The area between two curves is the integral of the top curve minus the bottom curve, taken between the x-values where they meet. The steps are: find the intersections, decide which curve is on top, then integrate the difference.
This lesson is part of SPM Additional Mathematics integration. It extends finding area between a curve and an axis.
Worked example: a line and a parabola
Find the area enclosed by y = x² and y = 2x + 3.
Step 1: intersection points. Set x² = 2x + 3, so x² − 2x − 3 = 0, which factorises as (x − 3)(x + 1) = 0. The limits are x = −1 and x = 3.
Step 2: which is on top. Test x = 0. The line gives y = 3 and the parabola gives y = 0, so the line is on top.
Step 3: integrate top minus bottom.
Area = ∫ (2x + 3 − x²) dx from −1 to 3 = [x² + 3x − x³/3] from −1 to 3.
At x = 3: 9 + 9 − 9 = 9. At x = −1: 1 − 3 + 1/3 = −5/3.
Area = 9 − (−5/3) = 9 + 5/3 = 32/3 square units.
The mistake that costs marks
The slip is to write the difference in the wrong order, or to use limits that come from the axis instead of the intersections. Here, using ∫ (x² − (2x + 3)) dx from −1 to 3 gives −32/3, which is negative and cannot be an area.
Another version is to integrate from 0 to 3 because “areas start at 0”.
| Step | Wrong | Right |
|---|---|---|
| Limits | 0 and 3 | −1 and 3, from x² = 2x + 3 |
| Integrand | x² − (2x + 3) | (2x + 3) − x² |
| Result | −32/3 or an incomplete area | 32/3 |
A test value settles which curve is on top before you integrate, and it takes ten seconds.
When one curve dips below the axis
The top-minus-bottom rule still holds, because subtracting a negative adds its size. You do not need to split the integral at the axis unless the top and bottom curves swap places inside the interval.
If the curves cross each other between the limits, split at that crossing, because the top curve changes there.
A method you can reuse
- Sketch both curves, or at least the intersection points.
- Solve the two equations simultaneously for the limits.
- Use a test value to decide the top curve.
- Integrate top minus bottom between the limits.
- Check the answer is positive and write the units.
You can view the region and the integrand in the polynomial integration and area explorer.
Check yourself
Find the area enclosed by y = x² and y = x + 2.
Answer
Set x² = x + 2, so x² − x − 2 = 0 and (x − 2)(x + 1) = 0. The limits are x = −1 and x = 2.
At x = 0 the line gives 2 and the parabola gives 0, so the line is on top.
Area = [x²/2 + 2x − x³/3] from −1 to 2. At x = 2: 2 + 4 − 8/3 = 10/3. At x = −1: 1/2 − 2 + 1/3 = −7/6.
Area = 10/3 + 7/6 = 20/6 + 7/6 = 27/6 = 9/2 square units.
What to study next
Rotate a region to make a solid. Continue with calculating volumes of revolution within syllabus scope, then try the integration practice set.
For a teacher to set up your regions with you, see online one-to-one Additional Mathematics tuition. The mistake log helps you see which step repeats.