| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard C4 techniques: calculator work for table completion, trapezium rule application with given ordinates, and volume of revolution integration. The integration in part (c) is routine (∫tan x dx = ln|sec x|) with no algebraic complications. All steps are textbook exercises requiring careful execution rather than problem-solving insight, making it slightly easier than average. |
| Spec | 1.09f Trapezium rule: numerical integration4.08d Volumes of revolution: about x and y axes |
| \(x\) | 0 | \(\frac { \pi } { 16 }\) | \(\frac { \pi } { 8 }\) | \(\frac { 3 \pi } { 16 }\) | \(\frac { \pi } { 4 }\) |
| \(y\) | 0 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = \frac{\pi}{16}\): \(y = 0.445995927...\) (awrt 0.44600) | B1 | 0 can be implied |
| \(x = \frac{\pi}{8}\): \(y = 0.643594252...\) (awrt 0.64359) | B1 | |
| \(x = \frac{3\pi}{16}\): \(y = 0.817421946...\) (awrt 0.81742) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Outside brackets: \(\frac{1}{2} \times \frac{\pi}{16}\) or \(\frac{\pi}{32}\) | B1 | For structure of trapezium rule |
| Area \(\approx \frac{1}{2} \times \frac{\pi}{16} \times \{0 + 2(0.44600 + 0.64359 + 0.81742) + 1\}\) | M1\(\checkmark\) | Correct rule \(\{\ldots\}\); correct expression inside brackets all multiplied by \(\frac{h}{2}\) |
| A1\(\checkmark\) | ||
| \(= \frac{\pi}{32} \times 4.81402... = 0.472615308... = \underline{0.4726}\) (4dp) | A1 cao | For seeing 0.4726 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area \(\approx \frac{\pi}{16} \times \left\{\frac{0+0.44600}{2} + \frac{0.44600+0.64359}{2} + \frac{0.64359+0.81742}{2} + \frac{0.81742+1}{2}\right\}\) | B1 | \(\frac{\pi}{16}\) and a divisor of 2 on all terms inside brackets |
| which is equivalent to: Area \(\approx \frac{1}{2} \times \frac{\pi}{16} \times \{0 + 2(0.44600 + 0.64359 + 0.81742) + 1\}\) | M1\(\checkmark\) | One of first and last ordinates, two of middle ordinates inside brackets ignoring the 2 |
| A1\(\checkmark\) | Correct expression inside brackets if \(\frac{1}{2}\) was to be factorised out | |
| \(= \frac{\pi}{16} \times 2.40701... = 0.472615308... = \underline{0.4726}\) | A1 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Volume \(= (\pi)\int_0^{\frac{\pi}{4}} (\sqrt{\tan x})^2\, dx = (\pi)\int_0^{\frac{\pi}{4}} \tan x\, dx\) | M1 | \(\int(\sqrt{\tan x})^2\,dx\) or \(\int \tan x\,dx\) can be implied; ignore limits and \((\pi)\) |
| \(= (\pi)[\ln\sec x]_0^{\frac{\pi}{4}}\) or \((\pi)[-\ln\cos x]_0^{\frac{\pi}{4}}\) | A1 | \(\tan x \to \ln\sec x\) or \(\tan x \to -\ln\cos x\) |
| \(= (\pi)[(\ln\sec\frac{\pi}{4}) - (\ln\sec 0)]\) or \(= (\pi)[(-\ln\cos\frac{\pi}{4}) - (\ln\cos 0)]\) | dM1 | Correct use of limits \(x=\frac{\pi}{4}\) minus \(x=0\); \(\ln(\sec 0)=0\) may be implied; ignore \((\pi)\) |
| \(= \pi\left[\ln\!\left(\frac{1}{\frac{1}{\sqrt{2}}}\right) - \ln\!\left(\frac{1}{1}\right)\right] = \pi\left[\ln\sqrt{2} - \ln 1\right]\) or \(= \pi\left[-\ln\!\left(\frac{1}{\sqrt{2}}\right) - \ln(1)\right]\) | ||
| \(= \pi\ln\sqrt{2}\) or \(\pi\ln\frac{2}{\sqrt{2}}\) or \(\frac{1}{2}\pi\ln 2\) or \(-\pi\ln\!\left(\frac{1}{\sqrt{2}}\right)\) or \(\frac{\pi}{2}\ln\!\left(\frac{1}{2}\right)\) | A1 aef | Must be exact |
# Question 7:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \frac{\pi}{16}$: $y = 0.445995927...$ (awrt 0.44600) | B1 | 0 can be implied |
| $x = \frac{\pi}{8}$: $y = 0.643594252...$ (awrt 0.64359) | B1 | |
| $x = \frac{3\pi}{16}$: $y = 0.817421946...$ (awrt 0.81742) | B1 | |
## Part (b) — Way 1
| Answer | Marks | Guidance |
|--------|-------|----------|
