C4 January 2013 Q2
2.
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{1}{x^3}\ln x\,\mathrm{d}x,\quad \left\{\begin{aligned} u &= \ln x &\Rightarrow\ \frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{1}{x} \\ \frac{\mathrm{d}v}{\mathrm{d}x} &= x^{-3} &\Rightarrow\ v &= \frac{x^{-2}}{-2} = \frac{-1}{2x^2} \end{aligned}\right\}\) | |
| In the form \(\dfrac{\pm\lambda}{x^2}\ln x \pm \displaystyle\int \mu\frac{1}{x^2}.\frac{1}{x}\) | M1 |
| \(= \underline{\dfrac{-1}{2x^2}\ln x} - \underline{\underline{\displaystyle\int \frac{-1}{2x^2}.\frac{1}{x}\,\mathrm{d}x}}\) \(\underline{\dfrac{-1}{2x^2}\ln x}\) simplified or un-simplified. | A1 |
| \(\underline{\underline{-\displaystyle\int \frac{-1}{2x^2}.\frac{1}{x}}}\) simplified or un-simplified. | A1 |
| \(\left\{= \dfrac{-1}{2x^2}\ln x + \dfrac{1}{2}\displaystyle\int \frac{1}{x^3}\,\mathrm{d}x\right\}\) | |
| \(= -\dfrac{1}{2x^2}\ln x + \dfrac{1}{2}\left(-\dfrac{1}{2x^2}\right)\ \{+\,c\}\) \(\pm\displaystyle\int \mu\frac{1}{x^2}.\frac{1}{x} \to \pm\beta x^{-2}\). | dM1 |
| Correct answer, with/without \(+\,c\) | A1 |
| (5) |
Notes
M1: Integration by parts is applied in the form \(\dfrac{\pm\lambda}{x^2}\ln x \pm \displaystyle\int \mu\frac{1}{x^2}.\frac{1}{x}\) or equivalent.
A1: \(\underline{\dfrac{-1}{2x^2}\ln x}\) simplified or un-simplified.
A1: \(\underline{\underline{-\displaystyle\int \frac{-1}{2x^2}.\frac{1}{x}}}\) or equivalent. You can ignore the \(\mathrm{d}x\).
dM1: Depends on the previous M1. \(\pm\displaystyle\int \mu\frac{1}{x^2}.\frac{1}{x} \to \pm\beta x^{-2}\).
A1: \(-\dfrac{1}{2x^2}\ln x + \dfrac{1}{2}\left(-\dfrac{1}{2x^2}\right)\ \{+\,c\}\) or \(= -\dfrac{1}{2x^2}\ln x - \dfrac{1}{4x^2}\ \{+\,c\}\) or \(\dfrac{x^{-2}}{-2}\ln x - \dfrac{x^{-2}}{4}\ \{+\,c\}\)
or \(\dfrac{-1 - 2\ln x}{4x^2}\ \{+\,c\}\) or equivalent.
You can ignore subsequent working after a correct stated answer.
Alternative Solution
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{1}{x^3}\ln x\,\mathrm{d}x,\quad \left\{\begin{aligned} u &= x^{-3} &\Rightarrow\ \frac{\mathrm{d}u}{\mathrm{d}x} &= -3x^{-4} \\ \frac{\mathrm{d}v}{\mathrm{d}x} &= \ln x &\Rightarrow\ v &= x\ln x - x \end{aligned}\right\}\) \(\displaystyle\int \frac{1}{x^3}\ln x\,\mathrm{d}x = \dfrac{1}{x^3}(x\ln x - x) - \displaystyle\int (x\ln x - x)\frac{-3}{x^4}\,\mathrm{d}x\) | |
| \(k\displaystyle\int \frac{1}{x^3}\ln x\,\mathrm{d}x = \frac{1}{x^3}(x\ln x - x) \pm \int \frac{\lambda}{x^3}\,\mathrm{d}x\) where \(k \neq 1\) | M1 |
| \(-2\displaystyle\int \frac{1}{x^3}\ln x\,\mathrm{d}x = \frac{1}{x^3}(x\ln x - x) - \int \frac{3}{x^3}\,\mathrm{d}x\) Any one of \(\dfrac{1}{x^3}(x\ln x - x)\) or \(-\displaystyle\int \frac{3}{x^3}\,\mathrm{d}x\) | A1 |
| \(\dfrac{1}{x^3}(x\ln x - x) - \displaystyle\int \frac{3}{x^3}\,\mathrm{d}x\) and \(k = -2\) | A1 |
| \(-2\displaystyle\int \frac{1}{x^3}\ln x\,\mathrm{d}x = \frac{1}{x^3}(x\ln x - x) + \frac{3}{2x^2}\ \{+\,c\}\) \(\pm\displaystyle\int \mu\frac{1}{x^3} \to \pm\beta x^{-2}\). | dM1 |
| \(\displaystyle\int \frac{1}{x^3}\ln x\,\mathrm{d}x = -\dfrac{1}{2x^3}(x\ln x - x) - \dfrac{3}{4x^2}\ \{+\,c\}\) \(-\dfrac{1}{2x^3}(x\ln x - x) - \dfrac{3}{4x^2}\) or equivalent with/without \(+\,c\). | A1 |
| \(= -\dfrac{1}{2x^2}\ln x - \dfrac{1}{4x^2}\ \{+\,c\}\) |
(This alternative is printed on the page headed “2. (b) ctd”; it is a method for part (a).)
| Scheme | Marks |
|---|---|
| \(\left\{\left[-\dfrac{1}{2x^2}\ln x - \dfrac{1}{4x^2}\right]_1^2\right\} = \left(-\dfrac{1}{2(2)^2}\ln 2 - \dfrac{1}{4(2)^2}\right) - \left(-\dfrac{1}{2(1)^2}\ln 1 - \dfrac{1}{4(1)^2}\right)\) Applies limits of 2 and 1 to their part (a) answer and subtracts the correct way round. | M1 |
| \(= \dfrac{3}{16} - \dfrac{1}{8}\ln 2\) or \(\dfrac{3}{16} - \ln 2^{\frac{1}{8}}\) or \(\dfrac{1}{16}(3 - 2\ln 2)\), etc, or awrt 0.1 or equivalent. | A1 |
| (2) | |
| (7 marks) |
Notes
M1: Some evidence of applying limits of 2 and 1 to their part (a) answer and subtracts the correct way round.
A1: Two term exact answer of either \(\dfrac{3}{16} - \dfrac{1}{8}\ln 2\) or \(\dfrac{3}{16} - \ln 2^{\frac{1}{8}}\) or \(\dfrac{1}{16}(3 - 2\ln 2)\) or \(\dfrac{\ln\left(\frac{1}{4}\right) + 3}{16}\)
or \(0.1875 - 0.125\ln 2\). Also allow awrt 0.1. Also note the fraction terms must be combined.
Note: Award the final A0 in part (b) for a candidate who achieves awrt 0.1 in part (b), when their answer to part (a) is incorrect.
Note: Decimal answer is 0.100856... in part (b).