Area between a curve and a line

Method

1. Find the x-coordinates where the graphs meet: these are the limits.

2. Decide which graph is on top in between.

3. Area = ∫ (top - bottom) dx.

Example

y = 6x - x² and y = 2x meet where 6x - x² = 2x, so x(4 - x) = 0: x = 0 and x = 4. The curve is on top, so

Area = ∫₀^{4} (4x - x²) dx = 32 - 64/3 = 32/3.

Another way

You can also find the area under the curve and subtract (or add) the area of a triangle or trapezium under the line. Both methods give the same answer.

Worked example

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

1. x² = x + 2 gives x = -1 and x = 2
2. The line is on top
3. ∫_-1^{2} (x + 2 - x²) dx: integrate to x²/2 + 2x - x³/3
4. (2 + 4 - 8/3) - (1/2 - 2 + 1/3) = 10/3 + 7/6

Answer: 9/2

Check yourself

1. Find the area enclosed between y = x² and the line y = 9.
   Show answer

   36. They meet at x = ±3. ∫_-3^{3} (9 - x²) dx. The integral is 9x - x³/3, giving 18 - (-18) = 36.

2. Find the x-coordinates where y = 4x - x² meets y = 2x.
   Show answer

   x = 0 or x = 2. 4x - x² = 2x gives x(2 - x) = 0.

3. Find the area enclosed between y = 4x - x² and y = 2x. Give your answer as a fraction in its simplest form.
   Show answer

   4/3. ∫₀^{2} (4x - x² - 2x) dx = ∫₀^{2} (2x - x²) dx = 4 - 8/3 = 4/3.

4. The line y = x + 2 is above y = x² for -1 < x < 2. Which integral gives the area between them?
   • ∫_-1^{2} (x² - x - 2) dx
   • ∫_-1^{2} (x² + x + 2) dx
   • ∫₀^{2} (x + 2 - x²) dx
   • ∫_-1^{2} (x + 2 - x²) dx
   Show answer

   ∫_-1^{2} (x + 2 - x²) dx. Top minus bottom between the intersection points.

5. Find the area enclosed between y = x² and y = 8 - x². Give your answer as a fraction in its simplest form.
   Show answer

   64/3. They meet where 2x² = 8, so x = ±2. ∫_-2^{2} (8 - 2x²) dx = 32 - 32/3 = 64/3.

6. y = 6x - x² and y = 2x meet at x = 0 and x = 4. The area under the curve from 0 to 4 is 80/3. What is the area of the triangle under the line from 0 to 4?
   Show answer

   16. The line reaches y = 8 at x = 4, so the triangle has area 0.5 x 4 x 8 = 16. (Then 80/3 - 16 = 32/3, matching the integral.)

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3)
• Pearson Edexcel IAL Pure Mathematics 2 scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.