| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Tangent or Normal Bounded Area |
| Difficulty | Challenging +1.2 This question requires finding a normal line equation, setting up an area integral with two parts (curve and line), and integrating x ln x by parts. While it involves multiple steps and integration by parts, these are all standard A-level techniques with clear structure. The 10 marks reflect length rather than conceptual difficulty, making it moderately above average but not requiring novel insight. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = \ln x + x\left(\frac{1}{x}\right) = 1 + \ln x\) | M1, A1 | 2.1, 1.1b — Product rule giving \(\ln x + x\cdot g'(x)\) where \(g(x)=\ln x\) |
| \(x=\mathrm{e},\ m_T = 2 \Rightarrow m_N = -\frac{1}{2} \Rightarrow y - \mathrm{e} = -\frac{1}{2}(x-\mathrm{e})\) | M1 | 3.1a — Complete strategy to find \(x\)-coordinate where normal meets \(x\)-axis |
| \(y=0 \Rightarrow -\mathrm{e} = -\frac{1}{2}(x-\mathrm{e}) \Rightarrow x = 3\mathrm{e}\) | A1 | 1.1b — (allow \(x = 2\mathrm{e}+\mathrm{e}\ln\mathrm{e}\), awrt 8.15) |
| Area under curve \(= \int_1^{\mathrm{e}} x\ln x\,\mathrm{d}x\) or Area under line \(= \frac{1}{2}(x_A - \mathrm{e})\mathrm{e}\) | M1 | 2.1 |
| \(\int x\ln x\,\mathrm{d}x = \frac{1}{2}x^2\ln x - \int\frac{1}{x}\cdot\frac{x^2}{2}\,\mathrm{d}x\) | M1 | 2.1 — Integration by parts correct way: \(Ax^2\ln x - \int B\left(\frac{x^2}{x}\right)\mathrm{d}x\) |
| \(= \frac{1}{2}x^2\ln x - \frac{1}{4}x^2\) | dM1, A1 | 1.1b — Integrates second term to give \(\pm\lambda x^2\) |
| \(\text{Area}(R_1) = \left[\frac{1}{2}x^2\ln x - \frac{1}{4}x^2\right]_1^{\mathrm{e}}\); \(\text{Area}(R_2) = \frac{1}{2}(3\mathrm{e}-\mathrm{e})\mathrm{e}\) | M1 | 3.1a — Complete strategy: sum of two areas |
| \(\text{Area}(R) = \frac{5}{4}\mathrm{e}^2 + \frac{1}{4}\) | A1 | 1.1b |
| (10) |
# Question 13:
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{\mathrm{d}y}{\mathrm{d}x} = \ln x + x\left(\frac{1}{x}\right) = 1 + \ln x$ | M1, A1 | 2.1, 1.1b — Product rule giving $\ln x + x\cdot g'(x)$ where $g(x)=\ln x$ |
| $x=\mathrm{e},\ m_T = 2 \Rightarrow m_N = -\frac{1}{2} \Rightarrow y - \mathrm{e} = -\frac{1}{2}(x-\mathrm{e})$ | M1 | 3.1a — Complete strategy to find $x$-coordinate where normal meets $x$-axis |
| $y=0 \Rightarrow -\mathrm{e} = -\frac{1}{2}(x-\mathrm{e}) \Rightarrow x = 3\mathrm{e}$ | A1 | 1.1b — (allow $x = 2\mathrm{e}+\mathrm{e}\ln\mathrm{e}$, awrt 8.15) |
| Area under curve $= \int_1^{\mathrm{e}} x\ln x\,\mathrm{d}x$ or Area under line $= \frac{1}{2}(x_A - \mathrm{e})\mathrm{e}$ | M1 | 2.1 |
| $\int x\ln x\,\mathrm{d}x = \frac{1}{2}x^2\ln x - \int\frac{1}{x}\cdot\frac{x^2}{2}\,\mathrm{d}x$ | M1 | 2.1 — Integration by parts correct way: $Ax^2\ln x - \int B\left(\frac{x^2}{x}\right)\mathrm{d}x$ |
| $= \frac{1}{2}x^2\ln x - \frac{1}{4}x^2$ | dM1, A1 | 1.1b — Integrates second term to give $\pm\lambda x^2$ |
| $\text{Area}(R_1) = \left[\frac{1}{2}x^2\ln x - \frac{1}{4}x^2\right]_1^{\mathrm{e}}$; $\text{Area}(R_2) = \frac{1}{2}(3\mathrm{e}-\mathrm{e})\mathrm{e}$ | M1 | 3.1a — Complete strategy: sum of two areas |
| $\text{Area}(R) = \frac{5}{4}\mathrm{e}^2 + \frac{1}{4}$ | A1 | 1.1b |
| | **(10)** | |13.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{580fc9b9-d78c-4a86-91fc-22638cb5186d-38_714_826_251_621}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve $C$ with equation $y = x \ln x , x > 0$\\
The line $l$ is the normal to $C$ at the point $P ( \mathrm { e } , \mathrm { e } )$\\
The region $R$, shown shaded in Figure 2, is bounded by the curve $C$, the line $l$ and the $x$-axis.\\
Show that the exact area of $R$ is $A e ^ { 2 } + B$ where $A$ and $B$ are rational numbers to be found.\\
(10)
\hfill \mbox{\textit{Edexcel Paper 2 2018 Q13 [10]}}