| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume requiring substitution or integration by parts |
| Difficulty | Challenging +1.2 This is a structured volumes of revolution question where part (a) provides the key integration result needed for part (b). Students must recognize that V = π∫y² dx leads to π∫x sin²(2x) dx, then apply the given substitution. While it requires connecting multiple steps and careful algebraic manipulation, the scaffolding in part (a) significantly reduces the problem-solving demand, making it moderately above average difficulty for C4. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{du}{dx} = \frac{1}{2} - \frac{1}{2}\cos 4x = \frac{1}{2} - \frac{1}{2}(1 - 2\sin^2 2x) = \sin^2 2x\) | M1 A1; M1 A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(V = \pi\int x\sin^2 2x\,dx\) | M1 | |
| \(= \pi\left[x\left(\frac{x}{2} - \frac{1}{8}\sin 4x\right) - \int\frac{x}{2} - \frac{1}{8}\sin 4x\,dx\right]_0^{\frac{\pi}{4}}\) | M1 A1 A1 | |
| \(= \pi\left[\frac{x^2}{2} - \frac{x}{8}\sin 4x - \left(\frac{x^2}{4} + \frac{1}{32}\cos 4x\right)\right]_0^{\frac{\pi}{4}}\) | M1 A1 | |
| \(= \pi\left[\frac{\pi^2}{64} + \frac{1}{32} + \frac{1}{32}\right] = \pi\left[\frac{\pi^2}{64} + \frac{1}{16}\right]\) | M1 A1 | (8 marks) |
# Question 7:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{du}{dx} = \frac{1}{2} - \frac{1}{2}\cos 4x = \frac{1}{2} - \frac{1}{2}(1 - 2\sin^2 2x) = \sin^2 2x$ | M1 A1; M1 A1 | (4 marks) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $V = \pi\int x\sin^2 2x\,dx$ | M1 | |
| $= \pi\left[x\left(\frac{x}{2} - \frac{1}{8}\sin 4x\right) - \int\frac{x}{2} - \frac{1}{8}\sin 4x\,dx\right]_0^{\frac{\pi}{4}}$ | M1 A1 A1 | |
| $= \pi\left[\frac{x^2}{2} - \frac{x}{8}\sin 4x - \left(\frac{x^2}{4} + \frac{1}{32}\cos 4x\right)\right]_0^{\frac{\pi}{4}}$ | M1 A1 | |
| $= \pi\left[\frac{\pi^2}{64} + \frac{1}{32} + \frac{1}{32}\right] = \pi\left[\frac{\pi^2}{64} + \frac{1}{16}\right]$ | M1 A1 | (8 marks) |
---7. (a) Given that $u = \frac { x } { 2 } - \frac { 1 } { 8 } \sin 4 x$, show that $\frac { \mathrm { d } u } { \mathrm {~d} x } = \sin ^ { 2 } 2 x$.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{0191bf56-a59e-44fe-af8c-bad796156f63-5_697_1239_587_367}
\end{center}
\end{figure}
Figure 2 shows the finite region bounded by the curve $y = x ^ { \frac { 1 } { 2 } } \sin 2 x$, the line $x = \frac { \pi } { 4 }$ and the $x$-axis. This region is rotated through $2 \pi$ radians about the $x$-axis.\\
(b) Using the result in part (a), or otherwise, find the exact value of the volume generated.\\
(8)\\
\hfill \mbox{\textit{Edexcel C4 Q7 [12]}}