Year 11 Maths Advanced Module 3 ⏱ ~40 min Lesson 12 of 15

Areas Between Curves

Not all regions are bounded by a single curve and the $x$-axis. Sometimes we need the area trapped between two curves — like the cross-section of a pipe, or the gap between two hill profiles. In this lesson, you will learn how to find these areas by integrating the difference between the upper function and the lower function.

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Think First

Imagine two curves on a graph — one always above the other — trapping a region between them. If you wanted to find the area of that trapped region, why might subtracting the lower curve from the upper curve at each point, then integrating, give you the exact area?

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📐

Formula Reference — This Lesson

Area between two curves
$$A = \int_a^b \bigl(\text{upper function} - \text{lower function}\bigr) \, dx$$
Finding limits
Solve $f(x) = g(x)$ to find the $x$-coordinates of intersection points
Vertical strips
Height of strip = $y_{\text{top}} - y_{\text{bottom}}$ Width of strip = $dx$ Area of strip = $(y_{\text{top}} - y_{\text{bottom}}) \, dx$
Key insight: Always sketch the graphs first. The upper and lower functions may swap places if the curves cross, in which case you must split the integral at the intersection point.
📖 Know

Key Facts

  • The formula for area between two curves
  • How to find intersection points to determine limits
  • That areas are always positive
💡 Understand

Concepts

  • Why the integrand is upper minus lower function
  • How vertical strips approximate the region
  • When and why to split an integral if curves cross
✅ Can Do

Skills

  • Find the area enclosed by two curves
  • Find intersection points algebraically
  • Split integrals when the upper function changes
  • Sketch regions to verify setup

Misconceptions to Fix

✗

Wrong: A stationary point is always a maximum or minimum.

✗

Right: Stationary points can also be horizontal points of inflection where the concavity changes.

📚 Core Content

Key Terms
DerivativeThe rate of change of a function at a point; the gradient of the tangent.
DifferentiationThe process of finding the derivative of a function.
Stationary PointA point where the derivative equals zero.
Chain RuleA rule for differentiating composite functions: dy/dx = dy/du × du/dx.
Product RuleA rule for differentiating products: d(uv)/dx = u(dv/dx) + v(du/dx).
IntegrationThe reverse process of differentiation; finding the area under a curve.
🏞️

Area Between Two Curves

If two curves $y = f(x)$ and $y = g(x)$ intersect at $x = a$ and $x = b$, and $f(x) \ge g(x)$ throughout $[a, b]$, then the area between them is:

$$A = \int_a^b \bigl(f(x) - g(x)\bigr) \, dx$$

Think of this as adding up infinitely many thin vertical strips, each with height $f(x) - g(x)$ and width $dx$.

What If the Curves Cross?

If the curves intersect at $x = c$ between $a$ and $b$, the upper and lower functions may swap. In that case, split the integral:

$$A = \int_a^c \bigl(f(x) - g(x)\bigr) \, dx + \int_c^b \bigl(g(x) - f(x)\bigr) \, dx$$

Or more simply, integrate the absolute difference:

$$A = \int_a^b \bigl|f(x) - g(x)\bigr| \, dx$$

Why this matters for urban planning. When two roads diverge or two land boundaries meet, planners need to know the exact area of the enclosed region for zoning, taxation, and development permits. Integrating the difference between boundary curves gives precise land areas that cannot be found with simple geometric formulas.
📝

Step-by-Step Procedure

  1. Sketch both curves on the same axes.
  2. Find intersection points by solving $f(x) = g(x)$.
  3. Determine which curve is on top in each interval.
  4. Set up the integral(s) as upper minus lower.
  5. Evaluate and state the area with correct units.

🧮 Worked Examples

Worked Example 1 — Area Between a Parabola and a Line

Stepwise
Find the area enclosed by $y = x^2$ and $y = 2x$.
  1. 1
    Find intersection points
    x^2 = 2x \quad \Rightarrow \quad x^2 - 2x = 0 \quad \Rightarrow \quad x(x - 2) = 0
    x = 0 \quad \text{or} \quad x = 2
  2. 2
    Identify upper and lower functions
    Between $x = 0$ and $x = 2$, the line $y = 2x$ is above the parabola $y = x^2$.
  3. 3
    Set up and evaluate the integral
    A = \int_0^2 (2x - x^2) \, dx = \Bigl[x^2 - \frac{x^3}{3}\Bigr]_0^2
    = \Bigl(4 - \frac{8}{3}\Bigr) - 0 = \frac{4}{3}
✓ Answer $\frac{4}{3}$ square units

