June 2023 Paper 1 Q5
5 The graph of \(y = \dfrac{5}{\mathrm{e}^x - 1}\) is shown in the diagram below.
The trapezium rule with 6 ordinates (5 strips) is to be used to find an approximation for the shaded area.
The values required to obtain this approximation are shown in the table below.
| \(x\) | 1 | 1.6 | 2.2 | 2.8 | 3.4 | 4 |
|---|---|---|---|---|---|---|
| \(y\) | 2.90988 | 1.26485 | 0.62305 | 0.32374 | 0.17263 | 0.09329 |
(a) Use the trapezium rule with 6 ordinates (5 strips) to find an approximate value for the shaded area.
Give your answer to four decimal places. [3 marks]
(b) Using your answer to part (a) deduce an estimate for \(\displaystyle\int_1^4 \frac{20}{\mathrm{e}^x - 1}\,\mathrm{d}x\) [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| States or uses \(h = 0.6\) OE Accept 0.3 OE as the multiplier. PI by correct answer | B1 | 1.1b |
| Substitutes given \(y\) values to achieve \(2.90988 + 0.09329 +\) \(2(1.26485 + 0.62305 + 0.32374 + 0.17263)\) Ignore \(h\). Accept correct exact values or values to more than 5 decimal places. PI by correct answer or 7.77171 | M1 | 1.1a |
| Obtains 2.3315 AWRT | A1 | 1.1b |
| (3) |
Typical solution
\[\frac{0.6}{2}\begin{pmatrix} 2.90988 + 0.09329 + \\ 2(1.26485 + 0.62305 + 0.32374 + 0.17263) \end{pmatrix}\]\[= 2.3315\]| Scheme | Marks | AO |
|---|---|---|
| Obtains 4x their answer to (a) correct to at least 2 significant figures. | R1F | 2.2a |
| (1) | ||
| (4 marks) |
Typical solution
9.3