A2 June 2019 Paper 2 Q3
3 In this question you must show detailed reasoning.
Show that \(\displaystyle\int_{5}^{\infty} (x - 1)^{-\frac{3}{2}}\,\mathrm{d}x = 1\). [5]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\displaystyle\int (x - 1)^{-\frac{3}{2}}\,\mathrm{d}x = -2(x - 1)^{-\frac{1}{2}}\ (+c)\) | B1 | 1.1a |
| \(\displaystyle\int_{5}^{N} (x - 1)^{-\frac{3}{2}}\,\mathrm{d}x = \left[-2(x - 1)^{-\frac{1}{2}}\right]_{5}^{N}\) | M1 | 2.1 |
| \(-\dfrac{2}{\sqrt{N - 1}} + \dfrac{2}{\sqrt{5 - 1}}\) | A1 | 1.1 |
| \(\displaystyle\lim_{N \to \infty} \frac{1}{\sqrt{N - 1}} = 0\) oe | B1 | 2.1 |
| \(\displaystyle\int_{5}^{\infty} (x - 1)^{-\frac{3}{2}}\,\mathrm{d}x = \lim_{N \to \infty}\left\{-\frac{2}{\sqrt{N - 1}} + \frac{2}{\sqrt{5 - 1}}\right\} = 1\) | A1 | 2.2a |
| [5] |
Notes
M1: Consideration of a finite upper limit
B1: Not just eg \(\dfrac{1}{\infty} = 0\). Can be seen as part of limit of both terms, but must be explicitly shown as zero
A1: AG. Convincing argument equating improper integral to solution