In a controlled biochemical reaction, the rate of mass accumulation R(t)R(t)R(t) in grams per hour is modeled by the function
R(t)=15−5t(t+2)(2t−1)2,t≥1 R(t) = \frac{15 - 5t}{(t + 2)(2t - 1)^2}, \quad t \ge 1 R(t)=(t+2)(2t−1)215−5t,t≥1where t t\,t is the time in hours since the start of the experiment.
Given that
15−5t(t+2)(2t−1)2≡At+2+B2t−1+C(2t−1)2 \frac{15 - 5t}{(t + 2)(2t - 1)^2} \equiv \frac{A}{t + 2} + \frac{B}{2t - 1} + \frac{C}{(2t - 1)^2} (t+2)(2t−1)215−5t≡t+2A+2t−1B+(2t−1)2Cfind the values of the constants AAA, B B\,B and CCC.
Hence find the exact value of the total mass accumulated between t=1t = 1t=1 and t=2t = 2t=2 hours, which is given by
∫1215−5t(t+2)(2t−1)2 dt \int_{1}^{2} \frac{15 - 5t}{(t + 2)(2t - 1)^2} \, \mathrm{d}t ∫12(t+2)(2t−1)215−5tdtgiving your answer in the form plnq+rp \ln q + rplnq+r where p,q p, q\,p,q and r r\,r are rational numbers.
864 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.