A chemical engineer monitors the rate of a reaction, R(t)R(t)R(t), measured in mmol L−1^{-1}−1 min−1^{-1}−1, at regular intervals during the first 8 minutes of a process. The following table records the measured rate at various times t t\,t in minutes. The values of R(t)R(t)R(t) are rounded to 3 decimal places.
| ttt | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| R(t)R(t)R(t) | 0.450 | 1.282 | 2.914 | 1.845 | 0.312 |
Using the trapezium rule with all the values of R(t)R(t)R(t) in the given table, obtain an estimate for the total yield of the substance, given by
∫08R(t) dt \int_{0}^{8} R(t) \, dt ∫08R(t)dtgiving your answer to 2 decimal places.
Use your answer to part (a) to estimate (i)
∫08(R(t)−0.2) dt \int_{0}^{8} (R(t) - 0.2) \, dt ∫08(R(t)−0.2)dt(ii)
∫311R(t−3) dt \int_{3}^{11} R(t-3) \, dt ∫311R(t−3)dt43 exam-style questions on OCR (MEI) A Level Maths 1.10.6 Trapezium rule (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.