June 2022 Paper 1 Q14
14 The region bounded by the curve
\[y = (2x - 8)\ln x\]and the \(x\)-axis is shaded in the diagram below.
(a) Use the trapezium rule with 5 ordinates to find an estimate for the area of the shaded region.
Give your answer correct to three significant figures. [3 marks]
(b) Show that the exact area is given by\[32\ln 2 - \frac{33}{2}\]
Fully justify your answer. [6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Finds positive or negative \(y\)-values for 5 \(x\)-values with \(h\) = 0.75 PI by AWRT 5.28 or AWRT \(-5.28\) or Uses 6 \(x\)-values and obtains AWRT 5.42 or AWRT \(-5.42\) In this case maximum mark is M1A0 A0 | M1 | 1.1a |
| Uses the trapezium rule correctly with \(h\) = 0.75 and correct \(y\)-values Accept rounded or truncated values to 3 significant figures. | A1 | 1.1b |
| Obtains AWRT 5.28 Condone AWRT \(-5.28\) | A1 | 3.2a |
| (3) |
Typical solution
| \(x_n\) | \(y_n\) |
|---|---|
| 1 | 0 |
| 1.75 | \(-2.51827\) |
| 2.5 | \(-2.74887\) |
| 3.25 | \(-1.76798\) |
| 4 | 0 |
Area \(\approx\) 5.28
| Scheme | Marks | AO |
|---|---|---|
| Sets up integration by parts Condone \(u\) and \(v^{\prime}\) in wrong order Must have expressions for \(u\), \(u^{\prime}\), \(v\) and \(v^{\prime}\) with evidence of some integration | M1 | 3.1a |
| Applies integration by parts correctly to \((2x - 8)\ln x\) to obtain either \((x^2 - 8x)\ln x - \int x - 8\,\mathrm{d}x\) OE or \(\dfrac{1}{4}(2x - 8)^2\ln x - \displaystyle\int x - 8 + \dfrac{16}{x}\,\mathrm{d}x\) Condone missing brackets or omission of \(\mathrm{d}x\) | M1 | 1.1a |
| Completes integration fully to obtain either \((x^2 - 8x)\ln x - \dfrac{x^2}{2} + 8x\) OE or \(\dfrac{1}{4}(2x - 8)^2\ln x - \dfrac{x^2}{2} + 8x - 16\ln x\) OE | A1 | 1.1b |
| Substitutes limits 1 and 4 into their integrated function and subtracts either way round | M1 | 1.1a |
| Completes reasoned argument to correctly obtain \(\dfrac{33}{2} - 32\ln 2\) or \(32\ln 2 - \dfrac{33}{2}\) AG Brackets must be correct throughout | R1 | 2.1 |
| Explains change of sign due to shaded region being below \(x\)-axis This could be at an earlier stage eg swap limits explained but must still refer to the shaded region being below \(x\)-axis | E1 | 2.4 |
| (6) | ||
| (9 marks) |
Typical solution
\[\int_1^4 (2x - 8)\ln x\,\mathrm{d}x\]\[u = \ln x \qquad u^{\prime} = \frac{1}{x}\]\[v^{\prime} = 2x - 8 \qquad v = x^2 - 8x\]\[(x^2 - 8x)\ln x - \int x - 8\,\mathrm{d}x\]\[= \left[(x^2 - 8x)\ln x - \frac{x^2}{2} + 8x\right]_1^4\]\[= \left((16 - 32)\ln 4 - \frac{4^2}{2} + 32\right) - \left((1 - 8)\ln 1 - \frac{1^2}{2} + 8\right)\]\[= -16\ln 2^2 + 24 - \frac{15}{2}\]\[= \frac{33}{2} - 32\ln 2\]Shaded region is below \(x\)-axis
area = \(32\ln 2 - \dfrac{33}{2}\)