Area Between Curves — Worked Examples
You know the formula. Now the real challenge: two curves can cross each other more than once, and when they swap positions mid-interval you must split the integral. This lesson drills the three problem types that appear most on HSC papers — single region, split region, and region specified by a given domain — so you can handle any variation that comes up.
$y = x^3$ and $y = x$ cross at $x = -1$, $x = 0$, and $x = 1$. Without computing — if someone blindly evaluated $\int_{-1}^{1}(x^3 - x)\,dx$, what would they get and why? Would that number equal the total enclosed area?
When two curves cross each other mid-interval, the roles of "upper" and "lower" swap. Integrating blindly across the crossing point causes the two sub-areas to cancel — giving zero or an undercount instead of the true total area.
If $f$ and $g$ cross at $x = c$ inside $[a,b]$, split the integral:
$A = \int_a^c |f(x)-g(x)|\,dx + \int_c^b |f(x)-g(x)|\,dx$
In practice: determine which is on top in each sub-interval, then write the correct subtraction order in each piece.
Key facts
- When curves cross mid-interval, the integral must be split at each crossing point
- Total area = sum of all sub-region areas (each taken positive)
- A sketch always reveals where splits are needed
Concepts
- Why a blind integral gives cancellation rather than total area
- How symmetry can halve the work in symmetric problems
- The difference between "area enclosed" and "area on a given domain"
Skills
- Identify all crossing points and split the region accordingly
- Evaluate multi-piece area problems with three or more intersection points
- Use symmetry to simplify area calculations where applicable
Before tackling split regions, lock in the setup strategy for any area problem:
- Sketch both curves on the same axes.
- Find all intersection points: solve $f(x) = g(x)$.
- Identify sub-regions: mark each interval between consecutive intersection points.
- Determine top/bottom in each sub-interval (test a point).
- Write an integral per sub-region with correct (upper − lower) order.
- Add all results (always positive quantities).
For a single enclosed region (two intersection points only), steps 3–5 reduce to one integral.
Setup strategy: (1) sketch curves; (2) find intersections; (3) identify upper/lower on each sub-interval; (4) compute $A=\int[\text{upper}-\text{lower}]\,dx$ over each sub-interval.
Pause — copy the single-region setup checklist: (1) identify upper/lower curve; (2) state the intersection limits; (3) write $A=\int_a^b[f-g]\,dx$ into your book.
Quick check: The curves $y = x^3$ and $y = x$ intersect at $x = -1$, $0$, $1$. To find the total enclosed area, a student should:
We just saw the setup strategy: identify the upper/lower curve on $[a,b]$, then apply $A=\int_a^b[f(x)-g(x)]\,dx$. That raises a question: if the curves swap which is upper at some interior crossing point $x=c$, and you ignore this, the positive and negative signed areas cancel — how do you split the integral to avoid this? This card answers it → $A=\int_a^c[f-g]\,dx+\int_c^b[g-f]\,dx$, so each sub-integral is non-negative.
When curves cross at an interior point $x = c$, the region splits. The cancellation problem in the hook: $\int_{-1}^{1}(x^3 - x)\,dx = 0$ because the two regions exactly cancel — each has area $\frac{1}{2}$ but opposite signs. The true total area is $\frac{1}{2} + \frac{1}{2} = 1$.
How to split:
- On $[-1, 0]$: test $x = -0.5$: $(-0.5)^3 = -0.125 > (-0.5)$? No, $-0.125 > -0.5$ — so $y=x^3$ is above $y=x$ here. Integrand: $x^3 - x$.
- On $[0, 1]$: test $x = 0.5$: $(0.5)^3 = 0.125 < 0.5$ — so $y=x$ is above $y=x^3$. Integrand: $x - x^3$.
By symmetry (odd functions on symmetric interval): $A = 2\int_0^1(x-x^3)\,dx$.
When curves cross at an interior point $x = c$, the region splits. The cancellation problem in the hook: $\int_{-1}^{1}(x^3 - x)\,dx = 0$ because the two regions exactly cancel — each has area $\frac{1}{2}$ but opposite signs. The true...
Pause — copy the split-region formula: $A=\int_a^c[f(x)-g(x)]\,dx+\int_c^b[g(x)-f(x)]\,dx$, with an explanation of why the signs flip at the crossing point, into your book.
Did you get this? True or false: $\displaystyle\int_{-1}^{1}(x^3 - x)\,dx = 0$, so the total area enclosed by $y=x^3$ and $y=x$ on $[-1,1]$ is zero.
Worked examples · 3 in a row, reveal as you go
Find the total area enclosed by $y = x^2 - 1$ and $y = 1 - x^2$.
Find the total area enclosed between $y = x^3 - x$ and $y = 0$ (the $x$-axis).
$A_1 = \int_{-1}^{0}(x^3-x)\,dx = \left[\dfrac{x^4}{4}-\dfrac{x^2}{2}\right]_{-1}^{0}$
$A_2 = \int_{0}^{1}(0-(x^3-x))\,dx = \int_0^1(x-x^3)\,dx = \left[\dfrac{x^2}{2}-\dfrac{x^4}{4}\right]_0^1 = \frac{1}{4}$.
$A_{\text{total}} = A_1 + A_2 = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}$.
