_a^b f(x)\,dx, which is a signed area that can be negative where ff lies below the axis. SCSA questions are explicit about which they want: an enclosed area is always positive, whereas a net signed value may be requested in a motion or accumulation context. Reading the command word carefully decides whether to split for positivity or to leave the integral signed.

Find the area enclosed between y=x2y = x^2 and y=2xy = 2x

Step 1: Find the intersection points

The enclosed region is bounded between the two curves, so we need to know where they meet. Setting the expressions equal and solving gives the limits of integration.

x2=2x  ⟹  x(x−2)=0  ⟹  x=0 and x=2.x^2 = 2x \implies x(x - 2) = 0 \implies x = 0 \text{ and } x = 2.

Step 2: Determine which curve is on top

Before integrating, we must identify the upper and lower boundaries on the interval [0,2][0, 2]. Testing a convenient interior point avoids any ambiguity about which curve is greater.

At x=1x = 1: the line gives y=2(1)=2y = 2(1) = 2 and the parabola gives y=12=1y = 1^2 = 1. So y=2xy = 2x lies above y=x2y = x^2 throughout [0,2][0, 2].

Step 3: Integrate upper minus lower over the interval

The enclosed area equals the integral of (top curve) minus (bottom curve). Subtracting in the correct order ensures a positive result.

A=∫02(2x−x2) dx=[x2−x33]02=(4−83)−0=43.A = \int_0^2 (2x - x^2)\,dx = \Big[x^2 - \frac{x^3}{3}\Big]_0^2 = \Big(4 - \frac{8}{3}\Big) - 0 = \frac{4}{3}.

Final answer: The enclosed area is 43\tfrac{4}{3} square units.

Common traps
Wrong order of subtraction
Always upper minus lower; reversing it gives a negative area.
Skipping a crossing
If the curves swap top and bottom inside the interval, split the integral; one signed integral over the whole interval can cancel area.
Guessing the limits
Find intersection points by solving f=gf = g; do not assume them from the sketch.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style7 marks
Calculator-assumed. (a) Find the points of intersection of y=x3−xy = x^3 - x and y=xy = x. (b) Hence find the total area enclosed between the two curves.
Show worked answer →

A curves-that-cross area, requiring a split.

(a) Solve x3−x=xx^3 - x = x, so x3−2x=0x^3 - 2x = 0, x(x2−2)=0x(x^2 - 2) = 0, giving x=0x = 0, x=2x = \sqrt{2} and x=−2x = -\sqrt{2}.

(b) Between −2-\sqrt{2} and 00 the cubic is above the line; between 00 and 2\sqrt{2} the line is above the cubic. By the symmetry of these odd functions the two pieces have equal area. On [0,2][0, \sqrt{2}] the integrand is x−(x3−x)=2x−x3x - (x^3 - x) = 2x - x^3, so ∫02(2x−x3) dx=[x2−x44]02=2−44=1\displaystyle\int_0^{\sqrt{2}} (2x - x^3)\,dx = \left[x^2 - \dfrac{x^4}{4}\right]_0^{\sqrt{2}} = 2 - \dfrac{4}{4} = 1. The total area is 2×1=22 \times 1 = 2 square units.

Markers reward the three intersection points, recognising the swap of upper and lower curve at x=0x = 0, the split (or symmetry argument), and the total area 22.

WACE 2024-style5 marks
Calculator-free. The region is bounded by the curve x=y2x = y^2 and the line x=y+2x = y + 2. Find its area by integrating with respect to yy.
Show worked answer →

An integrate-with-respect-to-yy question.

Intersections: y2=y+2y^2 = y + 2, so y2−y−2=0y^2 - y - 2 = 0, (y−2)(y+1)=0(y - 2)(y + 1) = 0, giving y=2y = 2 and y=−1y = -1. For yy between −1-1 and 22 the line x=y+2x = y + 2 is to the right of the parabola x=y2x = y^2.

Area =∫−12[(y+2)−y2] dy=[y22+2y−y33]−12= \displaystyle\int_{-1}^{2} \big[(y + 2) - y^2\big]\,dy = \left[\dfrac{y^2}{2} + 2y - \dfrac{y^3}{3}\right]_{-1}^{2}. At y=2y = 2: 2+4−83=1032 + 4 - \dfrac{8}{3} = \dfrac{10}{3}. At y=−1y = -1: 12−2+13=−76\dfrac{1}{2} - 2 + \dfrac{1}{3} = -\dfrac{7}{6}. Difference: 103+76=276=92\dfrac{10}{3} + \dfrac{7}{6} = \dfrac{27}{6} = \dfrac{9}{2}.

Markers reward the limits in yy, right-curve minus left-curve, and the area 92\dfrac{9}{2} square units.

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Sources & how we know this

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