Find areas between curves and across multiple regions
Locate intersections, integrate top minus bottom on each interval and add positive regions when a problem asks for total area.
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How do you find where two curves intersect?
What you will learn
- Find intersection limits for two curves
- Integrate the difference between upper and lower curves
- Add separate positive areas when the curves cross or the interval has multiple regions
Intersections define natural area boundaries
Set the curve equations equal to find where they meet. Between consecutive intersections their vertical order is fixed, so use upper curve minus lower curve as the integrand.
If the curves exchange positions at a crossing, split the area calculation at that input and reverse the subtraction on the next interval.
Total area adds magnitudes, not signed cancellations
A single definite integral can cancel positive and negative parts. For total geometric area, evaluate each region as positive and add the results.
A curve may cross the x-axis or another curve more than once. Mark the intersections and identify which graph is higher in each interval before integrating.
Pause and explain
If the upper curve changes at x = c, what should you do?
Worked example
Find the total area between y = x² − 1 and the x-axis on −2 ≤ x ≤ 2.
Show the worked solution
- The curve crosses the axis at x = −1 and x = 1. It is above the axis outside these roots and below between them.
- By symmetry, total area is 2[∫₀¹(1 − x²)dx + ∫₁²(x² − 1)dx].
- The two areas are 2/3 and 4/3, so twice their sum is 4.
Answer Total geometric area = 4 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 between y = 2x and y = x² on 0 ≤ x ≤ 2.
Give me a hint
Check which curve is higher inside the interval, then integrate their difference.
Compare my reasoning
- 2x − x² = x(2 − x) is non-negative on [0,2].
- Area = ∫₀²(2x − x²)dx.
- Evaluate [x² − x³/3]₀².
4/3 square units.
Look for these in your work
- Curve order justified
- Difference integrated
- Boundaries evaluated
Find the total area between y = x and y = 2x − x² on 0 ≤ x ≤ 2.
Give me a hint
Find intersections and determine which curve is higher on each side.
Compare my reasoning
- Solve x = 2x − x², giving x = 0 and x = 1.
- On [0,1], 2x − x² is above x; on [1,2], x is above 2x − x².
- Integrate each positive difference and add: each area is 1/6, so the total is 1/3.
Total area = 1/3 square units.
Look for these in your work
- Intersections located
- Top curve checked on each interval
- Two positive areas added
A plot compares f(x) = x² and g(x) = 2x on 0 ≤ x ≤ 2. Find the signed integral of f − g and explain why the total geometric area is different.
Give me a hint
The curves meet at both endpoints, with g above f inside the interval.
Compare my reasoning
- ∫₀²(f − g)dx = ∫₀²(x² − 2x)dx.
- Evaluate [x³/3 − x²]₀² = 8/3 − 4 = −4/3.
- The geometric area is positive 4/3 because g − f is positive throughout the interior.
Signed integral −4/3; total geometric area 4/3 square units.
Look for these in your work
- Signed difference order retained
- Definite integral evaluated
- Geometric magnitude distinguished
Common mistakes
- Using one signed integral when the question asks for total area
- Failing to find where curves intersect
- Keeping the same top-minus-bottom order after the curves cross
- Adding signed areas that cancel
What can you explain now?
What integrand gives area between upper y = f(x) and lower y = g(x)?
Compare with the explanation
f(x) − g(x)
The positive vertical distance is upper minus lower.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How do intersection points and a top-minus-bottom check organise an area between curves?
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.