The area enclosed between two graphs is the integral of the upper graph minus the lower graph, between the x-values where they meet. Find the intersections first to set the limits, check which graph is on top, and if the graphs cross inside the region, split at the crossing and add the areas.
Area between two graphs
- Area = ∫ (y upper − y lower) dx from a to b
- Limits: solve the two equations simultaneously
- Graphs crossing inside: split at each crossing and add the areas
- Equivalent method: area under the upper graph − area under the lower graph
How do you find the area between two curves?
Find every point where the graphs meet; these give the limits. Decide which graph is on top between them, by testing a point in the middle rather than assuming. Then integrate the upper graph minus the lower graph. Both boundaries can be curves, or one can be a straight line.
When the graphs cross inside the region
If two graphs meet at three points, they enclose two separate regions, and the upper graph swaps over at the middle crossing. Integrate each region separately with the correct order, and add the two areas. Integrating straight through would let the two signed pieces cancel.
Look for symmetry: a cubic and a line through the origin often enclose two regions of equal area.
The subtract-areas method
Sometimes it is easier to find the area under each graph separately and subtract: area between = area under the upper graph − area under the lower graph. This is the same calculation as integrating the difference, split into two integrals. Use whichever keeps the algebra tidy.
Regions with three boundaries
Some regions are bounded by a curve, a straight line and an axis, so the upper boundary changes partway along. Find where it changes, split the x-range there, and work out each part separately. A part bounded by a straight line is often just a triangle or trapezium, which needs no integration.
Exact answers
These questions usually have exact answers, as fractions or surds, so avoid rounding. Always confirm which graph is on top in each part before integrating.
Worked examples
Example 1. Find the area enclosed between y = x² and y = 2x.
- x² = 2x gives x = 0 or x = 2.
- Between them 2x is on top, so the area is ∫ (2x − x²) dx from 0 to 2.
- [x² − x³/3] from 0 to 2 = 4 − 8/3.
Answer: 4/3 square units.
Marks: M1 for the limits, M1 for the integral, A1 for 4/3.
Example 2. The curves y = x³ and y = 16x meet at x = −4, 0 and 4. Find the total area they enclose.
- On 0 < x < 4, 16x is above x³: ∫ (16x − x³) dx from 0 to 4 = [8x² − x⁴/4] from 0 to 4 = 128 − 64 = 64.
- By symmetry, the region for −4 < x < 0 also has area 64.
Answer: 128 square units.
Marks: M1 for splitting at x = 0, M1 for one region, A1 for 128.
Example 3. Find the area between y = x² + 4 and y = x from x = 0 to x = 3.
- Area = ∫ (x² + 4) dx − ∫ x dx, both from 0 to 3, which is ∫ (x² + 4 − x) dx.
- [x³/3 + 4x − x²/2] from 0 to 3 = 9 + 12 − 9/2.
Answer: 33/2 square units.
Marks: M1 for the difference of integrals, A1 for 33/2.
Example 4. A region is bounded by y = x² from x = 0 to x = 1, then the line from (1, 1) to (2, 0), and the x-axis. Find its area.
- Curve part: ∫ x² dx from 0 to 1 = 1/3.
- Line part: a triangle with base 1 and height 1, area 1/2.
Answer: 1/3 + 1/2 = 5/6 square units.
Marks: M1 for the curve area, M1 for the triangle, A1 for 5/6.
Example 5. Find the area enclosed between y = 4 − x² and y = x² − 4.
- 4 − x² = x² − 4 gives x = ±2.
- Area = ∫ ((4 − x²) − (x² − 4)) dx from −2 to 2 = ∫ (8 − 2x²) dx from −2 to 2.
- [8x − 2x³/3] from −2 to 2 = (16 − 16/3) − (−16 + 16/3).
Answer: 64/3 square units.
Marks: M1 for the limits, A1 for 64/3.
Common mistakes
- Assuming which graph is on top. Test a point between the limits.
- Integrating through a crossing point. Split where the graphs cross and add the separate areas.
- Not finding all intersections. A cubic and a line can meet three times; you need every crossing.
- Integrating a straight-line part unnecessarily. A triangle or trapezium can be found with a simple area formula.
- Rounding exact answers. Give fractions or surds unless a decimal is asked for.
How area between curves is examined
Areas between graphs are tested on WMA12, the Pure 2 paper for Edexcel International A-Level, and in Year 1 of Edexcel A-Level (9MA0). Typical tasks:
- find the coordinates of the points where two graphs meet
- find the area of a shaded region between a curve and a line
- show that an enclosed area equals a given exact value
- find the total area when graphs meet at three points
Questions with diagrams and mark schemes are in the Pure 2 integration exam questions. Try the integration worksheet too, or generate new regions in the integration revision engine.
Make sure you are confident with definite integrals and area under a curve first.
Frequently asked questions
How do you find the area between two curves?
Find where they intersect, then integrate (upper − lower) between those x-values.
What if the curves cross between the limits?
Split the region at each crossing, integrate each part with the correct graph on top, and add the areas.
Can I subtract the two areas instead?
Yes. The area under the upper graph minus the area under the lower graph gives the same result as integrating the difference.