| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Area under curve requiring parts |
| Difficulty | Standard +0.8 This question requires integration by parts on a product involving ln(x), finding limits of integration from a diagram, handling the area calculation with appropriate sign considerations, and algebraic manipulation to reach a specific exact form. While integration by parts with ln(x) is a standard technique, the multi-step nature (finding bounds, applying parts correctly, simplifying to the given answer) and the need for careful justification elevates this above a routine exercise. |
| Spec | 1.08i Integration by parts1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Finds positive or negative \(y\)-values for 5 \(x\)-values with \(h = 0.75\): \(x_n\): 1, 1.75, 2.5, 3.25, 4; \(y_n\): 0, −2.51827, −2.74887, −1.76798, 0 | M1 | PI by AWRT 5.28 or AWRT −5.28; if 6 \(x\)-values used, max mark M1A0A0 |
| \(\frac{0.75}{2}(0 + 0 + 2(-2.51827 - 2.74887 - 1.76798))\) | A1 | Trapezium rule correctly with \(h = 0.75\) and correct \(y\)-values; accept rounded/truncated to 3 s.f. |
| Area \(\approx 5.28\) | A1 | AWRT 5.28; condone AWRT −5.28 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets up integration by parts | M1 | Condone \(u\) and \(v'\) in wrong order; must have expressions for \(u\), \(u'\), \(v\), \(v'\) with evidence of some integration |
| \(u = \ln x\), \(u' = \frac{1}{x}\), \(v' = 2x - 8\), \(v = x^2 - 8x\); obtains \((x^2 - 8x)\ln x - \int x - 8 \, dx\) | M1 | Applies IBP correctly to \((2x-8)\ln x\) |
| \(= \left[(x^2 - 8x)\ln x - \frac{x^2}{2} + 8x\right]_1^4\) | A1 | Fully correct integration |
| \(= \left((16-32)\ln 4 - \frac{16}{2} + 32\right) - \left((1-8)\ln 1 - \frac{1}{2} + 8\right)\) | M1 | Substitutes limits 1 and 4, subtracts either way |
| \(= -16\ln 2^2 + 24 - \frac{15}{2} = \frac{33}{2} - 32\ln 2\) | R1 | Completes reasoned argument; \(\frac{33}{2} - 32\ln 2\) or \(32\ln 2 - \frac{33}{2}\) AG; brackets must be correct throughout |
| Shaded region is below \(x\)-axis; area \(= 32\ln 2 - \frac{33}{2}\) | E1 | Explains change of sign due to shaded region being below \(x\)-axis |
## Question 14(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds positive or negative $y$-values for 5 $x$-values with $h = 0.75$: $x_n$: 1, 1.75, 2.5, 3.25, 4; $y_n$: 0, −2.51827, −2.74887, −1.76798, 0 | M1 | PI by AWRT 5.28 or AWRT −5.28; if 6 $x$-values used, max mark M1A0A0 |
| $\frac{0.75}{2}(0 + 0 + 2(-2.51827 - 2.74887 - 1.76798))$ | A1 | Trapezium rule correctly with $h = 0.75$ and correct $y$-values; accept rounded/truncated to 3 s.f. |
| Area $\approx 5.28$ | A1 | AWRT 5.28; condone AWRT −5.28 |
## Question 14(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets up integration by parts | M1 | Condone $u$ and $v'$ in wrong order; must have expressions for $u$, $u'$, $v$, $v'$ with evidence of some integration |
| $u = \ln x$, $u' = \frac{1}{x}$, $v' = 2x - 8$, $v = x^2 - 8x$; obtains $(x^2 - 8x)\ln x - \int x - 8 \, dx$ | M1 | Applies IBP correctly to $(2x-8)\ln x$ |
| $= \left[(x^2 - 8x)\ln x - \frac{x^2}{2} + 8x\right]_1^4$ | A1 | Fully correct integration |
| $= \left((16-32)\ln 4 - \frac{16}{2} + 32\right) - \left((1-8)\ln 1 - \frac{1}{2} + 8\right)$ | M1 | Substitutes limits 1 and 4, subtracts either way |
| $= -16\ln 2^2 + 24 - \frac{15}{2} = \frac{33}{2} - 32\ln 2$ | R1 | Completes reasoned argument; $\frac{33}{2} - 32\ln 2$ or $32\ln 2 - \frac{33}{2}$ AG; brackets must be correct throughout |
| Shaded region is below $x$-axis; area $= 32\ln 2 - \frac{33}{2}$ | E1 | Explains change of sign due to shaded region being below $x$-axis |14 The region bounded by the curve
$$y = ( 2 x - 8 ) \ln x$$
and the $x$-axis is shaded in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{22ff390e-1360-43bd-8c7f-3d2b58627e91-26_867_908_543_566}
14
\begin{enumerate}[label=(\alph*)]
\item 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.\\
14
\item Show that the exact area is given by
$$32 \ln 2 - \frac { 33 } { 2 }$$
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2022 Q14 [9]}}