A2 June 2019 Paper 2 Q3
3.
\[\mathrm{f}(x) = \frac{1}{\sqrt{4x^2 + 9}}\]Way 1
| Scheme | Marks | AO |
|---|---|---|
| \(x = \dfrac{3}{2}\sinh u\) | B1 | 2.1 |
| \(\displaystyle\int \frac{\mathrm{d}x}{\sqrt{4x^2 + 9}} = \int \frac{1}{\sqrt{4\left(\frac{9}{4}\right)\sinh^2 u + 9}} \times \frac{3}{2}\cosh u\,\mathrm{d}u\) | M1 | 3.1a |
| \(\displaystyle = \int \frac{1}{2}\,\mathrm{d}u\) | A1 | 1.1b |
| \(\displaystyle = \int \frac{1}{2}\,\mathrm{d}u = \frac{1}{2}u = \frac{1}{2}\sinh^{-1}\left(\frac{2x}{3}\right) + c\) | A1 | 1.1b |
| (4) |
Notes
B1: Selects an appropriate substitution leading to an integrable form
M1: Demonstrates a fully correct method for the substitution that includes substituting into the function and dealing with the “\(\mathrm{d}x\)”. The substitution being substituted does not need to be “correct” for this mark but the substitution must be an attempt at \(\displaystyle\int \frac{1}{\sqrt{4\left[\mathrm{f}(u)\right]^2 + 9}} \times \mathrm{f}^{\prime}(u)\,\mathrm{d}u\) with the \(\mathrm{f}^{\prime}(u)\) correct for their substitution. E.g. if \(x = \dfrac{1}{2}u\) is used, must see \(\mathrm{d}x = \dfrac{1}{2}\mathrm{d}u\) not \(2\mathrm{d}u\).
A1: Correct simplified integral in terms of \(u\) from correct work and from a correct substitution
A1: Correct answer including “\(+ c\)”. Allow arcsinh or arsinh for \(\sinh^{-1}\) from correct work and from a correct substitution
Alternative: Way 2
| Scheme | Marks | AO |
|---|---|---|
| \(x = \dfrac{3}{2}\tan u\) | B1 | 2.1 |
| \(\displaystyle\int \frac{\mathrm{d}x}{\sqrt{4x^2 + 9}} = \int \frac{1}{\sqrt{4\left(\frac{9}{4}\right)\tan^2 u + 9}} \times \frac{3}{2}\sec^2 u\,\mathrm{d}u\) | M1 | 3.1a |
| \(\displaystyle = \int \frac{1}{2}\sec u\,\mathrm{d}u\) | A1 | 1.1b |
| \(= \dfrac{1}{2}\ln(\sec u + \tan u) = \dfrac{1}{2}\ln\left(\dfrac{2x}{3} + \sqrt{1 + \left(\dfrac{2x}{3}\right)^2}\right)\) \(= \dfrac{1}{2}\sinh^{-1}\left(\dfrac{2x}{3}\right) + c\) | A1 | 1.1b |
(Corrected from the printed mark scheme: the last line is printed as “\(u = \dfrac{1}{2}\sinh^{-1}\left(\dfrac{2x}{3}\right) + c\)”; it is the integral, not \(u\), that equals this.)
Alternative: Way 3
| Scheme | Marks | AO |
|---|---|---|
| \(x = \dfrac{1}{2}u\) or \(x = ku\) where \(k \gt 0\ \ k \neq 1\) | B1 | 2.1 |
| \(\displaystyle\int \frac{\mathrm{d}x}{\sqrt{4x^2 + 9}} = \int \frac{1}{\sqrt{4\left(\frac{1}{4}\right)u^2 + 9}} \times \frac{1}{2}\mathrm{d}u\) | M1 | 3.1a |
| \(\displaystyle = \frac{1}{2}\int \frac{1}{\sqrt{u^2 + 9}}\,\mathrm{d}u\quad \left(\text{or } \frac{1}{2}\int \frac{1}{\sqrt{u^2 + \frac{9}{4k^2}}}\,\mathrm{d}u \text{ for } x = ku\right)\) | A1 | 1.1b |
| \(= \dfrac{1}{2}\sinh^{-1}\dfrac{u}{3} = \dfrac{1}{2}\sinh^{-1}\dfrac{2x}{3} + c\) | A1 | 1.1b |
| Scheme | Marks | AO |
|---|---|---|
| Mean value \(=\) \(\dfrac{1}{3(-0)}\left[\dfrac{1}{2}\sinh^{-1}\left(\dfrac{2x}{3}\right)\right]_0^3 = \dfrac{1}{3} \times \dfrac{1}{2}\sinh^{-1}\left(\dfrac{2 \times 3}{3}\right)(-0)\) | M1 | 2.1 |
| \(= \dfrac{1}{6}\ln\left(2 + \sqrt{5}\right)\) (Brackets are required) | A1ft | 1.1b |
| (2) | ||
| (6 marks) |
Notes
M1: Correctly applies the method for the mean value for their integration which must be of the form specified in part (a) and substitutes the limits 0 and 3 but condone omission of 0
A1: Correct exact answer (follow through their \(A\) and \(B\)). Brackets are required if appropriate.