Area Between Curves

This lesson covers finding the area between two curves — a key application of definite integration required by the Edexcel A-Level Mathematics specification (9MA0). You need to identify which curve is on top, set up the correct integral, and handle regions that may need to be split.

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The Basic Principle

To find the area enclosed between two curves y = f(x) and y = g(x) from x = a to x = b, where f(x) ≥ g(x) throughout the interval:

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

In words: integrate the top curve minus the bottom curve.

This works because:

• ∫ from a to b of f(x) dx gives the area under f(x) down to the x-axis
• ∫ from a to b of g(x) dx gives the area under g(x) down to the x-axis
• Subtracting removes the unwanted region below g(x)

Key Point: You must always subtract the lower function from the upper function to get a positive area.

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Finding the Limits of Integration

The limits a and b are usually the x-coordinates where the two curves intersect. To find these:

1. Set f(x) = g(x)
2. Solve for x
3. The solutions give you the limits of integration

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

Step 1: Find intersection points. x² = 2x → x² - 2x = 0 → x(x - 2) = 0 → x = 0 or x = 2

Step 2: Determine which curve is on top for 0 ≤ x ≤ 2. At x = 1: y = 1² = 1 and y = 2(1) = 2. Since 2 > 1, the line y = 2x is on top.

Step 3: Set up and evaluate the integral. Area = ∫ from 0 to 2 of (2x - x²) dx = [x² - x³/3] from 0 to 2 = (4 - 8/3) - (0) = 12/3 - 8/3 = 4/3

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How to Determine Which Curve Is on Top

There are several methods:

Method 1: Substitute a Test Value

Choose any x-value between the limits and evaluate both functions. The larger value is the upper curve in that region.

Method 2: Sketch the Curves

A rough sketch helps you visualise which curve is above the other. This is especially useful when the curves swap positions.

Method 3: Evaluate f(x) - g(x)

If f(x) - g(x) > 0, then f(x) is above g(x). If f(x) - g(x) < 0, then g(x) is above f(x).

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Worked Examples

Example 1: Area Between a Quadratic and a Line

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 or x = 5

Step 2: Determine which is on top (test x = 1). Curve: y = 6(1) - 1 = 5. Line: y = 1. The curve is on top.

Step 3: Integrate. Area = ∫ from 0 to 5 of [(6x - x²) - x] dx = ∫ from 0 to 5 of (5x - x²) dx = [5x²/2 - x³/3] from 0 to 5 = (125/2 - 125/3) - 0 = 375/6 - 250/6 = 125/6

Example 2: Area Between Two Quadratics

Find the area enclosed between y = x² + 2 and y = 4x - x² + 2.

Step 1: Find intersections. x² + 2 = 4x - x² + 2 → 2x² - 4x = 0 → 2x(x - 2) = 0 → x = 0 or x = 2

Step 2: Determine which is on top (test x = 1). First curve: 1 + 2 = 3. Second curve: 4 - 1 + 2 = 5. Second is on top.

Step 3: Integrate. Area = ∫ from 0 to 2 of [(4x - x² + 2) - (x² + 2)] dx = ∫ from 0 to 2 of (4x - 2x²) dx = [2x² - 2x³/3] from 0 to 2 = (8 - 16/3) - 0 = 24/3 - 16/3 = 8/3

Example 3: Area Between a Curve and a Line (Given Limits)

Find the area between y = x³ and y = 4x for 0 ≤ x ≤ 2.

Step 1: Check which is on top (test x = 1). y = 1³ = 1 and y = 4(1) = 4. The line is on top.

Step 2: Integrate. Area = ∫ from 0 to 2 of (4x - x³) dx = [2x² - x⁴/4] from 0 to 2 = (8 - 4) - 0 = 4

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Splitting Into Regions

When the curves swap (i.e., the one that was on top becomes the bottom), you must split the integral into separate regions.

Example: Find the total area enclosed between y = x³ and y = x for -1 ≤ x ≤ 1.

Step 1: Find intersections in the interval. x³ = x → x³ - x = 0 → x(x² - 1) = 0 → x = -1, 0, or 1

Step 2: Determine which is on top in each region.

• For -1 ≤ x ≤ 0 (test x = -0.5): x³ = -0.125, x = -0.5. Since -0.125 > -0.5, y = x³ is on top.
• For 0 ≤ x ≤ 1 (test x = 0.5): x³ = 0.125, x = 0.5. Since 0.5 > 0.125, y = x is on top.

