Did you know? A student can integrate both functions correctly and still find the wrong area. Why? Because the question is not merely ‘find two integrals’. It is ‘find the space between the graphs‘.
That single change of perspective makes integration between curves much easier. For JC2 students taking H1 Mathematics 8865 in Bukit Timah, remember the short rule: area between curves = the integral of (upper curve − lower curve) across the correct interval, for the region where one curve stays above the other.
First, picture the region
Consider the curves y = 6x − x² and y = 2x. We want the area enclosed between them.
Before finding any antiderivative, work out where the curves meet. Set 6x − x² = 2x, giving 4x − x² = 0. Factorise: x(4 − x) = 0. So the intersection x-values are 0 and 4.
Next, choose a test value between 0 and 4—for example, x = 2. The parabola gives y = 8, while the line gives y = 4. Therefore the parabola is above the line in the enclosed region. That is the moment you know which function to subtract.
Worked example: find the enclosed area
- Step 1 — boundaries: the graphs meet at x = 0 and x = 4.
- Step 2 — upper curve: y = 6x − x² on this interval.
- Step 3 — lower curve: y = 2x.
- Step 4 — set up: A = ∫ from 0 to 4 of [(6x − x²) − 2x] dx = ∫ from 0 to 4 of (4x − x²) dx.
- Step 5 — integrate: an antiderivative is 2x² − x³/3. Evaluating from 0 to 4 gives 32 − 64/3 = 32/3 square units, approximately 10.67 square units.
I like this problem because you can check the sign without a calculator. Between 0 and 4, the difference is x(4 − x), which is positive. That is exactly what we expect when subtracting the lower curve from the upper one.
Why ‘upper minus lower’ matters
If you reverse the subtraction, the definite integral produces −32/3. That is the signed result of subtracting in the wrong order—not the positive geometric area of this enclosed region.
Here both curves are on or above the x-axis between the intersections, so the example also stays within the familiar H1 area setting. If graphs cross each other inside a larger interval, check which is upper in each sub-interval before integrating.
Six things to check before moving on
- Have I found the intersection x-values, rather than guessing the limits?
- Have I identified which curve is on top in the relevant interval?
- Did I subtract the whole lower function correctly, including its signs?
- Did I integrate the simplified expression term by term?
- Did I substitute the upper and lower limits in the correct order?
- Did I state the final answer in square units when it represents geometric area?
The SEAB 2026 H1 Mathematics 8865 syllabus covers integration as the reverse of differentiation, definite integrals, and areas between curves or between a curve and a line. It excludes the trapezium-rule approximation and area below the x-axis. Check your own year’s syllabus before preparing additional methods.
A better way to practise for JC2
I would start with a quick sketch and two questions: ‘Where do the graphs meet?’ and ‘Which one is above?’ Only then would I let the student integrate. Repeat with a straight line and a different parabola, then explain the setup before touching the calculator.
This is also a good example of why method marks depend on clear working. A tidy integral with the right limits is easier to follow, check and correct than a floating numerical answer copied from a screen.
Quick questions
Can the graphing calculator check my area? It can help evaluate a definite integral, but it should not replace identifying the region and forming the correct expression.
Why is the answer in square units? Because the integral represents an area on the coordinate plane, not simply a function value.
For a balanced schedule that includes both calculus and probability, see the JC2 H1 Mathematics revision map and Bukit Timah H1 probability and statistics overview. Parents can ask eduKate about current JC2 H1 Maths teaching options and share the student’s most recent integration question.