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).
Practise Edexcel A Level Maths 11.5 Integration by Substitution with exam-style questions for A Level Maths. 59 questions, matched to the Edexcel A Level Maths (9MA0) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.