| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Apply trapezium rule to given table |
| Difficulty | Moderate -0.3 This is a standard multi-part C3/C4 question covering routine techniques: trapezium rule application with given values (straightforward arithmetic), integration by substitution (standard u-substitution with ln), definite integration, and finding a tangent equation. All parts follow textbook procedures with no novel insight required, making it slightly easier than average due to its predictable structure, though the multiple parts and algebraic manipulation keep it near average difficulty. |
| Spec | 1.07l Derivative of ln(x): and related functions1.08h Integration by substitution1.09f Trapezium rule: numerical integration |
| \(x\) | e | 2 e | 3 e | 4 e | 5 e |
| \(y\) | 0.3114 | 0.2195 | 0.1712 | 0.1416 | 0.1215 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2} \times e \times \{\ldots\}\) | B1 | See \(\frac{1}{2} \times e \times\) as part of trapezium rule or \(h = e\) stated or used |
| \(\frac{1}{2} \times h \times \{0.3114 + 0.1215 + 2(0.2195 + 0.1712 + 0.1416)\}\) | M1 | Correct structure of terms inside \(\{\ldots\}\) with their \(h\) |
| \(= 2.04\) (3 sf) | A1 | awrt 2.04; condone \(\frac{599}{800}e\) or awrt 0.749e |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(u = \ln 2x\), then \(\frac{du}{dx} = \frac{2}{2x}\) | B1 | Finds \(\frac{du}{dx} = \frac{2}{2x}\) or exact equivalent |
| \(\int \frac{1}{2x} \ln 2x\, dx = \int \frac{1}{2}u\, du = \frac{1}{4}[\ln(2x)]^2\) | M1 A1 | M1: integrates as far as \(ku^2\) or \(k[\ln(2x)]^2\); A1: \(\frac{1}{4}[\ln(2x)]^2\) cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left[\frac{1}{4}(\ln 2x)^2\right]_e^{5e} = \frac{1}{4}(\ln 10e)^2 - \frac{1}{4}(\ln 2e)^2 = 2.01\) | M1 A1 | M1: correct limits used correct way; A1: awrt 2.01 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Way 1: \(\frac{dy}{dx} = \frac{1}{2x} \times \frac{2}{2x} - \frac{1}{2x^2}\ln 2x = \left(\frac{1}{2x^2} - \frac{1}{2x^2}\ln 2x\right)\) or Way 2: \(\frac{dy}{dx} = \frac{2x \times \frac{2}{2x} - (\ln 2x)\times 2}{(2x)^2} = \left(\frac{1 - \ln 2x}{2x^2}\right)\) | M1 A1 | M1: attempts product or quotient rule achieving form \(\frac{A}{x^2} - \frac{B\ln 2x}{x^2}\) or \(\frac{\frac{Px}{x} - Q(\ln 2x)}{(2x)^2}\); A1: correct answer |
| When \(x = \frac{e^2}{2}\), \(y = \frac{2}{e^2}\) | B1 | Correct simplified \(y\)-coordinate \(\frac{2}{e^2}\) |
| Uses \(\left(\frac{e^2}{2}, \frac{2}{e^2}\right)\) with their \(\frac{dy}{dx}\Big | _{x=\frac{e^2}{2}}\) to form equation of tangent | dM1 |
| \(y = -\frac{2}{e^4}x + \frac{3}{e^2}\) cao | A1 | Must be simplified |
## Question 13:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2} \times e \times \{\ldots\}$ | B1 | See $\frac{1}{2} \times e \times$ as part of trapezium rule or $h = e$ stated or used |
| $\frac{1}{2} \times h \times \{0.3114 + 0.1215 + 2(0.2195 + 0.1712 + 0.1416)\}$ | M1 | Correct structure of terms inside $\{\ldots\}$ with their $h$ |
| $= 2.04$ (3 sf) | A1 | awrt 2.04; condone $\frac{599}{800}e$ or awrt 0.749e |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $u = \ln 2x$, then $\frac{du}{dx} = \frac{2}{2x}$ | B1 | Finds $\frac{du}{dx} = \frac{2}{2x}$ or exact equivalent |
| $\int \frac{1}{2x} \ln 2x\, dx = \int \frac{1}{2}u\, du = \frac{1}{4}[\ln(2x)]^2$ | M1 A1 | M1: integrates as far as $ku^2$ or $k[\ln(2x)]^2$; A1: $\frac{1}{4}[\ln(2x)]^2$ cao |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left[\frac{1}{4}(\ln 2x)^2\right]_e^{5e} = \frac{1}{4}(\ln 10e)^2 - \frac{1}{4}(\ln 2e)^2 = 2.01$ | M1 A1 | M1: correct limits used correct way; A1: awrt 2.01 |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Way 1: $\frac{dy}{dx} = \frac{1}{2x} \times \frac{2}{2x} - \frac{1}{2x^2}\ln 2x = \left(\frac{1}{2x^2} - \frac{1}{2x^2}\ln 2x\right)$ **or** Way 2: $\frac{dy}{dx} = \frac{2x \times \frac{2}{2x} - (\ln 2x)\times 2}{(2x)^2} = \left(\frac{1 - \ln 2x}{2x^2}\right)$ | M1 A1 | M1: attempts product or quotient rule achieving form $\frac{A}{x^2} - \frac{B\ln 2x}{x^2}$ or $\frac{\frac{Px}{x} - Q(\ln 2x)}{(2x)^2}$; A1: correct answer |
| When $x = \frac{e^2}{2}$, $y = \frac{2}{e^2}$ | B1 | Correct simplified $y$-coordinate $\frac{2}{e^2}$ |
| Uses $\left(\frac{e^2}{2}, \frac{2}{e^2}\right)$ with their $\frac{dy}{dx}\Big|_{x=\frac{e^2}{2}}$ to form equation of tangent | dM1 | Dependent on M mark; accept $y - \frac{2}{e^2} = -\frac{2}{e^4}\left(x - \frac{e^2}{2}\right)$ |
| $y = -\frac{2}{e^4}x + \frac{3}{e^2}$ cao | A1 | Must be simplified |
---13.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7d07e1ad-d87a-4eb5-a15e-05b927892915-40_495_634_207_657}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of part of the curve $C$ with equation
$$y = \frac { 1 } { 2 x } \ln 2 x , \quad x > \frac { 1 } { 2 }$$
The finite region $R$, shown shaded in Figure 4, is bounded by the curve $C$, the $x$-axis and the lines with equations $x = \mathrm { e }$ and $x = 5 \mathrm { e }$.
The table below shows corresponding values of $x$ and $y$ for $y = \frac { 1 } { 2 x } \ln 2 x$. The values for $y$ are given to 4 significant figures.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & e & 2 e & 3 e & 4 e & 5 e \\
\hline
$y$ & 0.3114 & 0.2195 & 0.1712 & 0.1416 & 0.1215 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule with all the $y$ values in the table to find an approximate value for the area of $R$, giving your answer to 3 significant figures.
\item Using the substitution $u = \ln 2 x$, or otherwise, find $\int \frac { 1 } { 2 x } \ln 2 x \mathrm {~d} x$
\item Use your answer to part (b) to find the true area of $R$, giving your answer to 3 significant figures.
\item Using calculus, find an equation for the tangent to the curve at the point where $x = \frac { \mathrm { e } ^ { 2 } } { 2 }$, giving your answer in the form $y = m x + c$ where $m$ and $c$ are exact multiples of powers of e.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2018 Q13 [13]}}