| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 9 |
| 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 requiring standard integration techniques. Part (a) involves integrating (3x+1)^(-1/2) using substitution or recognition, and part (b) requires squaring the function and integrating 1/(3x+1), which gives a logarithm. The algebraic manipulation to reach the form k·π·ln 2 is routine. This is slightly easier than average as it's a textbook application of standard C4 techniques with no conceptual surprises. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(= \int_1^5 \frac{1}{\sqrt{3x+1}} \, dx = [\frac{2}{3}(3x+1)^{\frac{1}{2}}]_1^5\) | M1 A1 | |
| \(= \frac{2}{3}(4-2) = \frac{4}{3}\) | M1 A1 | |
| (b) \(= \pi \int_1^5 \frac{1}{3x+1} \, dx\) | M1 | |
| \(= \pi[\frac{1}{3}\ln | 3x+1 | ]_1^5\) |
| \(= \frac{1}{3}\pi(\ln 16 - \ln 4) = \frac{1}{3}\pi \ln 4 = \frac{2}{3}\pi \ln 2\) | M1 A1 | \([k = \frac{2}{3}]\) |
**(a)** $= \int_1^5 \frac{1}{\sqrt{3x+1}} \, dx = [\frac{2}{3}(3x+1)^{\frac{1}{2}}]_1^5$ | M1 A1 |
$= \frac{2}{3}(4-2) = \frac{4}{3}$ | M1 A1 |
**(b)** $= \pi \int_1^5 \frac{1}{3x+1} \, dx$ | M1 |
$= \pi[\frac{1}{3}\ln|3x+1|]_1^5$ | M1 A1 |
$= \frac{1}{3}\pi(\ln 16 - \ln 4) = \frac{1}{3}\pi \ln 4 = \frac{2}{3}\pi \ln 2$ | M1 A1 | $[k = \frac{2}{3}]$ | **(9)**5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b71c9832-e502-4a25-85fb-a49c03ea9209-08_663_899_146_495}
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\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the curve with equation $y = \frac { 1 } { \sqrt { 3 x + 1 } }$.\\
The shaded region is bounded by the curve, the $x$-axis and the lines $x = 1$ and $x = 5$.
\begin{enumerate}[label=(\alph*)]
\item Find the area of the shaded region.
The shaded region is rotated completely about the $x$-axis.
\item Find the volume of the solid formed, giving your answer in the form $k \pi \ln 2$, where $k$ is a simplified fraction.\\
5. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q5 [9]}}