C4 January 2008 Q4
4.
| Scheme | Marks |
|---|---|
| \(\displaystyle\int\ln\left(\tfrac{x}{2}\right)\,\mathrm{d}x = \int 1.\ln\left(\tfrac{x}{2}\right)\,\mathrm{d}x \Rightarrow \left\{\begin{aligned} u &= \ln\left(\tfrac{x}{2}\right) &&\Rightarrow\ \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{\frac{1}{2}}{\frac{x}{2}} = \frac{1}{x}\\ \frac{\mathrm{d}v}{\mathrm{d}x} &= 1 &&\Rightarrow\ v = x\end{aligned}\right.\) | |
| \(\displaystyle\int\ln\left(\tfrac{x}{2}\right)\,\mathrm{d}x = x\ln\left(\tfrac{x}{2}\right) - \int x.\tfrac{1}{x}\,\mathrm{d}x\) | M1 A1 |
| \(= x\ln\left(\tfrac{x}{2}\right) - \displaystyle\int\underline{1}\,\mathrm{d}x\) | dM1 |
| \(= x\ln\left(\tfrac{x}{2}\right) - x + c\) | A1 aef |
| (4) |
Notes
M1: Use of ‘integration by parts’ formula in the correct direction. A1: Correct expression.
Note: \(\displaystyle\int\ln\left(\tfrac{x}{2}\right)\,\mathrm{d}x = (\text{their } v)\ln\left(\tfrac{x}{2}\right) - \int(\text{their } v).(\text{their } \tfrac{\mathrm{d}u}{\mathrm{d}x})\,\mathrm{d}x\) for M1 in part (i).
dM1: An attempt to multiply \(x\) by a candidate’s \(\tfrac{a}{x}\) or \(\tfrac{1}{bx}\) or \(\tfrac{1}{x}\).
A1 aef: Correct integration with \(+\,c\)
Aliter 4. (i) Way 2
| Scheme | Marks |
|---|---|
| \(\displaystyle\int\ln\left(\tfrac{x}{2}\right)\,\mathrm{d}x = \int(\ln x - \ln 2)\,\mathrm{d}x = \int\ln x\,\mathrm{d}x - \int\ln 2\,\mathrm{d}x\) | |
| \(\displaystyle\int\ln x\,\mathrm{d}x = \int 1.\ln x\,\mathrm{d}x \Rightarrow \left\{\begin{aligned} u &= \ln x &&\Rightarrow\ \frac{\mathrm{d}u}{\mathrm{d}x} = \tfrac{1}{x}\\ \frac{\mathrm{d}v}{\mathrm{d}x} &= 1 &&\Rightarrow\ v = x\end{aligned}\right.\) | |
| \(\displaystyle\int\ln x\,\mathrm{d}x = x\ln x - \int x.\tfrac{1}{x}\,\mathrm{d}x\) | M1 |
| \(= x\ln x - x + c\) | A1 |
| \(\displaystyle\int\ln 2\,\mathrm{d}x = x\ln 2 + c\) | M1 |
| Hence, \(\displaystyle\int\ln\left(\tfrac{x}{2}\right)\,\mathrm{d}x = x\ln x - x - x\ln 2 + c\) | A1 aef |
| (4) |
M1: Use of ‘integration by parts’ formula in the correct direction. Note: \(\displaystyle\int\ln x\,\mathrm{d}x = (\text{their } v)\ln x - \int(\text{their } v).(\text{their } \tfrac{\mathrm{d}u}{\mathrm{d}x})\,\mathrm{d}x\) for M1 in part (i).
A1: Correct integration of \(\ln x\) with or without \(+\,c\). M1: Correct integration of \(\ln 2\) with or without \(+\,c\). A1 aef: Correct integration with \(+\,c\)
Aliter 4. (i) Way 3
| Scheme | Marks |
|---|---|
| \(\displaystyle\int\ln\left(\tfrac{x}{2}\right)\,\mathrm{d}x\) | |
| \(u = \tfrac{x}{2} \Rightarrow \dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{2}\) | |
| \(\displaystyle\int\ln\left(\tfrac{x}{2}\right)\,\mathrm{d}x = 2\int\ln u\,\mathrm{d}u\) | |
| \(\displaystyle\int\ln u\,\mathrm{d}x = \int 1.\ln u\,\mathrm{d}u\) | |
| \(\displaystyle\int\ln u\,\mathrm{d}x = u\ln u - \int u.\tfrac{1}{u}\,\mathrm{d}u\) | M1 |
| \(= u\ln u - u + c\) | A1 |
| Decide to award 2nd M1 here! | M1 |
| \(\displaystyle\int\ln\left(\tfrac{x}{2}\right)\,\mathrm{d}x = 2(u\ln u - u) + c\) | |
| Hence, \(\displaystyle\int\ln\left(\tfrac{x}{2}\right)\,\mathrm{d}x = x\ln\left(\tfrac{x}{2}\right) - x + c\) | A1 aef |
| (4) |
Applying substitution correctly to give \(\displaystyle\int\ln\left(\tfrac{x}{2}\right)\,\mathrm{d}x = 2\int\ln u\,\mathrm{d}u\): Decide to award 2nd M1 here!
