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.
480 exam-style questions on AQA A Level Maths 1.11 H: Integration, covering 1.11.1 Fundamental Theorem of Calculus, 1.11.2 Integrating standard functions, 1.11.3 Definite integrals and areas, 1.11.4 Integration as the limit of a sum (A-level only), 1.11.5 Integration by substitution and by parts (A-level only), 1.11.6 Integration using partial fractions (A-level only), 1.11.7 Differential equations with separable variables (A-level only), and 1.11.8 Interpreting solutions of differential equations (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.