The rate at which a specific chemical compound is produced during a reaction, in grams per hour, is modeled by the function
R(t)=(4t+1)2t+1 R(t) = (4t + 1)\sqrt{2t + 1} R(t)=(4t+1)2t+1where t t\,t is the time in hours, 0≤t≤50 \le t \le 50≤t≤5.
Use the substitution u=2t+1u = 2t + 1u=2t+1 to show that the total mass of the compound produced,
∫05(4t+1)2t+1 dt \int_{0}^{5} (4t + 1)\sqrt{2t + 1} \, dt ∫05(4t+1)2t+1dtcan be written as
12∫111(2u−1)u12 du=12∫k11(2u32−u12) du \frac{1}{2} \int_{1}^{11} (2u - 1)u^{\frac{1}{2}} \, du = \frac{1}{2} \int_{k}^{11} (2u^{\frac{3}{2}} - u^{\frac{1}{2}}) \, du 21∫111(2u−1)u21du=21∫k11(2u23−u21)duwhere k k\,k is a constant to be found.
Hence, or otherwise, show that the total mass of the compound produced in the first 5 hours is
115(67111−1) grams \frac{1}{15}(671\sqrt{11} - 1) \text{ grams} 151(67111−1) gramsA technician approximates the total mass produced between t=0t = 0t=0 and t=5t = 5t=5 using five rectangles of equal width, where the left-hand edge of each rectangle touches the curve y=R(t)y = R(t)y=R(t). The total area of these five rectangles is MMM.
The technician then decides to use ten rectangles of equal width, still using the left-hand edge method, to find a second approximation.
Explain why the value of this second approximation will be greater than MMM, but less than 115(67111−1)\displaystyle \frac{1}{15}(671\sqrt{11} - 1)151(67111−1).
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.