Step 3: Calculate each region. Region 1: ∫ from -1 to 0 of (x³ - x) dx = [x⁴/4 - x²/2] from -1 to 0 = (0) - (1/4 - 1/2) = 1/4

Region 2: ∫ from 0 to 1 of (x - x³) dx = [x²/2 - x⁴/4] from 0 to 1 = (1/2 - 1/4) - 0 = 1/4

Total area = 1/4 + 1/4 = 1/2

Exam Tip: If you do not split at the points where the curves cross, the positive and negative signed areas may partially cancel, giving an incorrect answer.

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Area Between a Curve and the y-axis

Sometimes you need the area between a curve and the y-axis. In this case, integrate with respect to y.

Area = ∫ from c to d of x dy

You need to express x as a function of y.

Example: Find the area between the curve x = y² and the y-axis from y = 0 to y = 3.

Area = ∫ from 0 to 3 of y² dy = [y³/3] from 0 to 3 = 27/3 = 9

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Common Mistakes

1. Subtracting in the wrong order — Always check which function is greater in the interval. If you get a negative answer for an area, you have the subtraction the wrong way round.
2. Not finding the correct limits — Make sure you solve the intersection equation correctly.
3. Not splitting when curves cross — If the curves swap which is on top, you must split the integral.
4. Confusing area with the value of the integral — The integral can be negative; the area is always positive.

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Summary

• The area between curves y = f(x) and y = g(x) is ∫ from a to b of [f(x) - g(x)] dx, where f(x) ≥ g(x).
• Find the limits by solving f(x) = g(x).
• Always check which curve is on top using a test point or sketch.
• Split the integral at any points where the curves cross within the interval.
• For areas between a curve and the y-axis, express x in terms of y and integrate with respect to y.

A-Level Deep Dive: Area Between Curves

Spec mapping

Edexcel 9MA0 specification section 10 — Integration covers integration to find the area between two curves (refer to the official specification document for exact wording). This sub-strand sits inside Paper 1 (Pure Mathematics) but feeds Paper 2 (Pure) wherever modelling questions ask for the size of a region, and Paper 3 (Statistics and Mechanics) whenever a velocity-time region encodes a displacement difference. The key formula $\text{Area} = \int_a^b \big[\,f(x) - g(x)\,\big]\,dx$Area=∫ab​[f(x)−g(x)]dx for $f(x) \geq g(x)$f(x)≥g(x) on $[a, b]$[a,b] is not in the Edexcel formula booklet — only the basic power rule for integration is given. The intersection points $a$a and $b$b must be solved for, and the order of subtraction (upper minus lower) must be justified.

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Worked example with full mark scheme

Question (8 marks): Find the exact area of the finite region enclosed by the curve $y = x^2 - 4x + 3$y=x2−4x+3 and the line $y = x - 1$y=x−1.

Solution with mark scheme:

Step 1 — find intersection points.

Set the curve equal to the line:

$x^2 - 4x + 3 = x - 1$x2−4x+3=x−1 $x^2 - 5x + 4 = 0$x2−5x+4=0 $(x - 1)(x - 4) = 0$(x−1)(x−4)=0

So $x = 1$x=1 and $x = 4$x=4 are the limits of integration.

M1 — equating the two expressions and rearranging to a quadratic in $x$x. A1 — both intersection points correct: $x = 1$x=1 and $x = 4$x=4.

Step 2 — identify upper and lower curves on $[1, 4]$[1,4].

Test $x = 2$x=2 (midpoint): line gives $y = 1$y=1; curve gives $y = 4 - 8 + 3 = -1$y=4−8+3=−1. So the line lies above the curve on $[1, 4]$[1,4], hence upper = $x - 1$x−1 and lower = $x^2 - 4x + 3$x2−4x+3.

B1 — explicit identification of upper and lower (testing a midpoint or sketching counts; an unjustified statement does not).

Step 3 — set up the integral.

$\text{Area} = \int_1^4 \big[\,(x - 1) - (x^2 - 4x + 3)\,\big]\,dx = \int_1^4 \big(-x^2 + 5x - 4\big)\,dx$Area=∫14​[(x−1)−(x2−4x+3)]dx=∫14​(−x2+5x−4)dx

M1 — integral set up with upper minus lower and correct limits.