Worked Example 2 — Area Between Two Parabolas

Stepwise
Find the area enclosed by $y = x^2$ and $y = 4 - x^2$.
  1. 1
    Find intersection points
    x^2 = 4 - x^2 \quad \Rightarrow \quad 2x^2 = 4 \quad \Rightarrow \quad x = \pm \sqrt{2}
  2. 2
    Identify upper function
    Between $-\sqrt{2}$ and $\sqrt{2}$, $y = 4 - x^2$ is above $y = x^2$.
  3. 3
    Set up and evaluate
    A = \int_{-\sqrt{2}}^{\sqrt{2}} \bigl((4 - x^2) - x^2\bigr) \, dx = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2x^2) \, dx
    = \Bigl[4x - \frac{2x^3}{3}\Bigr]_{-\sqrt{2}}^{\sqrt{2}}
    = \Bigl(4\sqrt{2} - \frac{4\sqrt{2}}{3}\Bigr) - \Bigl(-4\sqrt{2} + \frac{4\sqrt{2}}{3}\Bigr)
    = \frac{8\sqrt{2}}{3} + \frac{8\sqrt{2}}{3} = \frac{16\sqrt{2}}{3}
✓ Answer $\frac{16\sqrt{2}}{3}$ square units

Worked Example 3 — Curves That Cross

Stepwise
Find the total area enclosed by $y = x^3 - x$ and $y = 0$ between $x = -1$ and $x = 1$.
  1. 1
    Factor and find intercepts
    y = x(x^2 - 1) = x(x - 1)(x + 1)
    The curve meets $y = 0$ at $x = -1, 0, 1$.
  2. 2
    Determine sign in each interval
    For $x \in (-1, 0)$: $y > 0$ (curve above axis)
    For $x \in (0, 1)$: $y < 0$ (curve below axis)
  3. 3
    Set up two integrals for total area
    A = \int_{-1}^0 (x^3 - x) \, dx + \Bigl|\int_0^1 (x^3 - x) \, dx\Bigr|
  4. 4
    Evaluate
    \int (x^3 - x) \, dx = \frac{x^4}{4} - \frac{x^2}{2}
    \int_{-1}^0 = 0 - \Bigl(\frac{1}{4} - \frac{1}{2}\Bigr) = \frac{1}{4}
    \int_0^1 = \Bigl(\frac{1}{4} - \frac{1}{2}\Bigr) - 0 = -\frac{1}{4} \quad \Rightarrow \quad \Bigl|-\frac{1}{4}\Bigr| = \frac{1}{4}
    A = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}
✓ Answer $\frac{1}{2}$ square unit
⚠️

Common Mistakes — Don't Lose Easy Marks

Integrating lower minus upper
If you subtract the upper function from the lower function, you get a negative area.
✓ Fix: Always set up the integral as (upper) − (lower).
Forgetting to find intersection points
The limits of integration must be the $x$-coordinates where the curves meet. Using arbitrary limits gives meaningless answers.
✓ Fix: Always solve $f(x) = g(x)$ to find the correct limits.
Not splitting when curves cross
If the upper and lower functions swap places inside the interval, a single integral will incorrectly cancel positive and negative contributions.
✓ Fix: Check whether the curves cross between the limits. If they do, split the integral.
Giving a negative area
Area is a geometric quantity and must be positive. Negative answers indicate that upper and lower were reversed.
✓ Fix: If your area comes out negative, swap the functions and re-evaluate.

📓 Copy Into Your Books

▼

🏞️ Area between curves

  • $A = \int_a^b (y_{\text{top}} - y_{\text{bottom}}) \, dx$

🔍 Limits

  • Solve $f(x) = g(x)$ for $a$ and $b$

✂️ Split if needed

  • If curves cross at $c$, split at $c$

⚠️ Watch out

  • Area must be positive
  • Upper − lower, never the reverse

📝 How are you completing this lesson?

🧪 Activities

🔍 Activity 1 — Calculate

Find Areas Between Curves

Set up and evaluate the definite integral for each region.

  1. 1 Find the area enclosed by $y = x^2$ and $y = x$.

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    Answer in your workbook.

    Answer in your workbook

  2. 2 Find the area enclosed by $y = 2x + 3$ and $y = x^2$.

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    Answer in your workbook

  3. 3 Find the area enclosed by $y = x^2 - 2x$ and $y = 0$ between $x = 0$ and $x = 3$.

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    Answer in your workbook

  4. 4 Find the area enclosed by $y = 9 - x^2$ and $y = x + 3$.

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🎨 Activity 2 — Interpret

Analyse the Setup

Use words to explain your reasoning.

  1. 1 Why is the area between two curves found by integrating $(y_{\text{top}} - y_{\text{bottom}})$ rather than $(y_{\text{top}} + y_{\text{bottom}})$?

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    Answer in your workbook

  2. 2 A student calculates the area between two crossing curves using a single integral and gets zero. Explain why this happened and how to fix it.

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Revisit Your Thinking

Earlier you thought about why subtracting the lower curve from the upper curve gives the area between them.

At any given $x$, the vertical distance between the two curves is exactly $y_{\text{top}} - y_{\text{bottom}}$. By integrating this difference from one intersection point to the next, we are summing up infinitely many thin vertical strips, each with that exact height. This is precisely how calculus turns a geometric problem into an algebraic one — and why the method works for any pair of curves, no matter how complicated.

