A2 June 2023 Paper 1 Q15
15 In this question you must show detailed reasoning.
Evaluate \(\displaystyle\int_1^2 \frac{1}{\sqrt{1 + 2x - x^2}}\,\mathrm{d}x\), giving your answer in terms of \(\pi\). [5]
| Scheme | Marks | AO |
|---|---|---|
| DR \(1 + 2x - x^2 = -(x^2 - 2x) + 1 = -(x - 1)^2 + 2\) | M1 A1 | 3.1a 1.1 |
| \(\left[\arcsin\dfrac{x - 1}{k}\right]_1^2\) | M1 | 2.1 |
| \(= \left[\arcsin\dfrac{x - 1}{\sqrt{2}}\right]_1^2\) | A1 | 2.1 |
| \(= \arcsin\left(\dfrac{1}{\sqrt{2}}\right) - \arcsin 0 = \dfrac{\pi}{4}\) | A1 | 2.2a |
| [5] |
Notes
M1: (1st) completing the square
A1: (3rd) substitution or evaluation of both limits must be seen
Alternative method
| Scheme | Marks |
|---|---|
| \(1 + 2x - x^2 = -(x^2 - 2x) + 1 = -(x - 1)^2 + 2\) | M1 A1 |
| Let \(u = x - 1\) \(\displaystyle\int_0^1 \frac{1}{\sqrt{2 - u^2}}\,\mathrm{d}u\) | M1 |
| \(= \left[\arcsin\dfrac{u}{\sqrt{2}}\right]_0^1\) | A1 |
| \(= \arcsin\left(\dfrac{1}{\sqrt{2}}\right) - \arcsin 0 = \dfrac{\pi}{4}\) | A1 |
M1: (1st) completing the square
M1: (2nd) complete substitution including limits
A1: (3rd) substitution or evaluation of both limits must be seen