| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with logarithmic functions |
| Difficulty | Standard +0.3 Part (a) is a straightforward trapezium rule application with simple function evaluation. Part (b) requires setting up a volume of revolution integral and using integration by parts, which is standard C4 content. The integration, while requiring careful algebraic manipulation, follows a predictable pattern for this type of question. This is slightly easier than average due to being a routine textbook-style question with well-signposted methods. |
| Spec | 1.09f Trapezium rule: numerical integration4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(x\) | \(0\) |
| \(y\) | ||
| area \(= \frac{1}{2} \times 1 \times [0 + 3.144 + 2(1.665)] = 3.24\) (3sf) | B1 M1 A1 | |
| (b) volume \(= \pi\int_1^3 x^2\ln x \, dx\) | M1 | |
| \(u = \ln x, \quad u' = \frac{1}{x}, \quad v' = x^2, \quad v = \frac{1}{3}x^3\) | ||
| \(I = \frac{1}{3}x^3\ln x - \int \frac{1}{3}x^2 \, dx\) | M1 A2 | |
| \(= \frac{1}{3}x^3\ln x - \frac{1}{9}x^3 + c\) | A1 | |
| volume \(= \pi[\frac{1}{3}x^3\ln x - \frac{1}{9}x^3]_1^3\) | ||
| \(= \pi[(9\ln 3 - 3) - (0 - \frac{1}{9})]\) | M1 | |
| \(= \pi(9\ln 3 - \frac{26}{9})\) | A1 | (11) |
**(a)** | $x$ | $0$ | $1.665$ | $2$ | $3$ | $3.144$ |
| $y$ | | | | | |
area $= \frac{1}{2} \times 1 \times [0 + 3.144 + 2(1.665)] = 3.24$ (3sf) | B1 M1 A1 |
**(b)** volume $= \pi\int_1^3 x^2\ln x \, dx$ | M1 |
$u = \ln x, \quad u' = \frac{1}{x}, \quad v' = x^2, \quad v = \frac{1}{3}x^3$ |
$I = \frac{1}{3}x^3\ln x - \int \frac{1}{3}x^2 \, dx$ | M1 A2 |
$= \frac{1}{3}x^3\ln x - \frac{1}{9}x^3 + c$ | A1 |
volume $= \pi[\frac{1}{3}x^3\ln x - \frac{1}{9}x^3]_1^3$ |
$= \pi[(9\ln 3 - 3) - (0 - \frac{1}{9})]$ | M1 |
$= \pi(9\ln 3 - \frac{26}{9})$ | A1 | (11) |
---\includegraphics{figure_1}
Figure 1 shows the curve with equation $y = x\sqrt{\ln x}$, $x \geq 1$.
The shaded region is bounded by the curve, the $x$-axis and the line $x = 3$.
\begin{enumerate}[label=(\alph*)]
\item Using the trapezium rule with two intervals of equal width, estimate the area of the shaded region. [4]
\end{enumerate}
The shaded region is rotated through $360°$ about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the exact volume of the solid formed. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q4 [11]}}