| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | January |
| Marks | 13 |
| 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 multi-part integration question covering routine trapezium rule application and integration by parts. Part (a) is simple substitution, (b) is direct trapezium rule formula application, (c) requires integration by parts (a standard C4 technique), (d) is percentage error calculation, and (e) tests understanding of numerical methods. While it has multiple parts (5 marks total likely 10-12 marks), each component is textbook-standard with no novel problem-solving required. Slightly easier than average due to the structured, guided nature. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts1.09f Trapezium rule: numerical integration |
| \(x\) | 1 | 1.5 | 2 | 2.5 | 3 |
| \(y\) | 2 | 1.3041 | 0.9089 | 1.2958 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.9242\) exactly | B1 | Either in table or within trapezium rule in part (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Strip width \(= 0.5\) | B1 | |
| Area \(\approx \frac{0.5}{2}\big((2 + 1.2958 + 2\times(1.3041 + 0.9242 + 0.9089)\big)\) | M1 | Correct form of trapezium rule |
| \(= 2.393\) | A1 | awrt \(2.393\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Integration by parts correct way round on \(\frac{x^2 \ln x}{3}\); expression of form \(px^3 \ln x - \int \frac{qx^3}{x}\,dx\) | M1 | Accept on \(x^2 \ln x\) or multiples |
| \(\frac{x^3}{9}\ln x - \int \frac{x^3}{9}\times\frac{1}{x}\,dx\) | A1 | |
| Integrates \(-2x+4\) to \(-x^2 + 4x\,(+c)\) | B1 | Ignore constants |
| \(\frac{x^3}{9}\ln x - \frac{x^3}{27}\,(+c)\) correct for \(\frac{x^2\ln x}{3}\) part | A1 | Independent of \(-2x+4\) integral |
| Substitute \(x=3\) and \(x=1\) and subtract | dM1 | Dependent on M mark |
| \(= \ln 27 - \frac{26}{27}\) | A1 | Must be in this form; \(3\ln 3 \rightarrow \ln 27\) required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\%\text{ error} = \pm\frac{ | real - approx | }{real}\times 100\) |
| Accept awrt \(\pm 2.6\%\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Increase the number of strips | B1 | Accept: decrease width of strips, use more trapezia, more \(x\)/\(y\) values; do not accept "use more decimal places" |
# Question 12(a) - Trapezium Rule Table Value:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.9242$ exactly | B1 | Either in table or within trapezium rule in part (b) |
---
# Question 12(b) - Trapezium Rule:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Strip width $= 0.5$ | B1 | |
| Area $\approx \frac{0.5}{2}\big((2 + 1.2958 + 2\times(1.3041 + 0.9242 + 0.9089)\big)$ | M1 | Correct form of trapezium rule |
| $= 2.393$ | A1 | awrt $2.393$ |
---
# Question 12(c) - Integration by Parts:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Integration by parts correct way round on $\frac{x^2 \ln x}{3}$; expression of form $px^3 \ln x - \int \frac{qx^3}{x}\,dx$ | M1 | Accept on $x^2 \ln x$ or multiples |
| $\frac{x^3}{9}\ln x - \int \frac{x^3}{9}\times\frac{1}{x}\,dx$ | A1 | |
| Integrates $-2x+4$ to $-x^2 + 4x\,(+c)$ | B1 | Ignore constants |
| $\frac{x^3}{9}\ln x - \frac{x^3}{27}\,(+c)$ correct for $\frac{x^2\ln x}{3}$ part | A1 | Independent of $-2x+4$ integral |
| Substitute $x=3$ and $x=1$ and subtract | dM1 | Dependent on M mark |
| $= \ln 27 - \frac{26}{27}$ | A1 | Must be in this form; $3\ln 3 \rightarrow \ln 27$ required |
---
# Question 12(d) - Percentage Error:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\%\text{ error} = \pm\frac{|real - approx|}{real}\times 100$ | M1 | Uses answers from (b) and (c) |
| Accept awrt $\pm 2.6\%$ | A1 | |
---
# Question 12(e) - Improving Approximation:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Increase the number of strips | B1 | Accept: decrease width of strips, use more trapezia, more $x$/$y$ values; do not accept "use more decimal places" |12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{03548211-79cb-4629-b6ca-aa9dfcc77a33-21_615_732_233_605}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of part of the curve $C$ with equation
$$y = \frac { x ^ { 2 } \ln x } { 3 } - 2 x + 4 , \quad x > 0$$
The finite region $S$, shown shaded in Figure 4, is bounded by the curve $C$, the $x$-axis and the lines with equations $x = 1$ and $x = 3$
\begin{enumerate}[label=(\alph*)]
\item Complete the table below with the value of $y$ corresponding to $x = 2$. Give your answer to 4 decimal places.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 1 & 1.5 & 2 & 2.5 & 3 \\
\hline
$y$ & 2 & 1.3041 & & 0.9089 & 1.2958 \\
\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 $S$, giving your answer to 3 decimal places.
\item Use calculus to find the exact area of $S$.
Give your answer in the form $\frac { a } { b } + \ln c$, where $a , b$ and $c$ are integers.
\item Hence calculate the percentage error in using your answer to part (b) to estimate the area of $S$. Give your answer to one decimal place.
\item Explain how the trapezium rule could be used to obtain a more accurate estimate for the area of $S$.
\section*{Question 12 continued}
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13. (a) Express $10 \cos \theta - 3 \sin \theta$ in the form $R \cos ( \theta + \alpha )$, where $R > 0$ and $0 < \alpha < 90 ^ { \circ }$
Give the exact value of $R$ and give the value of $\alpha$ to 2 decimal places.
Alana models the height above the ground of a passenger on a Ferris wheel by the equation
$$H = 12 - 10 \cos ( 30 t ) ^ { \circ } + 3 \sin ( 30 t ) ^ { \circ }$$
where the height of the passenger above the ground is $H$ metres at time $t$ minutes after the wheel starts turning.\\
\includegraphics[max width=\textwidth, alt={}, center]{03548211-79cb-4629-b6ca-aa9dfcc77a33-23_419_567_516_1160}\\
(b) Calculate
\begin{enumerate}[label=(\roman*)]
\item the maximum value of $H$ predicted by this model,
\item the value of $t$ when this maximum first occurs.
Give each answer to 2 decimal places.\\
(c) Calculate the value of $t$ when the passenger is 18 m above the ground for the first time. Give your answer to 2 decimal places.\\
(d) Determine the time taken for the Ferris wheel to complete two revolutions.\\
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\section*{Question 13 continued}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2015 Q12 [13]}}