The velocity v(t)v(t)v(t) of a remote-controlled rover on a research mission, measured in m/s\text{m/s}m/s, is modeled by the function v(t)=10t2+5\displaystyle v(t) = \frac{10}{\sqrt{t^2+5}}v(t)=t2+510 for 2≤t≤42 \le t \le 42≤t≤4, where t t\,t is the time in seconds after activation. A technician records the rover's velocity at regular intervals as shown in the table below.
| ttt | 2 | 2.5 | 3 | 3.5 | 4 |
|---|---|---|---|---|---|
| v(t)v(t)v(t) | 3.33333 | 2.98142 | 2.67261 | 2.40772 | 2.18218 |
Use the trapezium rule with all the values in the table to find an approximate value for the total distance traveled by the rover between t=2t=2t=2 and t=4t=4t=4. Give your answer to four decimal places.
Using your answer to part (a), deduce an estimate for ∫2430t2+5 dt\displaystyle \int_{2}^{4} \frac{30}{\sqrt{t^2+5}} \, dt∫24t2+530dt.
43 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.