Find total areas by splitting at axis crossings
Locate sign changes, integrate on each interval and add positive magnitudes for a geometric area or total distance.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Section 1.8 includes rational and linear-composite powers except exponent −1, simple improper integrals, areas and volumes about either coordinate axis, including regions away from the rotation axis. Written practice is self-checked.
Before you begin
- Solve a polynomial's zero equation and determine its signs.
- Evaluate definite integrals and interpret signed accumulation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Where does y = x² − 1 cross the x-axis?
What you will learn
- Find the area between a curve, an axis and lines parallel to the axes.
- Split at crossings so total area or distance does not cancel.
Locate every relevant axis crossing before integrating
A geometric area is non-negative. Above the x-axis, the vertical height is y; below it, the height from the curve to the axis is −y. On a finite interval, solve y = 0 and split at each root inside the stated bounds where the sign can change. Integrate the positive height on each part or take the magnitude of each signed part. Taking the absolute value of a single net integral can still undercount an area with cancelling contributions.
For y = x² − 1 from x = −2 to 2, the roots are −1 and 1. The graph is above the axis on [−2, −1] and [1, 2], and below it on [−1, 1]. With F(x) = x³/3 − x, the three signed integrals are 4/3, −4/3 and 4/3. The total area is 4/3 + 4/3 + 4/3 = 4 square units, while the single signed integral is 4/3.
Choose height and units to match the request
For a curve wholly below the axis, integrate 0 − y rather than report a negative area. For y = −x² between x = 0 and 2, the signed integral is −8/3 but the geometric area is 8/3 square units. Bounds may be stated as vertical or horizontal lines; use a variable in which each strip has a clearly specified positive width or height, with the matching differential.
Signed velocity behaves the same way. For v(t) = 2t − 4 on 0 ≤ t ≤ 3, the direction changes at t = 2. The distance is −∫₀²(2t − 4) dt + ∫₂³(2t − 4) dt = 4 + 1 = 5 m. The displacement remains −3 m. An area in a velocity–time graph has distance units because (m/s) × s = m, rather than square metres.
Pause and explain
Three signed area pieces are 4/3, −4/3 and 4/3. What is their total geometric area?
Worked example
Find the total area between y = x² − 1 and the x-axis for −2 ≤ x ≤ 2, and compare the signed integral.
Show the worked solution
- Solve x² − 1 = 0: split at x = −1 and 1, with signs positive, negative and positive.
- Using F(x) = x³/3 − x gives signed pieces 4/3, −4/3 and 4/3.
- Add their magnitudes for total area 4; their signed sum is 4/3.
Answer Total area 4 square units; signed integral 4/3.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the area between y = 2x − 2 and the x-axis on 0 ≤ x ≤ 3, then compare its signed integral.
Give me a hint
Split at x = 1 and use F(x) = x² − 2x.
Compare my reasoning
- The graph is below the axis on [0, 1] and above it on [1, 3].
- The signed pieces are F(1) − F(0) = −1 and F(3) − F(1) = 4.
- The total area is 1 + 4 = 5 square units; the net integral is −1 + 4 = 3.
Total area 5 square units; signed integral 3.
Look for these in your work
- Split at the root within the bounds.
- Add magnitudes for total area and signs for the net integral.
Find the area enclosed by y = x(3 − x) and the x-axis.
Give me a hint
Find both roots and check which side of the axis contains the enclosed region.
Compare my reasoning
- The roots are x = 0 and 3; the parabola is non-negative between them.
- The area is ∫₀³(3x − x²) dx = [(3/2)x² − x³/3]₀³.
- Evaluation gives 27/2 − 9 = 9/2 square units, with no sign split inside the interval.
Area 9/2 square units.
Look for these in your work
- Use both intersections to determine the enclosed bounds.
- Check the sign before treating an integral as an area.
For signed velocity v(t) = 2t − 4 m/s on 0 ≤ t ≤ 3, find both total distance and displacement.
Give me a hint
Split at the direction change t = 2 for distance, while displacement uses signed contributions.
Compare my reasoning
- The zero velocity occurs at t = 2; the signed integral from 0 to 2 is −4 m.
- From 2 to 3 the signed integral is 1 m, so total distance is 4 + 1 = 5 m.
- Displacement is −4 + 1 = −3 m, describing the net position change.
Distance 5 m; displacement −3 m.
Look for these in your work
- Reverse the negative contribution for distance.
- Keep signed contributions for displacement and state metres.
Common mistakes
- The absolute value of the net integral is not generally total area.
- Use every sign-changing root inside the bounds.
- Read the axes' units before assigning area or distance units.
What can you explain now?
For y = −x² on 0 ≤ x ≤ 2, give the signed integral and the geometric area.
Compare with the explanation
Signed integral −8/3; geometric area 8/3 square units.
The curve is below the axis throughout; integrate −y or take the magnitude of this one signed part.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Given a curve crossing an axis, explain why sign splits must precede a total-area calculation. Use the same reasoning to distinguish distance and displacement.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.