Areas Between Curves

This lesson covers finding the area between two curves using definite integration. This extends the previous lesson's ideas and is a common question type on AQA A-Level papers.

The Key Principle

If two curves y = f(x) and y = g(x) satisfy f(x) ≥ g(x) on the interval [a, b], then the area between them is:

Area = ∫(from a to b) [f(x) − g(x)] dx

This works because:

∫ₐᵇ f(x) dx gives the area under the upper curve.
∫ₐᵇ g(x) dx gives the area under the lower curve.
Subtracting gives the area between them.

Key Point: Always subtract the lower curve from the upper curve. If you get this the wrong way round, the answer will be negative (just take the absolute value, but it is better practice to identify which curve is on top).

Step-by-Step Method

Find the intersection points by solving f(x) = g(x). These give the limits of integration.
Determine which curve is on top in each interval (substitute a test value).
Set up and evaluate the integral ∫ₐᵇ [upper − lower] dx.

Example 1: Area Between a Line and a Curve

Find the area enclosed between y = x² and y = 2x.

Step 1: Find intersections:

x² = 2x
x² − 2x = 0
x(x − 2) = 0
x = 0, x = 2

Step 2: Which is on top? Test x = 1: y = 1 (parabola) vs y = 2 (line). The line is on top.

Step 3:

Area = ∫₀² (2x − x²) dx
= [x² − x³/3]₀²
= (4 − 8/3) − 0
= 12/3 − 8/3
= 4/3

The area is 4/3 square units.

Example 2: Area Between Two Curves

Find the area enclosed between y = x² and y = 4x − x².

Step 1: Find intersections:

x² = 4x − x²
2x² − 4x = 0
2x(x − 2) = 0
x = 0, x = 2

Step 2: Test x = 1: y = 1 (first curve) vs y = 4 − 1 = 3 (second curve). The second curve is on top.

Step 3:

Area = ∫₀² [(4x − x²) − x²] dx
= ∫₀² (4x − 2x²) dx
= [2x² − 2x³/3]₀²
= (8 − 16/3) − 0
= 24/3 − 16/3
= 8/3

The area is 8/3 square units.

Example 3: Area Between Curves That Cross

Find the total area enclosed between y = x³ and y = x for −1 ≤ x ≤ 1.

Step 1: Find intersections:

x³ = x
x³ − x = 0
x(x² − 1) = 0
x = −1, 0, 1

Step 2: The curves cross at x = 0, so we split into two intervals.

For −1 < x < 0: test x = −0.5. y = −0.125 (cubic) vs y = −0.5 (line). The cubic is above the line.

For 0 < x < 1: test x = 0.5. y = 0.125 (cubic) vs y = 0.5 (line). The line is above the cubic.

Step 3:

Area = ∫₋₁⁰ (x³ − x) dx + ∫₀¹ (x − x³) dx

First integral:

∫₋₁⁰ (x³ − x) dx = [x⁴/4 − x²/2]₋₁⁰
= 0 − (1/4 − 1/2)
= 1/4

Second integral:

∫₀¹ (x − x³) dx = [x²/2 − x⁴/4]₀¹
= (1/2 − 1/4) − 0
= 1/4

Total area = 1/4 + 1/4 = 1/2 square units.

Example 4: Area Between a Curve and the x-axis (as a Special Case)

The area between y = f(x) and the x-axis is a special case of the area between two curves, where g(x) = 0 (the x-axis).

Area = ∫ₐᵇ |f(x)| dx

This reinforces the need to split at roots when f(x) changes sign.

Example 5: Area Between a Parabola and a Line (Fully Worked)

Find the area enclosed between y = 6x − x² and y = x.

Step 1: Find intersections:

6x − x² = x
5x − x² = 0
x(5 − x) = 0
x = 0, x = 5

Step 2: Test x = 2: y = 12 − 4 = 8 (parabola) vs y = 2 (line). Parabola is on top.

Step 3:

Area = ∫₀⁵ [(6x − x²) − x] dx
= ∫₀⁵ (5x − x²) dx
= [5x²/2 − x³/3]₀⁵
= (125/2 − 125/3) − 0
= 375/6 − 250/6
= 125/6

The area is 125/6 square units (approximately 20.83).

Example 6: Using Symmetry

Find the area enclosed between y = x² − 1 and y = 1 − x².

Step 1: Find intersections:

x² − 1 = 1 − x²
2x² = 2
x² = 1
x = ±1

Step 2: Test x = 0: y = −1 (first curve) vs y = 1 (second curve). The second curve is on top.

