| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and area |
| Difficulty | Standard +0.3 This is a straightforward volumes of revolution question with standard trapezium rule application. Part (a) requires routine numerical integration using given strip numbers, while part (b) involves a standard integral that simplifies nicely using substitution or partial fractions. The 'show that' format provides the target answer, making it easier than an open-ended calculation. All techniques are core C4 material with no novel problem-solving required. |
| Spec | 1.09f Trapezium rule: numerical integration4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \(x\) | 0 | 0.5 |
| \(y\) | 0 | 0.5774 |
| B2 | ||
| (i) \(= \frac{1}{3} \times 1 \times [0 + 0.8660 + 2(0.7071 + 0.8165)] = 1.96\) (3sf) | B1 M1 A1 | |
| (ii) \(= \frac{1}{6} \times 0.5 \times [0 + 0.8660 + 2(0.5774 + 0.7071 + 0.7746 + 0.8165 + 0.8452)]\) \(= 2.08\) (3sf) | M1 A1 | |
| (b) \(= \pi \int_0^3 \frac{x}{x+1} dx\) | M1 | |
| \(= \pi \int_0^3 \frac{x+1-1}{x+1} dx = \pi \int_0^3 \left(1 - \frac{1}{x+1}\right) dx\) | M1 | |
| \(= \pi[x - \ln | x+1 | ]_0^3\) |
| \(= \pi[(3 - \ln 4) - (0)] = \pi(3 - \ln 4)\) | M1 A1 | (13) |
**(a)**
| $x$ | 0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 |
| $y$ | 0 | 0.5774 | 0.7071 | 0.7746 | 0.8165 | 0.8452 | 0.8660 |
| B2 |
(i) $= \frac{1}{3} \times 1 \times [0 + 0.8660 + 2(0.7071 + 0.8165)] = 1.96$ (3sf) | B1 M1 A1 |
(ii) $= \frac{1}{6} \times 0.5 \times [0 + 0.8660 + 2(0.5774 + 0.7071 + 0.7746 + 0.8165 + 0.8452)]$ $= 2.08$ (3sf) | M1 A1 |
**(b)** $= \pi \int_0^3 \frac{x}{x+1} dx$ | M1 |
$= \pi \int_0^3 \frac{x+1-1}{x+1} dx = \pi \int_0^3 \left(1 - \frac{1}{x+1}\right) dx$ | M1 |
$= \pi[x - \ln|x+1|]_0^3$ | M1 A1 |
$= \pi[(3 - \ln 4) - (0)] = \pi(3 - \ln 4)$ | M1 A1 | (13)
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**Total** (75)8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{922f404e-12d5-490b-9c8d-509f3a304c1e-14_656_999_146_429}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the curve with equation $y = \sqrt { \frac { x } { x + 1 } }$.\\
The shaded region is bounded by the curve, the $x$-axis and the line $x = 3$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Use the trapezium rule with three strips to find an estimate for the area of the shaded region.
\item Use the trapezium rule with six strips to find an improved estimate for the area of the shaded region.
The shaded region is rotated through $2 \pi$ radians about the $x$-axis.
\end{enumerate}\item Show that the volume of the solid formed is $\pi ( 3 - \ln 4 )$.\\
8. continued\\
8. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q8 [13]}}