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)+Cwhere 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+1864 exam-style questions on Edexcel A Level Maths Integration, covering 11.1 Integrating Standard Functions, 11.2 Integrating f(ax + b), 11.3 Using Trigonometric Identities, 11.4 Reverse Chain Rule, 11.5 Integration by Substitution, 11.6 Integration by Parts, 11.7 Partial Fractions, 11.8 Finding Areas, 11.9 The Trapezium Rule, 11.10 Solving Differential Equations, and 11.11 Modelling with Differential Equations. Each one has a worked solution and a mark scheme showing where the marks go.