The concentration of a certain chemical reagent, CCC, measured in mmol/L, during a reaction is modelled by the equation
C(t)=3+4log10(cost)for 0≤t≤1 C(t) = 3 + 4 \log_{10}(\cos t) \quad \text{for } 0 \le t \le 1 C(t)=3+4log10(cost)for 0≤t≤1where ttt is the time in seconds after the reaction begins.
Complete the table below for the concentration at various times, giving values of CCC to 3 decimal places.
| ttt | 0 | 0.25 | 0.5 | 0.75 | 1 |
|---|---|---|---|---|---|
| CCC | 3.000 | 2.773 | 1.931 |
Use the trapezium rule with all the values in the completed table to find an estimate for ∫01C(t) dt\int_{0}^{1} C(t) \, dt∫01C(t)dt, giving your answer to 2 decimal places.
Hence, determine an estimate for the value of
∫01(1−2log10(cost)) dt \int_{0}^{1} (1 - 2 \log_{10}(\cos t)) \, dt ∫01(1−2log10(cost))dtgiving your answer to 2 decimal places.
137 exam-style questions on OCR (MEI) A Level Maths 1.10 Numerical Methods (A-level only), covering 1.10.1 Locate roots by change of sign (A-level only), 1.10.2 When change of sign methods fail (A-level only), 1.10.3 Fixed point iteration (A-level only), 1.10.4 Newton-Raphson method (A-level only), 1.10.5 Convergence of iterations (A-level only), 1.10.6 Trapezium rule (A-level only), 1.10.7 Upper and lower bounds using rectangles (A-level only), and 1.10.8 Numerical methods to solve problems (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.