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.
438 exam-style questions on OCR A Level Maths 1.8 Integration, covering 1.8.1 Fundamental theorem of calculus (A-level only), 1.8.2 Integrating x^n, 1.8.3 Integrating standard functions (A-level only), 1.8.4 Evaluating definite integrals, 1.8.5 Area between a curve and the x-axis, 1.8.6 Area between two curves, 1.8.7 Integration as the limit of a sum (A-level only), 1.8.8 Integration by substitution (A-level only), 1.8.9 Integration by parts (A-level only), 1.8.10 Use of partial fractions in integration (A-level only), 1.8.11 Differential equations with separable variables (A-level only), 1.8.12 Interpreting the solution of a differential equation (A-level only), and 1.8 Integration. Each one has a worked solution and a mark scheme showing where the marks go.