Given that y=e−2t(sin2t+cos2t)y = \mathrm{e}^{-2t}(\sin 2t + \cos 2t)y=e−2t(sin2t+cos2t), find dydt\frac{\mathrm{d}y}{\mathrm{d}t}dtdy. Simplify your answer.
Hence, find
∫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 aaa is a rational number.
The displacement sss (in μm\mu\text{m}μm) of a micro-mechanical resonator at time ttt (in seconds) is modeled by s(t)=e−2tsin2ts(t) = \mathrm{e}^{-2t} \sin 2ts(t)=e−2tsin2t for t≥0t \ge 0t≥0. The areas of the finite regions bounded by the curve and the ttt-axis are denoted by A1,A2,…,An,…A_1, A_2, \dots, A_n, \dotsA1,A2,…,An,… where A1A_1A1 is the area of the region from t=0t=0t=0 to the first positive root.
(i) Find the exact value of the area A1A_1A1.
(ii) Show that An+1An=e−π\frac{A_{n+1}}{A_n} = \mathrm{e}^{-\pi}AnAn+1=e−π.
(iii) Show that the exact value of the total area enclosed between the curve and the ttt-axis for t≥0t \ge 0t≥0 is
1+e−π4(1−e−π) \frac{1 + \mathrm{e}^{-\pi}}{4(1 - \mathrm{e}^{-\pi})} 4(1−e−π)1+e−πor equivalently eπ+14(eπ−1)\frac{\mathrm{e}^{\pi} + 1}{4(\mathrm{e}^{\pi} - 1)}4(eπ−1)eπ+1
Practise Edexcel A Level Maths Integration with exam-style questions for A Level Maths. 437 questions 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, 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.