Find the area between $y = \sin x$ and $y = \cos x$ on the interval $[0, \pi]$. (Hint: they intersect once on this interval.)
$= \left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\right)-(0+1) = \sqrt{2} - 1$
$= (1-0)-\left(-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}\right) = 1 + \sqrt{2}$
$A_{\text{total}} = (\sqrt{2}-1)+(1+\sqrt{2}) = 2\sqrt{2}$
Fill the gap: To find the total area enclosed by $y=x^3-x$ and the $x$-axis, you need separate integral(s) because the curves cross at $x=-1$, $0$, and $1$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the total area between two curves that cross multiple times is the sum (not difference) of all the individual sub-region areas.
Activities · practice with the ideas
Find the total area between $y = x^2 - 1$ and $y = 1 - x^2$ by evaluating $\int_{-1}^1 2(1-x^2)\,dx$ without using the symmetry shortcut.
Find all intersection points of $y = x^3$ and $y = x$, then find the total enclosed area.
Set up (do not evaluate) the integrals for the total area between $y = \sin 2x$ and $y = 0$ on $[0, \pi]$. Identify where to split.
Find the area between $y = x^2$ and $y = x$ on the domain $[-1, 2]$. Note: they cross inside this domain.
Explain in one sentence why $\int_{-1}^1 (x^3 - x)\,dx = 0$ even though the enclosed area is $\frac{1}{2}$.
Odd one out: Three of these are correct statements about area between curves. Which one is NOT correct?
Earlier you predicted what $\int_{-1}^{1}(x^3-x)\,dx$ would give. The answer is indeed $0$ — and the total enclosed area is $\frac{1}{2}$, not zero.
The key insight: a definite integral computes signed area. When regions above and below $y=0$ have equal size, they cancel perfectly. To get the true geometric area you must split at each crossing and add the absolute values. This is one of the most common exam traps in Module 9.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the total area enclosed by $y = x^2 - 1$ and $y = 1 - x^2$. (2 marks)
Q2. Find the total area between $y = x^3 - x$ and $y = 0$ by splitting the integral at the appropriate points. (3 marks)
Q3. A student evaluates $\int_{-1}^{1}(x^3-x)\,dx = 0$ and concludes the area is zero. Explain the error and find the correct total area. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. $\int_{-1}^1 2(1-x^2)\,dx = \left[2x-\frac{2x^3}{3}\right]_{-1}^1 = \left(2-\frac{2}{3}\right)-\left(-2+\frac{2}{3}\right) = \frac{8}{3}$. · 2. Intersections: $x^3=x \Rightarrow x=0,\pm1$. Total area $= 2\int_0^1(x-x^3)\,dx = 2\cdot\frac{1}{4}=\frac{1}{2}$. · 3. $\sin 2x=0$ at $x=0,\pi/2,\pi$. Split at $x=\pi/2$. $A=\int_0^{\pi/2}\sin 2x\,dx + \int_{\pi/2}^{\pi}(-\sin 2x)\,dx$. · 4. Cross at $x=0,1$. On $[-1,0]$: $x^2$ above $x$. $A_1=\int_{-1}^0(x^2-x)\,dx=\frac{1}{6}+\frac{1}{2}=\frac{2}{3}$. On $[0,1]$: $x$ above $x^2$. $A_2=\frac{1}{6}$. On $[1,2]$: $x^2$ above $x$. $A_3=\int_1^2(x^2-x)\,dx=\frac{7}{6}-1=\frac{5}{6}$. Wait: $\int_1^2(x^2-x)\,dx=[\frac{x^3}{3}-\frac{x^2}{2}]_1^2=(\frac{8}{3}-2)-(\frac{1}{3}-\frac{1}{2})=\frac{2}{3}+\frac{1}{6}=\frac{5}{6}$. Total $=\frac{2}{3}+\frac{1}{6}+\frac{5}{6}=\frac{4}{6}+\frac{1}{6}+\frac{5}{6}=\frac{10}{6}=\frac{5}{3}$. · 5. The definite integral sums signed areas; the two lobes are equal in size but opposite in sign, so they cancel. The geometric area counts each region positively.
Q1 (2 marks): Intersections: $x=\pm1$ [1]. $A=\int_{-1}^1 2(1-x^2)\,dx = 4\int_0^1(1-x^2)\,dx = 4\left[x-\frac{x^3}{3}\right]_0^1 = 4\cdot\frac{2}{3} = \dfrac{8}{3}$ [1].
Q2 (3 marks): Zeros at $x=-1,0,1$ [1]. Split: $A_1=\int_{-1}^0(x^3-x)\,dx=\frac{1}{4}$; $A_2=\int_0^1(x-x^3)\,dx=\frac{1}{4}$ [1]. Total $=\frac{1}{4}+\frac{1}{4}=\dfrac{1}{2}$ [1].
Q3 (3 marks): Error: the integral gives signed area; the two lobes cancel because the integrand changes sign at $x=0$ [1]. Correct method: split at $x=0$ [1]. $A=\left|\int_{-1}^0(x^3-x)\,dx\right|+\left|\int_0^1(x^3-x)\,dx\right|=\frac{1}{4}+\frac{1}{4}=\dfrac{1}{2}$ [1].
Five timed questions including split-region problems. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering area-between-curves questions. Lighter alternative to the boss.
Mark lesson as complete
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