A2 June 2024 Paper 1 Q6
6 In this question you must show detailed reasoning.
Determine the exact value of \(\displaystyle\int_9^{\infty} \frac{18}{x^2\sqrt{x}}\,\mathrm{d}x\). [4]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\displaystyle\int \frac{18}{x^2\sqrt{x}}\,\mathrm{d}x = 18\left(-\frac{2}{3}x^{-\frac{3}{2}}\right)(+c)\) | M1* | 1.1 |
| \(\displaystyle = 18\lim_{k \to \infty}\left(-\frac{2}{3}k^{-\frac{3}{2}} - \left(-\frac{2}{3} \times 9^{-\frac{3}{2}}\right)\right)\) | M1 | 1.1 |
| \(k^{-\frac{3}{2}} \to 0\) as \(k \to \infty\) | B1dep* | 2.1 |
| \(= \dfrac{4}{9}\) | A1 | 2.2a |
| [4] |
Notes
M1*: For obtaining \(ax^{-\frac{3}{2}}\) where \(a \neq 0\).
M1: Correct use of 9 as a lower limit and any letter (except \(x\)) for the upper limit (so must be considering a finite upper limit) in their integrated expression (indicated by their power increased by 1). Need not see mention of limiting process for this mark.
B1dep*: Taking limit as \(k \to \infty\) for their expression of the form \(ax^{-\frac{3}{2}}\) (so \(\dfrac{1}{\sqrt{\infty^3}} = 0\) oe is B0). Implied by, for example, \(\displaystyle\lim_{k \to \infty}\left[-\frac{2}{3}k^{-\frac{3}{2}} - \ldots\right] = 0 - \ldots\) but not, for example, for \(-\dfrac{2}{3}k^{-\frac{3}{2}} - \ldots = 0 - \ldots\) without clear use of limiting process.
A1: cao from correct integrated expression and finite upper limit (so dependent on both previous M marks but not the B mark). Accept equivalent exact forms e.g. \(\frac{12}{27}\).