| Outside brackets: $\frac{1}{2} \times \frac{\pi}{16}$ or $\frac{\pi}{32}$ | B1 | For structure of trapezium rule |
| Area $\approx \frac{1}{2} \times \frac{\pi}{16} \times \{0 + 2(0.44600 + 0.64359 + 0.81742) + 1\}$ | M1$\checkmark$ | Correct rule $\{\ldots\}$; correct expression inside brackets all multiplied by $\frac{h}{2}$ |
| | A1$\checkmark$ | |
| $= \frac{\pi}{32} \times 4.81402... = 0.472615308... = \underline{0.4726}$ (4dp) | A1 cao | For seeing 0.4726 |
## Part (b) — Way 2 (Aliter)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area $\approx \frac{\pi}{16} \times \left\{\frac{0+0.44600}{2} + \frac{0.44600+0.64359}{2} + \frac{0.64359+0.81742}{2} + \frac{0.81742+1}{2}\right\}$ | B1 | $\frac{\pi}{16}$ and a divisor of 2 on all terms inside brackets |
| which is equivalent to: Area $\approx \frac{1}{2} \times \frac{\pi}{16} \times \{0 + 2(0.44600 + 0.64359 + 0.81742) + 1\}$ | M1$\checkmark$ | One of first and last ordinates, two of middle ordinates inside brackets ignoring the 2 |
| | A1$\checkmark$ | Correct expression inside brackets if $\frac{1}{2}$ was to be factorised out |
| $= \frac{\pi}{16} \times 2.40701... = 0.472615308... = \underline{0.4726}$ | A1 cao | |
> Note: Area $= \frac{1}{2} \times \frac{\pi}{20} \times \{0 + 2(0.44600 + 0.64359 + 0.81742) + 1\} = 0.3781$ gains B0M1A1A0
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Volume $= (\pi)\int_0^{\frac{\pi}{4}} (\sqrt{\tan x})^2\, dx = (\pi)\int_0^{\frac{\pi}{4}} \tan x\, dx$ | M1 | $\int(\sqrt{\tan x})^2\,dx$ or $\int \tan x\,dx$ can be implied; ignore limits and $(\pi)$ |
| $= (\pi)[\ln\sec x]_0^{\frac{\pi}{4}}$ or $(\pi)[-\ln\cos x]_0^{\frac{\pi}{4}}$ | A1 | $\tan x \to \ln\sec x$ or $\tan x \to -\ln\cos x$ |
| $= (\pi)[(\ln\sec\frac{\pi}{4}) - (\ln\sec 0)]$ or $= (\pi)[(-\ln\cos\frac{\pi}{4}) - (\ln\cos 0)]$ | dM1 | Correct use of limits $x=\frac{\pi}{4}$ minus $x=0$; $\ln(\sec 0)=0$ may be implied; ignore $(\pi)$ |
| $= \pi\left[\ln\!\left(\frac{1}{\frac{1}{\sqrt{2}}}\right) - \ln\!\left(\frac{1}{1}\right)\right] = \pi\left[\ln\sqrt{2} - \ln 1\right]$ or $= \pi\left[-\ln\!\left(\frac{1}{\sqrt{2}}\right) - \ln(1)\right]$ | | |
| $= \pi\ln\sqrt{2}$ or $\pi\ln\frac{2}{\sqrt{2}}$ or $\frac{1}{2}\pi\ln 2$ or $-\pi\ln\!\left(\frac{1}{\sqrt{2}}\right)$ or $\frac{\pi}{2}\ln\!\left(\frac{1}{2}\right)$ | A1 aef | Must be exact |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b73fe78-cc47-4615-9cfb-0b8d9ec0ffda-09_627_606_244_667}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows part of the curve with equation $y = \sqrt { } ( \tan x )$. The finite region $R$, which is bounded by the curve, the $x$-axis and the line $x = \frac { \pi } { 4 }$, is shown shaded in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Given that $y = \sqrt { } ( \tan x )$, complete the table with the values of $y$ corresponding to $x = \frac { \pi } { 16 } , \frac { \pi } { 8 }$ and $\frac { 3 \pi } { 16 }$, giving your answers to 5 decimal places.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & $\frac { \pi } { 16 }$ & $\frac { \pi } { 8 }$ & $\frac { 3 \pi } { 16 }$ & $\frac { \pi } { 4 }$ \\
\hline
$y$ & 0 & & & & 1 \\
\hline
\end{tabular}
\end{center}
\item Use the trapezium rule with all the values of $y$ in the completed table to obtain an estimate for the area of the shaded region $R$, giving your answer to 4 decimal places.
The region $R$ is rotated through $2 \pi$ radians around the $x$-axis to generate a solid of revolution.
\item Use integration to find an exact value for the volume of the solid generated.
\section*{LO}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2007 Q7 [11]}}