| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Normal/tangent then area with parts |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining standard integration by parts (∫ln x dx is a textbook example), finding a normal line (routine differentiation and perpendicular gradient), and calculating an area using the result from part (a). All techniques are standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{1}{x}\) | B1 | Differentiates \(\ln x\) to give \(\frac{1}{x}\) |
| Gradient of \(C\) at \((e, 1) = \frac{1}{e}\), so gradient of \(l = -e\); \(l: y = -ex + c\); substituting \((e,1)\): \(1 = -e(e) + c \Rightarrow c = 1 + e^2\) | M1 | Complete strategy to find the x-coordinate where the normal to \(C\) at \(P(e,1)\) meets the x-axis |
| Setting \(y = 0\): \(0 = -ex + 1 + e^2 \Rightarrow x = \frac{e^2+1}{e}\) | A1 | \(l\) meets x-axis at \(x = \frac{e^2+1}{e}\) or \(e + \frac{1}{e}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(R = \int_1^e \ln x \, dx + \text{triangle } PQ(e,0)\) | M1 | Complete strategy of finding area \(R\) by finding the sum of two key areas |
| \(R = \int_1^e \ln x \, dx + \frac{\frac{1}{e} \times 1}{2}\); \(= (e\ln e - e) - (1\ln 1 - 1) + \frac{\frac{1}{e}\times 1}{2}\) | M1 | Some evidence of applying limits of \(e\) and 1 and subtracts the correct way round |
| \(= 1 + \frac{1}{2e}\) | A1 | Correct solution only |
## Question 13:
### Part b: Find the exact x-coordinate of Q
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{1}{x}$ | B1 | Differentiates $\ln x$ to give $\frac{1}{x}$ |
| Gradient of $C$ at $(e, 1) = \frac{1}{e}$, so gradient of $l = -e$; $l: y = -ex + c$; substituting $(e,1)$: $1 = -e(e) + c \Rightarrow c = 1 + e^2$ | M1 | Complete strategy to find the x-coordinate where the normal to $C$ at $P(e,1)$ meets the x-axis |
| Setting $y = 0$: $0 = -ex + 1 + e^2 \Rightarrow x = \frac{e^2+1}{e}$ | A1 | $l$ meets x-axis at $x = \frac{e^2+1}{e}$ or $e + \frac{1}{e}$ |
**(3 marks)**
### Part c: Find the exact area of R
| Working/Answer | Mark | Guidance |
|---|---|---|
| $R = \int_1^e \ln x \, dx + \text{triangle } PQ(e,0)$ | M1 | Complete strategy of finding area $R$ by finding the sum of two key areas |
| $R = \int_1^e \ln x \, dx + \frac{\frac{1}{e} \times 1}{2}$; $= (e\ln e - e) - (1\ln 1 - 1) + \frac{\frac{1}{e}\times 1}{2}$ | M1 | Some evidence of applying limits of $e$ and 1 and subtracts the correct way round |
| $= 1 + \frac{1}{2e}$ | A1 | Correct solution only |
**(3 marks)**
---\begin{enumerate}
\item a. Find $\int \ln x \mathrm {~d} x$
\end{enumerate}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f9dcb521-6aaa-4496-86e8-2dcd07838e10-22_919_1139_276_456}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve with equation
$$y = \ln x , \quad x > 0$$
The point P lies on $C$ and has coordinate $( e , 1 )$.\\
The line 1 is a normal to $C$ at $P$. The line $l$ cuts the $x$-axis at the point $Q$.\\
b. Find the exact value of the $x$-coordinate of $Q$.
The finite region $\mathbf { R }$, shown shaded in figure 3, is bounded by the curve, the line $l$ and the $x$-axis.\\
c. Find the exact area of $\mathbf { R }$.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q13 [10]}}