June 2022 Paper 1 Q11
11 Given that \(k\) is a positive constant, show that \(\displaystyle\int_{k}^{2k} \frac{2}{(2x+k)^2}\,\mathrm{d}x\) is inversely proportional to \(k\). [6]
| Scheme | Marks | AO |
|---|---|---|
| Let \(u = 2x + k\), \(2\,\mathrm{d}x = \mathrm{d}u\) | M1 | 2.1 |
| \(\displaystyle\int \frac{2}{(2x+k)^2}\,\mathrm{d}x = \int \frac{1}{u^2}\,\mathrm{d}u\) | A1 | 2.1 |
| \(= -\dfrac{1}{u}\ [+c]\) | A1 | 2.1 |
| \(\displaystyle\int_{k}^{2k} \frac{2}{(2x+k)^2}\,\mathrm{d}x = \int_{3k}^{5k} \left(\frac{1}{u^2}\right)\mathrm{d}u = -\frac{1}{5k} + \frac{1}{3k}\) | M1 | 2.1 |
| \(= \dfrac{2}{15k}\) | A1 | 2.1 |
| This is inversely proportional to \(k\) [with constant of proportionality \(\frac{2}{15}\)] | E1 | 2.2a |
| [6] |
Notes
M1: Substituting \(u = 2x + k\)
Allow for a different substitution giving an integral in \(u\) only
Allow for \(\displaystyle\int \frac{a}{u^2}\,\mathrm{d}u\) for any constant seen
A1: Correct integrand in terms of \(u\)
Ignore limits
A1: correct indefinite integral
constant need not be seen
M1: substituting correct new limits into their integrated expression, or substituting in terms of \(x\) and using original limits
A1: Allow \(\left(-\dfrac{1}{5} + \dfrac{1}{3}\right)\dfrac{1}{k}\) seen
E1: FT their definite integral
Must use phrase “inversely proportional” to \(k\) or indicates \(\propto \dfrac{1}{k}\)
Allow if \(\dfrac{a}{k}\) required at the start of the argument
Alternatively, by inspection
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_{k}^{2k} \frac{2}{(2x+k)^2}\,\mathrm{d}x = \left[-(2x+k)^{-1}\right]_{k}^{2k}\) | M1 A2 |
| \(-\dfrac{1}{5k} + \dfrac{1}{3k}\) | M1 |
M1: Integrating by inspection to obtain any multiple of \((2x+k)^{-1}\)
A2: Fully correct indefinite integral – need not be simplified.
M1: substituting limits into their integrated expression