Step 3: By symmetry (both curves are even functions), the area from −1 to 1 is twice the area from 0 to 1:

Area = 2 ∫₀¹ [(1 − x²) − (x² − 1)] dx
= 2 ∫₀¹ (2 − 2x²) dx
= 2 [2x − 2x³/3]₀¹
= 2 (2 − 2/3)
= 2 × 4/3
= 8/3

The area is 8/3 square units.

Three or More Intersection Points

When curves intersect at three or more points, the area between them may consist of multiple separate regions. You must set up a separate integral for each region and determine which curve is on top in each region.

Example 7

Find the total area enclosed between y = x³ − 3x and y = x.

Intersections:

x³ − 3x = x
x³ − 4x = 0
x(x² − 4) = 0
x = −2, 0, 2

Region 1: [−2, 0]

Test x = −1: cubic = −1 + 3 = 2, line = −1. Cubic is on top.

∫₋₂⁰ [(x³ − 3x) − x] dx = ∫₋₂⁰ (x³ − 4x) dx
= [x⁴/4 − 2x²]₋₂⁰ = 0 − (4 − 8) = 4

Region 2: [0, 2]

Test x = 1: cubic = 1 − 3 = −2, line = 1. Line is on top.

∫₀² [x − (x³ − 3x)] dx = ∫₀² (4x − x³) dx
= [2x² − x⁴/4]₀² = (8 − 4) − 0 = 4

Total area = 4 + 4 = 8 square units.

Summary

The area between y = f(x) and y = g(x) from a to b is ∫ₐᵇ |f(x) − g(x)| dx.
Find intersection points to determine limits of integration.
Identify which curve is on top in each interval.
If the curves cross between the limits, split into separate integrals.
Use symmetry where possible to simplify calculations.

Exam Tip: Always start by finding the intersection points — these are your limits. Then test a value in each interval to determine which curve is on top. Set up your integral as ∫(upper − lower). A very common error is to integrate (f − g) without checking that f is actually above g throughout the interval — if they swap, you must split the integral.

A-Level Deep Dive: Areas Between Curves

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section H: Integration. Areas between curves sit within the integration sub-strand, which covers evaluating definite integrals and using definite integrals to find areas under and between curves (refer to the official AQA 7357 specification document for exact wording). The formula $A = \int_a^b [f(x) - g(x)]\,dx$ , where $f(x) \geq g(x)$ on $[a, b]$ , is the engine of the topic. The limits $a$ and $b$ are the $x$ -coordinates of the intersection points, found by solving $f(x) = g(x)$ algebraically. Although examined in Paper 1 directly, the technique is reused in Paper 2 (Mechanics, work-done as area under a force-displacement curve) and Paper 3 (Statistics, probability density functions, where $P(a \leq X \leq b) = \int_a^b f(x)\,dx$ ). The AQA formula booklet does not list the area-between-curves formula — it must be derived from the standard area integral.

Worked example with full mark scheme

Question (8 marks):

The curve $C_1$ has equation $y = 6x - x^2$ and the line $C_2$ has equation $y = x + 4$ .

(a) Find the coordinates of the points where $C_1$ and $C_2$ intersect. (3)

(b) Find the exact area of the region enclosed between $C_1$ and $C_2$ . (5)

Solution with mark scheme:

(a) Step 1 — set the equations equal.

$6x - x^2 = x + 4$

Rearranging: $x^2 - 5x + 4 = 0$ .

M1 — equating $y$ -values and rearranging to a quadratic in standard form. Common error: a sign slip on the constant. Always collect to the side that keeps $x^2$ positive.

Step 2 — factorise.

$(x - 1)(x - 4) = 0$ , so $x = 1$ or $x = 4$ .

M1 — correct factorisation or quadratic-formula application.

Step 3 — find $y$ -coordinates.

Substituting back into $y = x + 4$ (the simpler equation): at $x = 1$ , $y = 5$ ; at $x = 4$ , $y = 8$ .

A1 — both intersection points stated as coordinates: $(1, 5)$ and $(4, 8)$ .

(b) Step 1 — identify upper and lower curve.

On $[1, 4]$ , test $x = 2$ : parabola gives $6(2) - 4 = 8$ , line gives $2 + 4 = 6$ . So the parabola is the upper curve on this interval.

B1 — explicit identification of upper curve, with justification (substituting an interior point is the standard method).

Step 2 — set up the integral.

$A = \int_1^4 \left[(6x - x^2) - (x + 4)\right]\,dx = \int_1^4 (5x - x^2 - 4)\,dx$

M1 — correct subtraction $\text{upper} - \text{lower}$ and limits matching the intersection $x$ -values.

Step 3 — integrate.

$\int (5x - x^2 - 4)\,dx = \dfrac{5x^2}{2} - \dfrac{x^3}{3} - 4x + C$

M1 — correct term-by-term integration of all three terms.

Step 4 — evaluate at limits.

