June 2022 Paper 1 Q9
9 Use the substitution \(x = 2\sin\theta\) to show that \(\displaystyle\int_1^{\sqrt{3}} \sqrt{4 - x^2}\,\mathrm{d}x = \tfrac{1}{3}\pi\). [7]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{d}x = 2\cos\theta\,\mathrm{d}\theta\) | M1 | 1.1a |
| \(\displaystyle\int \sqrt{4 - x^2}\,\mathrm{d}x = \int \sqrt{4 - 4\sin^2\theta}\,.\,2\cos\theta\,\mathrm{d}\theta\) | M1 | 3.1a |
| \(\displaystyle = \int \sqrt{4\cos^2\theta}\,.\,2\cos\theta\,\mathrm{d}\theta\) \(\displaystyle = \int 4\cos^2\theta\,\mathrm{d}\theta\) | A1 | 1.1 |
| \(\displaystyle = \int (2\cos 2\theta + 2)\,\mathrm{d}\theta\) | M1 | 2.1 |
| \(= \sin 2\theta + 2\theta\) | A1FT | 1.1 |
| \(\left[\sin 2\theta + 2\theta\right]_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} = \left(\sin\tfrac{2}{3}\pi + \tfrac{2}{3}\pi\right) - \left(\sin\tfrac{2}{6}\pi + \tfrac{2}{6}\pi\right)\) \(= \left(\tfrac{1}{2}\sqrt{3} + \tfrac{2}{3}\pi\right) - \left(\tfrac{1}{2}\sqrt{3} + \tfrac{1}{3}\pi\right)\) | M1 | 2.1 |
| \(= \tfrac{1}{3}\pi\) A.G. | A1 | 2.1 |
| [7] |
Notes
M1: Attempt to link \(\mathrm{d}x\) and \(\mathrm{d}\theta\)
Allow sign error only
M1: Attempt to write integrand in terms of \(\theta\)
Must substitute for both function and \(\mathrm{d}x\)
Can follow M0 but do not allow just \(\mathrm{d}x = \mathrm{d}\theta\)
A1: Obtain correct integrand in terms of \(\cos\theta\) only
Condone no \(\mathrm{d}\theta\), as long as previously seen
M1: Attempt use of double angle formula
Using \(\cos 2\theta = \pm 2\cos^2\theta \pm 1\)
Integrand must be of form \(k\cos^2\theta\), which must have come from correct method with coefficient errors only
A1FT: Integrate to obtain \(\sin 2\theta + 2\theta\)
FT on \(a\cos 2\theta + b\) only
M1: Attempt use of limits
Must be correct limits (either \(x\) or \(\theta\), as long as consistent with their integral), correct order and subtraction
Allow M1 for use of limits in any integration attempt in terms of \(\theta\). Allow M1 for either expressions that still involve sin, or exact equivs
M0 for decimal values, even if then stated to be the same as \(\frac{1}{3}\pi\)
Condone eg \(\frac{1}{2}\sqrt{3}\) from \(\sin 120^\circ\), but M0 if degrees used in linear term
A1: Obtain given answer of \(\frac{1}{3}\pi\)
Must see both surd values, or an explanation as to why \(\sin\frac{2}{3}\pi = \sin\frac{2}{6}\pi\)