Find areas between curves with the correct boundary order
Solve intersections, identify the upper and lower boundary on each interval and integrate their positive difference.
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 simultaneous curve equations to find intersections.
- Evaluate definite integrals and check a function's sign on intervals.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
On 0 < x < 1, which expression is the positive height between y = x and y = x²?
What you will learn
- Find areas between a curve and a line or between two curves.
- Split where boundary order changes rather than allow separate regions to cancel.
Integrate a strip height, not two unrelated areas
For vertical strips between curves y = u(x) and y = l(x), the height is upper minus lower: u(x) − l(x). Find the relevant intersections by equating the boundary expressions and check which is above on the interval. An area is ∫[u(x) − l(x)] dx where that ordering holds. It is not necessary for either curve to meet the x-axis; the difference measures the actual space between them.
The curves y = x and y = x² meet where x² = x, so x = 0 or 1. On 0 < x < 1 the line is above the parabola, giving area ∫₀¹(x − x²) dx = [x²/2 − x³/3]₀¹ = 1/6. Integrating the reverse difference would produce −1/6, signalling the wrong height order for a geometric area.
Intersections can change which curve is above
For y = x and y = x³, the intersections are x = −1, 0 and 1. On (−1, 0), x³ is above x; on (0, 1), x is above x³. The two regions have areas ∫₋₁⁰(x³ − x) dx = 1/4 and ∫₀¹(x − x³) dx = 1/4. Their total is 1/2. One integral of x − x³ over [−1, 1] is zero because it cancels two signed contributions, not because the regions are absent.
A sketch and a test input on each interval help prevent an incorrect boundary order. Use exact intersection values, including roots when necessary, until final evaluation. If a region is supplied with other vertical or horizontal boundaries, respect those bounds rather than assume all roots belong to the requested region. Check the dimensions of the strip and differential before reporting square units.
Pause and explain
For y = x and y = x³ on −1 ≤ x ≤ 1, why split at zero?
Worked example
Find the area enclosed by y = x and y = x².
Show the worked solution
- Solve x² = x to find intersections at x = 0 and 1.
- Between the intersections, x ≥ x², so integrate the height x − x².
- [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6 square units.
Answer Area 1/6 square units.
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 enclosed by y = 4 − x² and y = 0.
Give me a hint
Find the roots and integrate the positive parabola between them.
Compare my reasoning
- The intersections satisfy 4 − x² = 0, giving x = −2 and 2.
- The positive height is 4 − x² and its antiderivative is 4x − x³/3.
- Endpoint subtraction gives (8 − 8/3) − (−8 + 8/3) = 32/3 square units.
Area 32/3 square units.
Look for these in your work
- Use both roots as bounds.
- Subtract the full lower endpoint value, including its signs.
Find the area enclosed by y = 2x and y = x².
Give me a hint
Find intersections, then use a test input to choose the upper boundary.
Compare my reasoning
- Solve x² = 2x to obtain x = 0 and 2.
- On that interval the line is above the parabola, giving height 2x − x².
- The area is [x² − x³/3]₀² = 4 − 8/3 = 4/3 square units.
Area 4/3 square units.
Look for these in your work
- Solve the correct intersection equation.
- Integrate upper minus lower on the enclosed interval.
Find the combined area of the two regions enclosed by y = x and y = x³ on −1 ≤ x ≤ 1. Explain why the single integral of x − x³ is insufficient.
Give me a hint
Find all three intersections and check ordering on both sides of zero.
Compare my reasoning
- The intersections solve x(x² − 1) = 0: x = −1, 0 and 1.
- Use x³ − x on [−1, 0] and x − x³ on [0, 1]; each integral is 1/4.
- The combined area is 1/2 square units, while the unsplit signed integral of x − x³ is zero by cancellation.
Combined area 1/2 square units; the unsplit signed integral cancels to zero.
Look for these in your work
- Change the boundary order at the interior intersection.
- Explain the difference between two positive regions and a cancelling net value.
Common mistakes
- Use the positive boundary difference on each interval.
- An intersection inside the bounds can require a change of ordering.
- Do not replace the area between curves with an area to an axis unless that axis is a boundary.
What can you explain now?
Find the area between y = 4 − x² and y = x².
Compare with the explanation
16√2/3 square units.
They meet at x = ±√2. The upper-minus-lower height is 4 − 2x², whose integral on those bounds is 16√2/3.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain how you find the bounds and decide which curve is above on each part. Show why one interior intersection may require two integrals.
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.