M1: Use of ‘integration by parts’ formula in the correct direction. A1: Correct integration of \(\ln u\) with or without \(+\,c\). M1: Decide to award 2nd M1 here! A1 aef: Correct integration with \(+\,c\)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin^2 x\,\mathrm{d}x\) | |
| \(\left[\text{NB: } \underline{\cos 2x = \pm 1 \pm 2\sin^2 x} \text{ or } \underline{\sin^2 x = \tfrac{1}{2}(\pm 1 \pm \cos 2x)}\right]\) | M1 |
| \(= \displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1 - \cos 2x}{2}\,\mathrm{d}x = \frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\underline{(1 - \cos 2x)}\,\mathrm{d}x\) | |
| \(= \dfrac{1}{2}\Big[\underline{x - \tfrac{1}{2}\sin 2x}\Big]_{\frac{\pi}{4}}^{\frac{\pi}{2}}\) | dM1 A1 |
| \(= \tfrac{1}{2}\left[\left(\tfrac{\pi}{2} - \tfrac{\sin(\pi)}{2}\right) - \left(\tfrac{\pi}{4} - \tfrac{\sin(\frac{\pi}{2})}{2}\right)\right]\) | ddM1 |
| \(= \tfrac{1}{2}\left[\left(\tfrac{\pi}{2} - 0\right) - \left(\tfrac{\pi}{4} - \tfrac{1}{2}\right)\right]\) | |
| \(= \tfrac{1}{2}\left(\tfrac{\pi}{4} + \tfrac{1}{2}\right) = \tfrac{\pi}{8} + \tfrac{1}{4}\) | A1 aef, cso |
| (5) | |
| (9 marks) |
Notes
M1: Consideration of double angle formula for \(\cos 2x\)
dM1: Integrating to give \(\underline{\pm ax \pm b\sin 2x}\); \(a, b \ne 0\). A1: Correct result of anything equivalent to \(\tfrac{1}{2}x - \tfrac{1}{4}\sin 2x\)
ddM1: Substitutes limits of \(\tfrac{\pi}{2}\) and \(\tfrac{\pi}{4}\) and subtracts the correct way round.
A1 aef, cso: \(\underline{\tfrac{1}{2}\left(\tfrac{\pi}{4} + \tfrac{1}{2}\right)}\) or \(\underline{\tfrac{\pi}{8} + \tfrac{1}{4}}\) or \(\underline{\tfrac{\pi}{8} + \tfrac{2}{8}}\). Candidate must collect their \(\pi\) term and constant term together for A1. No fluked answers, hence cso.
Note \(\tfrac{\pi}{8} + \tfrac{1}{4} = 0.64269\ldots\)
Aliter 4. (ii) Way 2
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin^2 x\,\mathrm{d}x = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin x.\sin x\,\mathrm{d}x\) and \(I = \displaystyle\int\sin^2 x\,\mathrm{d}x\) | |
| \(\left\{\begin{aligned} u &= \sin x &&\Rightarrow\ \tfrac{\mathrm{d}u}{\mathrm{d}x} = \cos x\\ \tfrac{\mathrm{d}v}{\mathrm{d}x} &= \sin x &&\Rightarrow\ v = -\cos x\end{aligned}\right\}\) | |
| \(\therefore I = \underline{\left\{-\sin x\cos x + \displaystyle\int\cos^2 x\,\mathrm{d}x\right\}}\) | M1 |
| \(\therefore I = \left\{-\sin x\cos x + \displaystyle\int(1 - \sin^2 x)\,\mathrm{d}x\right\}\) | |
| \(\displaystyle\int\sin^2 x\,\mathrm{d}x = \left\{-\sin x\cos x + \int 1\,\mathrm{d}x - \int\sin^2 x\,\mathrm{d}x\right\}\) | |
| \(2\displaystyle\int\sin^2 x\,\mathrm{d}x = \left\{-\sin x\cos x + \int 1\,\mathrm{d}x\right\}\) | dM1 |
| \(2\displaystyle\int\sin^2 x\,\mathrm{d}x = \{-\sin x\cos x + x\}\) | |
| \(\displaystyle\int\sin^2 x\,\mathrm{d}x = \underline{\left\{-\tfrac{1}{2}\sin x\cos x + \tfrac{x}{2}\right\}}\) | A1 |
| \(\therefore \displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin^2 x\,\mathrm{d}x = \left[\left(-\tfrac{1}{2}\sin(\tfrac{\pi}{2})\cos(\tfrac{\pi}{2}) + \tfrac{(\frac{\pi}{2})}{2}\right) - \left(-\tfrac{1}{2}\sin(\tfrac{\pi}{4})\cos(\tfrac{\pi}{4}) + \tfrac{(\frac{\pi}{4})}{2}\right)\right]\) \(= \left[\left(0 + \tfrac{\pi}{4}\right) - \left(-\tfrac{1}{4} + \tfrac{\pi}{8}\right)\right]\) | ddM1 |
| \(= \tfrac{\pi}{8} + \tfrac{1}{4}\) | A1 aef cso |
| (5) |
M1: An attempt to use the correct by parts formula. dM1: For the LHS becoming \(2I\). A1: Correct integration. ddM1: Substitutes limits of \(\tfrac{\pi}{2}\) and \(\tfrac{\pi}{4}\) and subtracts the correct way round.
A1 aef, cso: \(\underline{\tfrac{1}{2}\left(\tfrac{\pi}{4} + \tfrac{1}{2}\right)}\) or \(\underline{\tfrac{\pi}{8} + \tfrac{1}{4}}\) or \(\underline{\tfrac{\pi}{8} + \tfrac{2}{8}}\). Candidate must collect their \(\pi\) term and constant term together for A1. No fluked answers, hence cso.
Note \(\tfrac{\pi}{8} + \tfrac{1}{4} = 0.64269\ldots\)