In a bio-reactor, the rate of oxygen consumption V(t)V(t)V(t), in milligrams per hour, is modelled by the equation
V(t)=(10t+13)2t+3 V(t) = (10t + 13)\sqrt{2t + 3} V(t)=(10t+13)2t+3where t t\,t is the time in hours since the start of an experiment, 0≤t≤30 \le t \le 30≤t≤3.
Use the substitution u=2t+3u = 2t + 3u=2t+3 to show that
∫03(10t+13)2t+3 dt \int_{0}^{3} (10t + 13)\sqrt{2t + 3} \, dt ∫03(10t+13)2t+3dtcan be written as
12∫39(5u−2u0)u12 du=12∫a9(5u32−2u12) du \frac{1}{2} \int_{3}^{9} (5u - 2u^0)u^{\frac{1}{2}} \, du = \frac{1}{2} \int_{a}^{9} (5u^{\frac{3}{2}} - 2u^{\frac{1}{2}}) \, du 21∫39(5u−2u0)u21du=21∫a9(5u23−2u21)duwhere a a\,a is a constant to be found.
Hence, or otherwise, show that the total oxygen consumed over the 3-hour period is
225−73 mg 225 - 7\sqrt{3} \text{ mg} 225−73 mgA scientist uses three rectangles of equal width to approximate the total oxygen consumed, VtotalV_{total}Vtotal, using the left-hand edge method. The total area of these three rectangles is RRR.
The scientist decides to improve the approximation by increasing the number of rectangles used (still using the left-hand edge method).
Explain why the value of this improved approximation will be greater than RRR, but less than 225−73225 - 7\sqrt{3}225−73.
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.