A2 June 2020 Paper 2 Q12
12
(a) Given that \(I = \displaystyle\int_a^b \mathrm{e}^{2t}\sin t\,\mathrm{d}t\), show that\[I = \Big[q\mathrm{e}^{2t}\sin t + r\mathrm{e}^{2t}\cos t\Big]_a^b\]
where \(q\) and \(r\) are rational numbers to be found. [6 marks]
(b) A small object is initially at rest. The subsequent motion of the object is modelled by the differential equation\[\frac{\mathrm{d}v}{\mathrm{d}t} + v = 5\mathrm{e}^t\sin t\]
where \(v\) is the velocity at time \(t\).
Find the speed of the object when \(t = 2\pi\), giving your answer in exact form. [6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Selects a method to find the required result by integrating by parts. | M1 | 3.1a |
| Obtains correct result of integration by parts. | A1 | 1.1b |
| Uses integration by parts a second time, consistent with their choice of \(u\) and \(v^{\prime}\) in their first integration by parts. | M1 | 1.1a |
| Obtains correct result of second integration by parts FT their first integration by parts. | A1 | 1.1b |
| Deduces that the second integration by parts gives an equation in \(I\) which can be solved | M1 | 2.2a |
| Completes rigorous argument to show that the result of the second integration by parts gives \(I = \left[\dfrac{2}{5}\mathrm{e}^{2t}\sin t - \dfrac{1}{5}\mathrm{e}^{2t}\cos t\right]_a^b\) | R1 | 2.1 |
Typical solution
\[I = \left[\frac{1}{2}\mathrm{e}^{2t}\sin t\right]_a^b - \frac{1}{2}\int_a^b \mathrm{e}^{2t}\cos t\,\mathrm{d}t\]\[I = \left[\frac{1}{2}\mathrm{e}^{2t}\sin t\right]_a^b - \frac{1}{2}\left\{\left[\frac{1}{2}\mathrm{e}^{2t}\cos t\right]_a^b + \frac{1}{2}\int_a^b \mathrm{e}^{2t}\sin t\,\mathrm{d}t\right\}\]\[I = \left[\frac{1}{2}\mathrm{e}^{2t}\sin t - \frac{1}{4}\mathrm{e}^{2t}\cos t\right]_a^b - \frac{1}{4}I\]\[I = \left[\frac{2}{5}\mathrm{e}^{2t}\sin t - \frac{1}{5}\mathrm{e}^{2t}\cos t\right]_a^b\]| Scheme | Marks | AO |
|---|---|---|
| Selects a method to solve the differential equation by finding an integrating factor. | M1 | 3.1a |
| Multiplies the differential equation by their integrating factor | M1 | 1.1a |
| Integrates LHS correctly to obtain \(v\mathrm{e}^t\) | A1 | 1.1b |
| Integrates their RHS correctly with or without limits FT their \(q\) and \(r\) from (a) | B1F | 1.1b |
| Uses the initial conditions when evaluating a definite integral or determining the constant of integration. | M1 | 3.4 |
| States the correct exact value of the speed, which must be positive FT their \(q\) and \(r\) from (a) provided \(v \lt 0\) when \(t = 2\pi\) | A1 | 3.2a |
| (12 marks) |
Typical solution
\[\frac{\mathrm{d}v}{\mathrm{d}t} + v = 5\mathrm{e}^t\sin t\]IF: \(\mathrm{e}^{\int 1\,\mathrm{d}t} = \mathrm{e}^t\)
\[\frac{\mathrm{d}}{\mathrm{d}t}\left(v\mathrm{e}^t\right) = 5\mathrm{e}^{2t}\sin t\]\[v\mathrm{e}^t = 2\mathrm{e}^{2t}\sin t - \mathrm{e}^{2t}\cos t + c\]When \(t = 0\), \(v = 0\) \(\therefore c = 1\)
\[v\mathrm{e}^t = 2\mathrm{e}^{2t}\sin t - \mathrm{e}^{2t}\cos t + 1\]\[\begin{aligned} v\mathrm{e}^{2\pi} &= 2\mathrm{e}^{4\pi}\sin 2\pi - \mathrm{e}^{4\pi}\cos 2\pi + 1 \\ &= -\mathrm{e}^{4\pi} + 1 \end{aligned}\]\[v = -\mathrm{e}^{2\pi} + \mathrm{e}^{-2\pi}\]Speed \(= \mathrm{e}^{2\pi} - \mathrm{e}^{-2\pi}\)