An electronic sensor measures a damped oscillation signal given by the function V(t)=e−2tsin(2t)V(t) = \mathrm{e}^{-2t} \sin(2t)V(t)=e−2tsin(2t) for t≥0t \ge 0t≥0, where t t\,t is time in milliseconds.
Given that y=e−2t(sin2t+cos2t)y = \mathrm{e}^{-2t}(\sin 2t + \cos 2t)y=e−2t(sin2t+cos2t), find dydt\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}dtdy. Simplify your answer.
Hence, show that ∫e−2tsin2t dt=ae−2t(sin2t+cos2t)+C\int \mathrm{e}^{-2t} \sin 2t \, \mathrm{d}t = a \mathrm{e}^{-2t}(\sin 2t + \cos 2t) + C∫e−2tsin2tdt=ae−2t(sin2t+cos2t)+C where a a\,a is a rational number to be determined.
The areas of the finite regions bounded by the signal curve and the ttt-axis are denoted by A1,A2,…,An,… A_1, A_2, \dots, A_n, \dots\,A1,A2,…,An,… where A1 A_1\,A1 is the area of the first pulse (the region between t=0t=0t=0 and the first root of V(t)=0V(t) = 0V(t)=0 for t>0t > 0t>0).
Find the exact value of the area A1A_1A1.
Show that the ratio of successive areas An+1An\displaystyle \frac{A_{n+1}}{A_n}AnAn+1 is constant and find its value in terms of e\mathrm{e}e.
Show that the exact value of the total area enclosed between the signal curve and the ttt-axis for all t≥0 t \ge 0\,t≥0 is 1+e−π4(1−e−π) or equivalently eπ+14(eπ−1)\frac{1 + \mathrm{e}^{-\pi}}{4(1 - \mathrm{e}^{-\pi})} \text{ or equivalently } \frac{\mathrm{e}^{\pi} + 1}{4(\mathrm{e}^{\pi} - 1)}4(1−e−π)1+e−π or equivalently 4(eπ−1)eπ+1
Practise Edexcel A Level Maths 11.6 Integration by Parts with exam-style questions for A Level Maths. 52 questions, matched to the Edexcel A Level Maths (9MA0) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.