A research drone monitors the rate of data transmission, D(t)D(t)D(t), in gigabits per hour, from a remote sensor over a 4-hour period. The transmission rate is modelled by the function
D(t)=1.25t0.5t2+4,0≤t≤4 D(t) = \frac{1.25^t}{\sqrt{0.5t^2 + 4}}, \quad 0 \le t \le 4 D(t)=0.5t2+41.25t,0≤t≤4Complete the table below, giving the values of D(t)D(t)D(t) to 3 decimal places.
| ttt | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| D(t)D(t)D(t) | 0.500 | 0.638 | 0.705 |
Use the trapezium rule, with all the values of D(t)D(t)D(t) from the completed table, to find an approximate value for the total data transmitted,
∫041.25t0.5t2+4 dt \int_{0}^{4} \frac{1.25^t}{\sqrt{0.5t^2 + 4}} \, dt ∫040.5t2+41.25tdt43 exam-style questions on CCEA A Level Maths 3.7.3 Numerical methods (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.