A definite integral gives signed area: positive for parts of the curve above the x-axis and negative for parts below. To find an actual area, find where the curve crosses the axis, integrate each part separately and add the magnitudes. Area between a curve and the y-axis is found by integrating x with respect to y, and the area under a rate graph gives an accumulated quantity such as distance.
Area under a curve
- Above the x-axis: area = ∫ y dx from a to b
- Below the x-axis: area = −∫ y dx from a to b
- Crossing the axis: split at each root and add the magnitudes
- Area to the y-axis: ∫ x dy from c to d, with x written in terms of y
- Distance = ∫ speed dt
Signed area and regions below the axis
When a curve lies above the x-axis, the definite integral equals the area. When it lies below, the integral comes out negative, and the area is the magnitude of that value.
An answer of 0 does not mean there is no area. For a function like y = x³ from −2 to 2, the parts above and below the axis cancel exactly.
Curves that cross the x-axis
If a curve crosses the axis between the limits, a single integral gives the net value, with positive and negative parts partly cancelling. For the total area, find the roots, split the interval at each one, integrate each part separately, and add the magnitudes. Integrating straight through is the most common way to lose marks here.
Area between a curve and the y-axis
To find the area between a curve and the y-axis, integrate x with respect to y. Rearrange the equation to make x the subject, such as y = x² becoming x = √y, and use the y-values as limits, not x-values.
Accumulation: area under a rate graph
The area under a rate graph is the total amount accumulated over that time. The area under a speed–time graph is distance travelled; the area under a flow rate graph is volume. If a velocity changes sign, total distance needs the split-and-add method, while displacement does not.
Working backwards to a limit
Some questions give the area and ask for an unknown limit. Integrate with the unknown as a limit, set the result equal to the given area, and solve. For example, ∫ 3x² dx from 0 to k = k³, so an area of 8 gives k = 2.
Worked examples
Example 1. Find the area between y = x² − 4 and the x-axis, between the roots.
- The roots are x = ±2, and the curve is below the axis between them.
- ∫ (x² − 4) dx from −2 to 2 = [x³/3 − 4x] from −2 to 2 = (8/3 − 8) − (−8/3 + 8) = −32/3.
- The region is below the axis, so the area is the magnitude.
Answer: 32/3 square units.
Marks: M1 for recognising the region is below the axis, A1 for 32/3.
Example 2. The curve y = x² − 2x crosses the x-axis at x = 2. Find the total area between the curve and the axis from x = 0 to x = 3.
- Split at the root x = 2.
- ∫ (x² − 2x) dx from 0 to 2 = [x³/3 − x²] from 0 to 2 = 8/3 − 4 = −4/3, so that part has area 4/3.
- ∫ (x² − 2x) dx from 2 to 3 = (9 − 9) − (8/3 − 4) = 4/3.
Answer: 4/3 + 4/3 = 8/3 square units. Integrating straight from 0 to 3 would wrongly give 0.
Marks: M1 for splitting at x = 2, A1 for the two parts, A1 for the total.
Example 3. Find the area between y = x² (x ≥ 0) and the y-axis, from y = 0 to y = 1.
- Rearrange: x = √y = y^(1/2).
- Area = ∫ y^(1/2) dy from 0 to 1 = [⅔y^(3/2)] from 0 to 1.
Answer: ⅔ square units.
Marks: M1 for x = √y, M1 for integrating with respect to y, A1 for ⅔.
Example 4. A particle moves with speed v = 6t m/s. Find the distance travelled in the first 3 seconds.
- Distance = ∫ 6t dt from 0 to 3 = [3t²] from 0 to 3.
- = 27 − 0.
Answer: 27 m.
Marks: M1 for the integral, A1 for 27 m.
Example 5. The area under y = 3x² from x = 0 to x = k is 8. Find k.
- ∫ 3x² dx from 0 to k = [x³] from 0 to k = k³.
- k³ = 8.
Answer: k = 2.
Marks: M1 for k³, A1 for k = 2.
Common mistakes
- Integrating straight across a root. Positive and negative parts cancel. Split at each root first.
- Giving a negative area. Area is always positive; take the magnitude of a negative integral.
- Using x-limits for a y-axis region. When integrating with respect to y, the limits are y-values.
- Not rearranging for x. Write x in terms of y before integrating with respect to y.
- Confusing distance with displacement. Distance needs the split-and-add method if the velocity changes sign.
How area under a curve is examined
Signed area and splitting at roots come up regularly on WMA12 (Pure 2, Edexcel International A-Level) and in Year 1 of Edexcel A-Level (9MA0). You will typically be asked to:
- find where a curve meets the x-axis, then the area of a region
- explain why an integral is not equal to an area
- find the total area of a region that crosses the axis
- find an area between a curve and the y-axis
Questions like these, with full mark schemes and diagrams, are among the Pure 2 integration exam questions. There is also the Pure 2 integration worksheet, and the revision engine for A-Level integration gives unlimited practice.
This follows on from definite integrals. When you cannot integrate exactly, the trapezium rule estimates the area.
Frequently asked questions
Why is the integral negative when the curve is below the x-axis?
Because y is negative there, so each thin strip contributes a negative amount. The area itself is the magnitude.
How do you find the total area when a curve crosses the axis?
Find the roots, integrate each section separately, make each result positive, and add them.
How do you find the area between a curve and the y-axis?
Rearrange the equation for x in terms of y, then integrate x with respect to y between the y-limits.