| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Multi-part: volume and area |
| Difficulty | Moderate -0.3 This is a straightforward C4 volumes of revolution question with standard integration techniques. Part (a) requires a simple u-substitution for the area, and part (b) uses the standard formula V = π∫y² dx with the same substitution. Both integrations are routine with no conceptual challenges, making this slightly easier than average for A-level. |
| 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 |
|---|---|---|
| Answer | Mark | Guidance |
| Using \(u = 1+4x \Rightarrow \frac{du}{dx} = 4\); limits: \(x=0, u=1\) & \(x=2, u=9\) | — | |
| \(\int_1^9 3u^{-\frac{1}{2}} \cdot \frac{1}{4}\,du\) | — | |
| Integrating \(\pm\lambda u^{-\frac{1}{2}}\) to give \(\pm ku^{\frac{1}{2}}\) | M1 | Ignore limits |
| \(\left[\frac{3}{2}u^{\frac{1}{2}}\right]_1^9\) | A1 | Correct integration |
| \(= \left(\frac{3}{2}\sqrt{9}\right) - \left(\frac{3}{2}(1)\right)\) | M1 | Substitutes limits \(u=9\) and \(u=1\) (or \(x=2\) and \(x=0\)) into changed function and subtracts |
| \(= \frac{9}{2} - \frac{3}{2} = 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using \(u^2 = 1+4x \Rightarrow 2u\,du = dx\); limits: \(x=0, u=1\) & \(x=2, u=3\) | — | |
| \(\int_1^3 \frac{3}{2}\,du\) | — | |
| Integrating \(\pm\lambda\) to give \(\pm ku\) | M1 | Ignore limits |
| \(\left[\frac{3}{2}u\right]_1^3\) | A1 | Correct integration |
| \(= \left(\frac{3}{2}(3)\right) - \left(\frac{3}{2}(1)\right)\) | M1 | Substitutes limits \(u=3\) and \(u=1\) (or \(x=2\) and \(x=0\)) into changed function, subtracts |
| \(= \frac{9}{2} - \frac{3}{2} = 3\) | A1 |
# Question 2 (Appendix):
## Part (a) — Aliter Way 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| Using $u = 1+4x \Rightarrow \frac{du}{dx} = 4$; limits: $x=0, u=1$ & $x=2, u=9$ | — | |
| $\int_1^9 3u^{-\frac{1}{2}} \cdot \frac{1}{4}\,du$ | — | |
| Integrating $\pm\lambda u^{-\frac{1}{2}}$ to give $\pm ku^{\frac{1}{2}}$ | M1 | Ignore limits |
| $\left[\frac{3}{2}u^{\frac{1}{2}}\right]_1^9$ | A1 | Correct integration |
| $= \left(\frac{3}{2}\sqrt{9}\right) - \left(\frac{3}{2}(1)\right)$ | M1 | Substitutes limits $u=9$ and $u=1$ (or $x=2$ and $x=0$) into changed function and subtracts |
| $= \frac{9}{2} - \frac{3}{2} = 3$ | A1 | |
## Part (a) — Aliter Way 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| Using $u^2 = 1+4x \Rightarrow 2u\,du = dx$; limits: $x=0, u=1$ & $x=2, u=3$ | — | |
| $\int_1^3 \frac{3}{2}\,du$ | — | |
| Integrating $\pm\lambda$ to give $\pm ku$ | M1 | Ignore limits |
| $\left[\frac{3}{2}u\right]_1^3$ | A1 | Correct integration |
| $= \left(\frac{3}{2}(3)\right) - \left(\frac{3}{2}(1)\right)$ | M1 | Substitutes limits $u=3$ and $u=1$ (or $x=2$ and $x=0$) into changed function, subtracts |
| $= \frac{9}{2} - \frac{3}{2} = 3$ | A1 | |
---\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a5579938-e202-4543-8513-6483ede49850-03_410_552_205_694}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows part of the curve $y = \frac { 3 } { \sqrt { } ( 1 + 4 x ) }$. The region $R$ is bounded by the curve, the $x$-axis, and the lines $x = 0$ and $x = 2$, as shown shaded in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Use integration to find the area of $R$.
The region $R$ is rotated $360 ^ { \circ }$ about the $x$-axis.
\item Use integration to find the exact value of the volume of the solid formed.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2009 Q2 [9]}}