| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Apply trapezium rule to given table |
| Difficulty | Moderate -0.3 Part (a) is a straightforward trapezium rule application with values provided in a table—pure mechanical calculation. Part (b) requires setting up and evaluating a volume of revolution integral, but the function simplifies nicely when squared (y² = x/(x²+1)), leading to a standard logarithmic integral. This is slightly easier than average due to the algebraic simplification and routine integration techniques. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.09f Trapezium rule: numerical integration |
| \(x\) | 2 | 3 | 4 | 5 | 6 | 7 |
| \(y\) | 0.6325 | 0.5477 | 0.4851 | 0.4385 | 0.4027 | 0.3742 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Strip width \(= 1\) | B1 | B1: Strip width = 1, may be implied by \(\frac{1}{2}\) in trapezium rule |
| Area \(\approx \frac{1}{2}(0.6325 + 0.3742 + 2\times(0.5477 + 0.4851 + 0.4385 + 0.4027))\) \(\left(= \frac{1}{2} \times 4.7547\right)\) | M1 | M1: Correct attempt at trapezium rule. Look for \(\frac{1}{2}h((y \text{ at } x=2)+(y \text{ at } x=7)+2(\text{sum of other } y\text{ values}))\). Must be correct with no missing and no extra values. |
| Awrt \(= 2.377\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Volume \(= (\pi)\int_2^7 \frac{x}{x^2+1}\,dx = (\pi)\left[\frac{1}{2}\ln(x^2+1)\right]_2^7\) | M1 A1 | M1: Attempts to find \(C\int \frac{x}{x^2+1}\,dx\) giving form \(D\left[\ln k(x^2+1)\right]\). A1: \((\pi)\cdot\frac{1}{2}[\ln(x^2+1)]\) correct expression with or without \(\pi\). Ignore any limits. Do not allow missing brackets around \(x^2+1\) unless implied by later work. |
| \(= \frac{(\pi)}{2}(\ln 50 - \ln 5)\) | dM1 | dM1: Dependent on previous M. Substituting \(x=7\) and \(x=2\) and subtracting either way round. |
| \(= \frac{\pi}{2}\ln 10\) | A1 | A1: \(V = \frac{\pi}{2}\ln 10\) or exact equivalent e.g. \(\pi\ln\sqrt{10}\). Allow \(V = \frac{\pi}{2}\ln |
## Question 8:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| Strip width $= 1$ | B1 | B1: Strip width = 1, may be implied by $\frac{1}{2}$ in trapezium rule |
| Area $\approx \frac{1}{2}(0.6325 + 0.3742 + 2\times(0.5477 + 0.4851 + 0.4385 + 0.4027))$ $\left(= \frac{1}{2} \times 4.7547\right)$ | M1 | M1: Correct attempt at trapezium rule. Look for $\frac{1}{2}h((y \text{ at } x=2)+(y \text{ at } x=7)+2(\text{sum of other } y\text{ values}))$. Must be correct with no missing and no extra values. |
| Awrt $= 2.377$ | A1 | |
**(3 marks)**
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| Volume $= (\pi)\int_2^7 \frac{x}{x^2+1}\,dx = (\pi)\left[\frac{1}{2}\ln(x^2+1)\right]_2^7$ | M1 A1 | M1: Attempts to find $C\int \frac{x}{x^2+1}\,dx$ giving form $D\left[\ln k(x^2+1)\right]$. A1: $(\pi)\cdot\frac{1}{2}[\ln(x^2+1)]$ correct expression **with or without** $\pi$. Ignore any limits. Do not allow missing brackets around $x^2+1$ unless implied by later work. |
| $= \frac{(\pi)}{2}(\ln 50 - \ln 5)$ | dM1 | dM1: Dependent on previous M. Substituting $x=7$ and $x=2$ and subtracting either way round. |
| $= \frac{\pi}{2}\ln 10$ | A1 | A1: $V = \frac{\pi}{2}\ln 10$ or exact equivalent e.g. $\pi\ln\sqrt{10}$. Allow $V = \frac{\pi}{2}\ln|10|$ |
**(4 marks) — (7 marks total)**8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{29b56d51-120a-4275-a761-8b8aed7bca54-24_560_1029_219_463}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation $y = \sqrt { \frac { x } { x ^ { 2 } + 1 } } , \quad x \geqslant 0$\\
The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the line with equation $x = 2$, the $x$-axis and the line with equation $x = 7$\\
The table below shows corresponding values of $x$ and $y$ for $y = \sqrt { \frac { x } { x ^ { 2 } + 1 } }$
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
$y$ & 0.6325 & 0.5477 & 0.4851 & 0.4385 & 0.4027 & 0.3742 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule, with all the values of $y$ in the table, to find an estimate for the area of $R$, giving your answer to 3 decimal places.
The region $R$ is rotated $360 ^ { \circ }$ about the $x$-axis to form a solid of revolution.
\item Use calculus to find the exact volume of the solid of revolution formed. Write your answer in its simplest form.\\
\includegraphics[max width=\textwidth, alt={}, center]{29b56d51-120a-4275-a761-8b8aed7bca54-24_2255_47_314_1979}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q8 [7]}}