Step 4 — integrate.

$\int \big(-x^2 + 5x - 4\big)\,dx = -\dfrac{x^3}{3} + \dfrac{5x^2}{2} - 4x \,(+ C)$∫(−x2+5x−4)dx=−3x3​+25x2​−4x(+C)

M1 — power rule applied to each term, at least two terms correct. A1 — fully correct antiderivative.

Step 5 — evaluate.

At $x = 4$x=4: $-\dfrac{64}{3} + \dfrac{5 \cdot 16}{2} - 16 = -\dfrac{64}{3} + 40 - 16 = -\dfrac{64}{3} + 24 = \dfrac{-64 + 72}{3} = \dfrac{8}{3}$−364​+25⋅16​−16=−364​+40−16=−364​+24=3−64+72​=38​.

At $x = 1$x=1: $-\dfrac{1}{3} + \dfrac{5}{2} - 4 = -\dfrac{1}{3} - \dfrac{3}{2} = \dfrac{-2 - 9}{6} = -\dfrac{11}{6}$−31​+25​−4=−31​−23​=6−2−9​=−611​.

Difference: $\dfrac{8}{3} - \left(-\dfrac{11}{6}\right) = \dfrac{16}{6} + \dfrac{11}{6} = \dfrac{27}{6} = \dfrac{9}{2}$38​−(−611​)=616​+611​=627​=29​.

M1 — both endpoints substituted. A1 — exact area $\dfrac{9}{2}$29​ (square units).

Total: 8 marks (B1 M4 A3, split as shown).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): The curve $C$C has equation $y = 6x - x^2$y=6x−x2 and the line $\ell$ℓ has equation $y = 2x$y=2x.

(a) Find the coordinates of the points where $C$C and $\ell$ℓ intersect. (2)

(b) Find the exact area of the finite region enclosed between $C$C and $\ell$ℓ. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — equating $6x - x^2 = 2x$6x−x2=2x, rearranging to $x^2 - 4x = 0$x2−4x=0, factorising $x(x - 4) = 0$x(x−4)=0.
• A1 (AO1.1b) — intersection points $(0, 0)$(0,0) and $(4, 8)$(4,8).

(b)

• M1 (AO1.1b) — set up integral $\int_0^4 \big[(6x - x^2) - 2x\big]\,dx = \int_0^4 (4x - x^2)\,dx$∫04​[(6x−x2)−2x]dx=∫04​(4x−x2)dx, with the curve correctly identified as the upper boundary on $[0, 4]$[0,4].
• M1 (AO1.1b) — integrate to $2x^2 - \dfrac{x^3}{3}$2x2−3x3​.
• A1 (AO1.1b) — substitute limits: $\left[2(16) - \dfrac{64}{3}\right] - 0 = 32 - \dfrac{64}{3} = \dfrac{32}{3}$[2(16)−364​]−0=32−364​=332​.
• A1 (AO2.5) — exact answer $\dfrac{32}{3}$332​ (square units), presentation matching the "exact" instruction.

Total: 6 marks split AO1 = 5, AO2 = 1. Edexcel reserves the AO2 mark for the final exact-form presentation; a decimal answer (10.67) loses it even when arithmetically correct.

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Synoptic links

1. Section 7 — Differentiation: finding stationary points of the difference function $h(x) = f(x) - g(x)$h(x)=f(x)−g(x) locates where the gap between the curves is maximised, a common follow-up part to area questions ("find the maximum vertical separation").
2. Section 2 — Algebra and functions: intersection points are roots of $f(x) - g(x) = 0$f(x)−g(x)=0, so factorisation, the discriminant, and the quadratic formula are prerequisites. A cubic difference may need the factor theorem.
3. Section 5 — Trigonometry: areas between $y = \sin x$y=sinx and $y = \cos x$y=cosx on $[0, 2\pi]$[0,2π] require finding intersection points from $\tan x = 1$tanx=1 and splitting at the crossover.
4. Section 8 — Differentiation as a rate of change / Mechanics: the area between two velocity-time curves is the difference in displacement between two particles — a Paper 3 modelling staple.
5. Section 11 — Numerical methods: when intersection points have no closed form, the trapezium rule (or Simpson's rule, in some specifications) approximates the same area, with the limits found by Newton-Raphson.

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Mark-scheme literacy

Area-between-curves questions on 9MA0 split AO marks roughly: