The area between a curve and the x-axis is the integral of y with respect to x, taken in sections wherever the curve crosses the axis. A section below the axis gives a negative integral, so its positive size is used for the area.
This lesson is part of SPM Additional Mathematics integration. It uses evaluating definite integrals.
Worked example: the curve crosses the axis
Find the area enclosed by the curve y = x² − 4x + 3, the x-axis and the lines x = 0 and x = 3.
Step 1: find where it crosses. y = (x − 1)(x − 3), so the curve meets the x-axis at x = 1 and x = 3. Between 1 and 3 it lies below the axis.
Step 2: integrate each section. F(x) = x³/3 − 2x² + 3x.
F(0) = 0, F(1) = 1/3 − 2 + 3 = 4/3 and F(3) = 9 − 18 + 9 = 0.
- From 0 to 1: F(1) − F(0) = 4/3, above the axis.
- From 1 to 3: F(3) − F(1) = −4/3, below the axis, so the area is 4/3.
Step 3: add the areas. Total area = 4/3 + 4/3 = 8/3 square units.
The mistake that costs marks
The slip is to integrate over the whole interval in one go. Here that gives F(3) − F(0) = 0 − 0 = 0. The positive and negative parts cancel, so the number 0 looks like a result but is not an area.
| Step | Wrong | Right |
|---|---|---|
| Interval | One integral, 0 to 3 | Split at x = 1 |
| Value | 0 | 4/3 and −4/3 |
| Area | 0 | 4/3 + 4/3 = 8/3 |
Whenever the curve crosses the x-axis inside the interval, split the integral there. A quick sketch shows this immediately.
Area with the y-axis
When the region sits against the y-axis, integrate x with respect to y. Find the area bounded by y = x², the y-axis and the line y = 4.
Rewrite as x = √y, which is y^(1/2). Then:
Area = ∫ y^(1/2) dy from 0 to 4 = [(2/3) y^(3/2)] from 0 to 4 = (2/3)(8) − 0 = 16/3 square units.
Check against the x form: the region is the square [0, 2] × [0, 4] (area 8) minus the area under y = x² from 0 to 2, which is 8/3. So 8 − 8/3 = 16/3.
A method you can reuse
- Sketch the curve and mark where it meets the axis.
- Shade the region the question describes.
- Split at every axis crossing inside the limits.
- Integrate each section and take its positive size.
- Add the areas and write the units.
The polynomial integration and area explorer lets you see how the signed integral and the true area differ.
Check yourself
Find the area enclosed by y = 6x − x² and the x-axis.
Answer
y = x(6 − x), so the curve meets the x-axis at x = 0 and x = 6, and lies above the axis between them.
Area = [3x² − x³/3] from 0 to 6 = (108 − 72) − 0 = 36 square units.
Check: the maximum is y = 9 at x = 3, so the region is inside a 6 by 9 rectangle of area 54. The area 36 is exactly two thirds of it, which is correct for a parabola.
What to study next
Two curves need a top minus bottom approach. Continue with finding area between two curves, then test the chapter with the integration practice set.
For a teacher to sketch and set up regions with you, see online one-to-one Additional Mathematics tuition. The word problem structure worksheet helps with longer questions.