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S6 Maths: Definite Integrals – Areas under Curves (Paper 2: Intermediate)
S6 Maths: Definite Integrals – Areas under Curves (Paper 2: Intermediate)
Ten items on areas under curves and between curves, with intersections used as limits.
Direct answers Answers in order: 8/3, 1/6, 4/3, 2, 32/3, 4/3, 1, 4/3, 1/2, 1/6.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S6 Maths: Definite Integrals and Area |
| Level | S6 |
02 Questions (10 items)
- FoundationFind the area under y = x squared from x = 0 to x = 2.
- FoundationFind the area enclosed by y = x squared and the line y = x.
- FoundationEvaluate the integral of x squared plus 1 from 0 to 1.
- IntermediateEvaluate the integral of 1 over the square root of x from 1 to 4.
- IntermediateFind the area enclosed by y = 4 minus x squared and the x-axis.
- IntermediateFind the area enclosed by y = x squared and the line y = 2x.
- ChallengeEvaluate the integral of cosine x from 0 to pi over 2.
- ChallengeFind the area between y = x squared minus 4x plus 3 and the axis from x = 1 to 3.
- ChallengeFind the area under y = 1 over x squared from x = 1 to x = 2.
- Error redoError replay: a student integrated x squared minus x from 0 to 1, got minus 1/6 and gave that as the area.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 8/3 (about 2.67) | The area is the integral of x squared, and x cubed over 3 gives 8/3. |
| 2 | 1/6 (about 0.17) | The curves meet at 0 and 1; integrating x minus x squared gives 1/2 minus 1/3. |
| 3 | 4/3 | The antiderivative is x cubed over 3 plus x, giving 1/3 plus 1. |
| 4 | 2 | The antiderivative is twice the square root, which is 4 at 4 and 2 at 1. |
| 5 | 32/3 (about 10.67) | The curve cuts the axis at plus and minus 2; by symmetry the area is twice the integral to 2. |
| 6 | 4/3 | They meet at 0 and 2; integrating 2x minus x squared gives 4 minus 8/3. |
| 7 | 1 | The antiderivative is sine, which is 1 at pi over 2 and 0 at 0. |
| 8 | 4/3 | The curve lies below the axis here, so the integral is minus 4/3 and the area is 4/3. |
| 9 | 1/2 | The antiderivative is minus 1 over x; the difference between the limits is 1/2. |
| 10 | 1/6 | Area is the absolute value; the minus sign only shows the curve lies below the axis. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Sketch first to see which curve is on top, then decide which one is subtracted.Tutor Registration →
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