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S6 Maths: Definite Integrals – Mixed Applications (Paper 3: Challenge)
S6 Maths: Definite Integrals – Mixed Applications (Paper 3: Challenge)
Ten mixed items with unknown limits, average values and symmetric-interval area traps.
Direct answers Answers in order: 4, 2 plus pi, 9/2, 3, 4, 3, 8, 32/3, 0.807, 1/2.
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 cubed from x = 0 to x = 2.
- FoundationEvaluate the integral of sine x plus 1 from 0 to pi.
- FoundationFind the area enclosed by the line y = 3x and the curve y = x squared.
- IntermediateGiven the integral of x squared from 0 to a equals 9 with a positive, find a.
- IntermediateGiven the integral of 2x from 0 to k equals 16 with k positive, find k.
- IntermediateFind the average value of f(x) = x squared on the interval from 0 to 3.
- ChallengeFind the total area enclosed by y = x cubed and the line y = 4x.
- ChallengeFind the area enclosed by y = x squared and the line y = 2x plus 3.
- ChallengeEvaluate the integral of x minus 1 over x from 1 to 2, to 3 decimal places.
- Error redoError replay: a student got 0 for the integral of x cubed from minus 1 to 1 and called the area 0.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 4 | The area is the integral of x cubed, and x to the fourth over 4 gives 4. |
| 2 | 2 + pi (about 5.14) | The sine part gives 2 and the constant part gives pi, so the total is 2 plus pi. |
| 3 | 9/2 (that is 4.5) | They meet at 0 and 3; integrating 3x minus x squared gives 13.5 minus 9. |
| 4 | 3 | The antiderivative gives a cubed over 3 equal to 9, so a cubed is 27 and a is 3. |
| 5 | 4 | The antiderivative gives k squared equal to 16, so the positive root is 4. |
| 6 | 3 | The average is the integral divided by the interval length, so 9 divided by 3. |
| 7 | 8 | The curves meet at 0 and plus or minus 2; two symmetric pieces of 4 give 8. |
| 8 | 32/3 (about 10.67) | They meet at minus 1 and 3; integrating the line minus the curve gives 32/3. |
| 9 | 0.807 | The antiderivative is x squared over 2 minus the natural log, which gives about 0.807. |
| 10 | 1/2 | An odd function cancels over a symmetric interval, but the two areas add to 1/2. |
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.
With an unknown limit, write the integral first and solve the equation instead of guessing.Tutor Registration →
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