Now revisit your initial response. What did you get right? What has changed in your thinking?

Look back at your initial response in your book. Annotate it with what you now understand differently.

Annotate your initial response in your book
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Interactive: Area Between Curves Visualiser
Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

MC

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

✍️ Short Answer

📝

Extended Questions

ApplyBand 4

6. Find the area enclosed by the curves $y = x^2$ and $y = 3 - 2x$. Show all working, including finding the points of intersection. 4 MARKS

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✏️ Answer in your workbook
ApplyBand 4

7. Find the total area enclosed by $y = x^3 - x$ and the $x$-axis between $x = -1$ and $x = 1$. Show all working, including any splitting of integrals. 4 MARKS

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AnalyseBand 5

8. The curves $y = x^2$ and $y = kx$ (where $k > 0$) enclose a region of area $\frac{9}{2}$ square units. Find the value of $k$. 4 MARKS

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✅ Comprehensive Answers

▼

🔍 Activity 1 — Calculate Model Answers

1. Intersections at $x = 0, 1$. Area = $\int_0^1 (x - x^2) \, dx = \bigl[\frac{x^2}{2} - \frac{x^3}{3}\bigr]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$.

2. Intersections at $x = -1, 3$. Area = $\int_{-1}^3 (2x + 3 - x^2) \, dx = \bigl[x^2 + 3x - \frac{x^3}{3}\bigr]_{-1}^3 = (9 + 9 - 9) - (1 - 3 + \frac{1}{3}) = 9 + \frac{5}{3} = \frac{32}{3}$.

3. Intercepts at $x = 0, 2$. Area = $\bigl|\int_0^2 (x^2 - 2x) \, dx\bigr| + \int_2^3 (x^2 - 2x) \, dx = \bigl|\frac{8}{3} - 4\bigr| + (9 - 9 - \frac{8}{3} + 4) = \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$.

4. Intersections at $x = -3, 2$. Area = $\int_{-3}^2 (9 - x^2 - x - 3) \, dx = \int_{-3}^2 (6 - x - x^2) \, dx = \bigl[6x - \frac{x^2}{2} - \frac{x^3}{3}\bigr]_{-3}^2 = (12 - 2 - \frac{8}{3}) - (-18 - \frac{9}{2} + 9) = \frac{22}{3} + \frac{27}{2} = \frac{44 + 81}{6} = \frac{125}{6}$.

🎨 Activity 2 — Interpret Model Answers

1. We want the vertical distance between the curves at each $x$. Adding them would give the sum of their heights from the axis, not the gap between them. Subtracting gives the exact height of the region.

2. When curves cross, one integral part is positive and the other negative. If not split, they cancel out. The fix is to split at the intersection point and ensure each part is positive.

❓ Multiple Choice

1. B — Intersections at $x = 0$ and $x = 2$. Area = $int_0^2 (2x - x^2) , dx = \frac{4}{3}$.

2. C — Setting $x^2 = 4 - x^2$ gives $2x^2 = 4$, so $x = \\pm \\sqrt{2}$.

3. B — Area = $int_a^b (y_{ ext{top}} - y_{ ext{bottom}}) , dx$.

4. A — Intersections at $x = -1$ and $x = 3$. Area = $int_{-1}^3 (x + 3 - x^2) , dx = \frac{9}{2}$.

5. B — When curves cross, the upper and lower functions swap. You must split the integral at the crossing point.

6. A — The curve is below the axis on $(0, 2)$. Area = $\bigl|int_0^2 (x^2 - 2x) , dx\bigr| = \frac{4}{3}$.

7. C — $int_0^k (kx - x^2) , dx = \bigl[\frac{kx^2}{2} - \frac{x^3}{3}\bigr]_0^k = \frac{k^3}{6}$.

📝 Short Answer Model Answers

Q6 (4 marks): Intersections: $x^2 = 3 - 2x \Rightarrow x^2 + 2x - 3 = 0 \Rightarrow (x + 3)(x - 1) = 0$, so $x = -3$ or $x = 1$ [1]. Area $= \int_{-3}^{1} (3 - 2x - x^2) \, dx = \bigl[3x - x^2 - \frac{x^3}{3}\bigr]_{-3}^{1} = \frac{32}{3}$ [3].

Q7 (4 marks): Factor: $y = x(x - 1)(x + 1)$ [1]. Intercepts at $x = -1, 0, 1$. For $x \in (-1, 0)$, $y > 0$; for $x \in (0, 1)$, $y < 0$ [1]. Area $= \int_{-1}^{0} (x^3 - x) \, dx + \bigl|\int_{0}^{1} (x^3 - x) \, dx\bigr| = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ [2].

Q8 (4 marks): Intersections at $x = 0$ and $x = k$ [1]. Area $= \int_0^k (kx - x^2) \, dx = \bigl[\frac{kx^2}{2} - \frac{x^3}{3}\bigr]_0^k = \frac{k^3}{6}$ [3].

Consolidation Game

Areas Between Curves

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