| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Applied context: real-world solid |
| Difficulty | Standard +0.8 This is a standard volumes of revolution question requiring integration by parts twice (for the exponential-polynomial product), but the algebraic manipulation is substantial and error-prone. Part (a) is routine setup, but part (b) requires careful repeated integration by parts with e^(2x) terms. The final answer format adds minor complexity. Slightly above average difficulty for P4/Further Maths. |
| Spec | 1.08i Integration by parts4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(V = (\pi)\int_{0}^{4}\left(\frac{1}{4}(4-x)e^x\right)^2\,dx\) | M1 | Sets up volume using volume formula; limits, \(\pi\) and \(dx\) may be missing as long as intent to square \(f(x)\) is clear |
| \(\int(f(x))^2\,dx = \frac{1}{16}\int(4-x)^2e^{2x}\,dx = \ldots\) | M1 | Attempts to square \(f(x)\); power of \(e\) should be correct and bracket expanded. Allow if \(\frac{1}{4}\) is not squared |
| \(V = \frac{\pi}{16}\int_0^4\left(x^2-8x+16\right)e^{2x}\,dx\) | A1 | Correct expression; \(\pi\) and limits must be included, no errors, \(dx\) must be present throughout |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(= \left(\frac{\pi}{16}\right)\left((x^2-8x+16)\frac{e^{2x}}{2} - \int(2x-8)\times\frac{e^{2x}}{2}\,dx\right)\) | M1 A1 | M1: Attempts IBP correct way at least once to obtain \(\alpha(x^2-8x+16)e^{2x} \pm \beta\int(ax-b)e^{2x}\,dx\). A1: Correct first application |
| \(= \left(\frac{\pi}{16}\right)\left((x^2-8x+16)\frac{e^{2x}}{2} - \left[(2x-8)\frac{e^{2x}}{4} - \int 2\times\frac{e^{2x}}{4}\,dx\right]\right)\) \(= \left(\frac{\pi}{16}\right)\left((x^2-8x+16)\frac{e^{2x}}{2} - (2x-8)\frac{e^{2x}}{4} + \frac{e^{2x}}{4}\right)\) | dM1 | Attempts IBP again in same direction on \(\int(2x-8)\times\frac{e^{2x}}{2}\,dx\) and proceeds to complete integration. Depends on first M |
| \(V = \frac{\pi}{16}\left[(x^2-8x+16)\frac{e^{2x}}{2}-(2x-8)\frac{e^{2x}}{4}+\frac{e^{2x}}{4}\right]_0^4 = \frac{\pi}{16}\left(0+0+\frac{e^8}{4}-\frac{16}{2}-\frac{8}{4}-\frac{1}{4}\right)\) | ddM1 | Applies limits and subtracts; \(K\) or \(\pi\) not needed for this mark. May be done in stages. Depends on both previous M marks |
| \(= \frac{\pi}{64}(e^8-41)\) (cm³) | A1 | Correct answer; \(\pi\) must have been introduced before final step. Condone lack of units |
## Question 7(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $V = (\pi)\int_{0}^{4}\left(\frac{1}{4}(4-x)e^x\right)^2\,dx$ | M1 | Sets up volume using volume formula; limits, $\pi$ and $dx$ may be missing as long as intent to square $f(x)$ is clear |
| $\int(f(x))^2\,dx = \frac{1}{16}\int(4-x)^2e^{2x}\,dx = \ldots$ | M1 | Attempts to square $f(x)$; power of $e$ should be correct and bracket expanded. Allow if $\frac{1}{4}$ is not squared |
| $V = \frac{\pi}{16}\int_0^4\left(x^2-8x+16\right)e^{2x}\,dx$ | A1 | Correct expression; $\pi$ and limits must be included, no errors, $dx$ must be present throughout |
---
## Question 7(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $= \left(\frac{\pi}{16}\right)\left((x^2-8x+16)\frac{e^{2x}}{2} - \int(2x-8)\times\frac{e^{2x}}{2}\,dx\right)$ | M1 A1 | M1: Attempts IBP correct way at least once to obtain $\alpha(x^2-8x+16)e^{2x} \pm \beta\int(ax-b)e^{2x}\,dx$. A1: Correct first application |
| $= \left(\frac{\pi}{16}\right)\left((x^2-8x+16)\frac{e^{2x}}{2} - \left[(2x-8)\frac{e^{2x}}{4} - \int 2\times\frac{e^{2x}}{4}\,dx\right]\right)$ $= \left(\frac{\pi}{16}\right)\left((x^2-8x+16)\frac{e^{2x}}{2} - (2x-8)\frac{e^{2x}}{4} + \frac{e^{2x}}{4}\right)$ | dM1 | Attempts IBP again in same direction on $\int(2x-8)\times\frac{e^{2x}}{2}\,dx$ and proceeds to complete integration. Depends on first M |
| $V = \frac{\pi}{16}\left[(x^2-8x+16)\frac{e^{2x}}{2}-(2x-8)\frac{e^{2x}}{4}+\frac{e^{2x}}{4}\right]_0^4 = \frac{\pi}{16}\left(0+0+\frac{e^8}{4}-\frac{16}{2}-\frac{8}{4}-\frac{1}{4}\right)$ | ddM1 | Applies limits and subtracts; $K$ or $\pi$ not needed for this mark. May be done in stages. Depends on both previous M marks |
| $= \frac{\pi}{64}(e^8-41)$ (cm³) | A1 | Correct answer; $\pi$ must have been introduced before final step. Condone lack of units |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fe07afad-9cfc-48c0-84f1-5717f81977d4-20_473_313_244_350}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fe07afad-9cfc-48c0-84f1-5717f81977d4-20_390_627_246_970}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 3 shows the design of a doorknob.\\
The shape of the doorknob is formed by rotating the curve shown in Figure 4 through $360 ^ { \circ }$ about the $x$-axis, where the units are centimetres.
The equation of the curve is given by
$$\mathrm { f } ( x ) = \frac { 1 } { 4 } ( 4 - x ) \mathrm { e } ^ { x } \quad 0 \leqslant x \leqslant 4$$
\begin{enumerate}[label=(\alph*)]
\item Show that the volume, $V \mathrm {~cm} ^ { 3 }$, of the doorknob is given by
$$V = K \int _ { 0 } ^ { 4 } \left( x ^ { 2 } - 8 x + 16 \right) \mathrm { e } ^ { 2 x } \mathrm {~d} x$$
where $K$ is a constant to be found.
\item Hence, find the exact value of the volume of the doorknob.
Give your answer in the form $p \pi \left( \mathrm { e } ^ { q } + r \right) \mathrm { cm } ^ { 3 }$ where $p , q$ and $r$ are simplified rational numbers to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2022 Q7 [8]}}