C4 January 2006 Q2
2.
(a) Given that \(y = \sec x\), complete the table with the values of \(y\) corresponding to \(x = \dfrac{\pi}{16}, \dfrac{\pi}{8}\) and \(\dfrac{\pi}{4}\).
(2)
| \(x\) | 0 | \(\dfrac{\pi}{16}\) | \(\dfrac{\pi}{8}\) | \(\dfrac{3\pi}{16}\) | \(\dfrac{\pi}{4}\) |
| \(y\) | 1 | 1.20269 |
(b) Use the trapezium rule, with all the values for \(y\) in the completed table, to obtain an estimate for \(\displaystyle\int_0^{\frac{\pi}{4}} \sec x\,\mathrm{d}x\). Show all the steps of your working, and give your answer to 4 decimal places. (3)
The exact value of \(\displaystyle\int_0^{\frac{\pi}{4}} \sec x\,\mathrm{d}x\) is \(\ln(1 + \surd 2)\).
(c) Calculate the % error in using the estimate you obtained in part (b). (2)
| \(x\) | 0 | \(\dfrac{\pi}{16}\) | \(\dfrac{\pi}{8}\) | \(\dfrac{3\pi}{16}\) | \(\dfrac{\pi}{4}\) |
| \(y\) | 1 | 1.01959 | 1.08239 | 1.20269 | 1.41421 |
| Scheme | Marks |
|---|---|
| M1 for one correct, A1 for all correct | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Integral \(= \dfrac{1}{2} \times \dfrac{\pi}{16} \times \left\{1 + 1.4142 + 2(1.01959 + \ldots + 1.20269)\right\}\) | M1 A1ft |
| \(\left(= \dfrac{\pi}{32} \times 9.02355\right) = 0.8859\) | A1 cao |
| (3) |
| Scheme | Marks |
|---|---|
| Percentage error \(= \dfrac{\textit{approx} - 0.88137}{0.88137} \times 100 = 0.51\ \%\) (allow 0.5% to 0.54% for A1) | M1 A1 |
| (2) | |
| (7 marks) |
Notes
M1 gained for \((\pm)\ \dfrac{\textit{approx} - \ln(1 + \sqrt{2})}{\ln(1 + \sqrt{2})}\)