At $x = 4$ : $\dfrac{5 \cdot 16}{2} - \dfrac{64}{3} - 16 = 40 - \dfrac{64}{3} - 16 = 24 - \dfrac{64}{3} = \dfrac{72 - 64}{3} = \dfrac{8}{3}$ .

At $x = 1$ : $\dfrac{5}{2} - \dfrac{1}{3} - 4 = \dfrac{15 - 2 - 24}{6} = -\dfrac{11}{6}$ .

Subtract: $\dfrac{8}{3} - \left(-\dfrac{11}{6}\right) = \dfrac{16}{6} + \dfrac{11}{6} = \dfrac{27}{6} = \dfrac{9}{2}$ .

A1 — exact answer $A = \dfrac{9}{2}$ .

Total: 8 marks (M5 A2 B1).

Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): The curves $y = x^2$ and $y = 2x - x^2$ intersect at two points.

(a) Find the coordinates of both intersection points. (2)

(b) Find the exact area of the region enclosed between the two curves. (4)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — setting $x^2 = 2x - x^2$ and rearranging to $2x^2 - 2x = 0$ , i.e. $2x(x - 1) = 0$ .
A1 (AO1.1b) — coordinates $(0, 0)$ and $(1, 1)$ .

(b)

M1 (AO1.1b) — testing an interior point (say $x = \tfrac{1}{2}$ ) to identify upper curve: $2(0.5) - 0.25 = 0.75$ vs $0.25$ , so $y = 2x - x^2$ is upper.
M1 (AO1.1b) — setting up $A = \int_0^1 [(2x - x^2) - x^2]\,dx = \int_0^1 (2x - 2x^2)\,dx$ .
A1 (AO1.1b) — antiderivative $x^2 - \dfrac{2x^3}{3}$ .
A1 (AO2.5) — evaluation: $\left[1 - \dfrac{2}{3}\right] - 0 = \dfrac{1}{3}$ .

Total: 6 marks split AO1 = 5, AO2 = 1. Areas-between-curves questions are AO1-dominated, with AO2 marks reserved for the simplification step or for justifying which curve is upper.

Synoptic links

Connects to:

Definite integration (Section H earlier sub-strands): the area-between-curves formula is the definite integral $\int_a^b [f(x) - g(x)]\,dx$ , applied to a difference of functions. All standard integration techniques (power rule, integration of trigonometric functions, integration by substitution) carry over directly.
Finding intersection points algebraically (Section B — Algebra and functions): every area-between-curves problem starts with solving $f(x) = g(x)$ . This may produce a linear equation, a quadratic (most commonly), a cubic, or a transcendental equation requiring numerical methods. Confidence with quadratic factorisation, the quadratic formula, and polynomial division is prerequisite.
Coordinate geometry of curves (Section G): sketching the curves before integrating clarifies which is upper. Recognising standard shapes — parabolas, cubics, reciprocal curves, sinusoids — and marking turning points and intersections converts an abstract integral into a concrete picture.
Centroid and centre of mass (Mechanics applications, Section M): the $x$ -coordinate of the centroid of a region between two curves is $\bar{x} = \dfrac{\int_a^b x[f(x) - g(x)]\,dx}{\int_a^b [f(x) - g(x)]\,dx}$ . The denominator is the area; the numerator is its first moment. Areas-between-curves is the foundational integral underneath all centroid calculations.
Modelling (Sections N, O): real-world contexts include "the area between supply and demand curves represents consumer plus producer surplus" (economics), "the area between velocity-time graphs of two vehicles is the relative displacement" (kinematics), and "the area between probability density functions measures distributional difference" (statistics). Each context dresses the same integral in different language.

Mark-scheme literacy

Areas-between-curves questions on 7357 split AO marks heavily toward AO1:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	70–80%	Solving the simultaneous equations, identifying upper curve, setting up the integral, integrating term by term, evaluating at limits
AO2 (reasoning / interpretation)	15–25%	Justifying which curve is upper, splitting the integral when curves cross within the region, presenting answers in the requested exact form
AO3 (problem-solving)	0–10%	Open-ended modelling appears in Year 2 within trickier composite-region or sideways-integration contexts

Areas Between Curves

Areas Between Curves

The Key Principle

Step-by-Step Method

Example 1: Area Between a Line and a Curve

Example 2: Area Between Two Curves

Example 3: Area Between Curves That Cross

Example 4: Area Between a Curve and the x-axis (as a Special Case)

Example 5: Area Between a Parabola and a Line (Fully Worked)

Example 6: Using Symmetry

Three or More Intersection Points

Example 7

Summary

A-Level Deep Dive: Areas Between Curves

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 1 format

Synoptic links

Mark-scheme literacy

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