| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Standard +0.3 This is a standard C4 integration question combining numerical methods (trapezium rule) with substitution integration. Part (a) requires simple function evaluation, (b) is routine trapezium rule application, (c) is a guided substitution (with 'show that' making it easier), and (d) compares numerical to exact answer. All techniques are core C4 syllabus with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08h Integration by substitution1.09f Trapezium rule: numerical integration |
| \(x\) | 0 | \(\frac { \pi } { 8 }\) | \(\frac { \pi } { 4 }\) | \(\frac { 3 \pi } { 8 }\) | \(\frac { \pi } { 2 }\) |
| \(y\) | 0 | 1.17157 | 1.02280 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.73508\) | B1 cao | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Outside brackets: \(\frac{1}{2} \times \frac{\pi}{8}\) | B1 | oe, or awrt \(0.196\) |
| \(\times\left[0 + 2(\text{their } 0.73508 + 1.17157 + 1.02280) + 0\right]\) | M1 | Structure of trapezium rule; 0 can be implied |
| \(= \frac{\pi}{16} \times 5.8589... = 1.150392325... \approx 1.1504\) (4 dp) | A1 | awrt \(1.1504\) [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u = 1 + \cos x \Rightarrow \frac{du}{dx} = -\sin x\) | B1 | |
| \(\sin 2x = 2\sin x\cos x\) applied | B1 | |
| After substitution: \(\pm k\int\frac{(u-1)}{u}\,du\) or \(\pm k\int\frac{(1-u)}{u}\,du\) | M1 | Allow "invisible" brackets |
| \(= 4\int\left(\frac{1}{u} - 1\right)du \rightarrow \pm k(\ln u - u)\) | dM1 | Dependent on previous M1; divide each term by \(u\) |
| \(= 4\ln(1+\cos x) - 4\cos x + k\) | A1 cso | Must combine \(+c\) and \(-4\) to give \(4\ln(1+\cos x) - 4\cos x + k\) [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Apply limits \(x = \frac{\pi}{2}\) and \(x = 0\) to \(\left\{4\ln(1+\cos x) - 4\cos x\right\}\) | M1 | Either way round |
| \(= \left[4\ln 1 - 0\right] - \left[4\ln 2 - 4\right]\) | ||
| \(= 4 - 4\ln 2\) \(\{= 1.227411278...\}\) | A1 | \(\pm 4(1-\ln 2)\) or \(\pm(4-4\ln 2)\) or awrt \(\pm 1.2\), however found |
| Error \(= \ | (4 - 4\ln 2) - 1.1504...\ | = 0.0770112776... \approx 0.077\) (2sf) |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.73508$ | B1 cao | **[1]** |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Outside brackets: $\frac{1}{2} \times \frac{\pi}{8}$ | B1 | oe, or awrt $0.196$ |
| $\times\left[0 + 2(\text{their } 0.73508 + 1.17157 + 1.02280) + 0\right]$ | M1 | Structure of trapezium rule; 0 can be implied |
| $= \frac{\pi}{16} \times 5.8589... = 1.150392325... \approx 1.1504$ (4 dp) | A1 | awrt $1.1504$ **[3]** |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = 1 + \cos x \Rightarrow \frac{du}{dx} = -\sin x$ | B1 | |
| $\sin 2x = 2\sin x\cos x$ applied | B1 | |
| After substitution: $\pm k\int\frac{(u-1)}{u}\,du$ or $\pm k\int\frac{(1-u)}{u}\,du$ | M1 | Allow "invisible" brackets |
| $= 4\int\left(\frac{1}{u} - 1\right)du \rightarrow \pm k(\ln u - u)$ | dM1 | Dependent on previous M1; divide each term by $u$ |
| $= 4\ln(1+\cos x) - 4\cos x + k$ | A1 cso | Must combine $+c$ and $-4$ to give $4\ln(1+\cos x) - 4\cos x + k$ **[5]** |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Apply limits $x = \frac{\pi}{2}$ and $x = 0$ to $\left\{4\ln(1+\cos x) - 4\cos x\right\}$ | M1 | Either way round |
| $= \left[4\ln 1 - 0\right] - \left[4\ln 2 - 4\right]$ | | |
| $= 4 - 4\ln 2$ $\{= 1.227411278...\}$ | A1 | $\pm 4(1-\ln 2)$ or $\pm(4-4\ln 2)$ or awrt $\pm 1.2$, however found |
| Error $= \|(4 - 4\ln 2) - 1.1504...\| = 0.0770112776... \approx 0.077$ (2sf) | A1 cso | awrt $\pm 0.077$ or awrt $\pm 6.3(\%)$ **[3]** |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8c963567-d751-4898-b7a7-7095d90514f0-09_639_1179_246_386}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the curve with equation $y = \frac { 2 \sin 2 x } { ( 1 + \cos x ) } , 0 \leqslant x \leqslant \frac { \pi } { 2 }$.\\
The finite region $R$, shown shaded in Figure 3, is bounded by the curve and the $x$-axis.
The table below shows corresponding values of $x$ and $y$ for $y = \frac { 2 \sin 2 x } { ( 1 + \cos x ) }$.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
$x$ & 0 & $\frac { \pi } { 8 }$ & $\frac { \pi } { 4 }$ & $\frac { 3 \pi } { 8 }$ & $\frac { \pi } { 2 }$ \\
\hline
$y$ & 0 & & 1.17157 & 1.02280 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Complete the table above giving the missing value of $y$ to 5 decimal places.
\item Use the trapezium rule, with all the values of $y$ in the completed table, to obtain an estimate for the area of $R$, giving your answer to 4 decimal places.
\item Using the substitution $u = 1 + \cos x$, or otherwise, show that
$$\int \frac { 2 \sin 2 x } { ( 1 + \cos x ) } d x = 4 \ln ( 1 + \cos x ) - 4 \cos x + k$$
where $k$ is a constant.
\item Hence calculate the error of the estimate in part (b), giving your answer to 2 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2012 Q6 [12]}}