| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume requiring substitution or integration by parts |
| Difficulty | Standard +0.3 This is a standard P4 volumes of revolution question with routine integration by parts. Part (a) requires simple substitution u = x/2 to transform the integral. Part (b) is textbook integration by parts applied twice. Part (c) involves straightforward arithmetic with unit conversion. All techniques are standard for this module with no novel insights required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.08i Integration by parts4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V = \pi\int_0^{10}\left(10xe^{-\frac{1}{2}x}\right)^2 dx = 100\pi\int_0^{10}x^2 e^{-x}\,dx\) | M1, A1 | M1: correct unsimplified volume expression; A1: achieves \(100\pi\int_0^{10}x^2e^{-x}dx\); condone missing \(dx\) but limits must be present |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int x^2 e^{-x}\,dx = -x^2 e^{-x} + 2\int xe^{-x}\,dx\) | M1 | Integration by parts giving \(...x^2e^{-x}\pm...\int xe^{-x}dx\); condone missing \(dx\) |
| \(= -x^2e^{-x}+2\left\{-xe^{-x}+\int e^{-x}dx\right\}\) | dM1 | Second integration by parts giving \(...x^2e^{-x}\pm...xe^{-x}\pm...\int e^{-x}dx\); dependent on previous M |
| \(= -x^2e^{-x}-2xe^{-x}-2e^{-x}\ (+c)\) | A1 | With or without constant of integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Total volume \(= 2\times 100\pi\int_0^{10}x^2e^{-x}\,dx\) | M1 | Correct strategy with their \(k\) |
| \(\int_0^{10}x^2e^{-x} = \left[-x^2e^{-x}-2xe^{-x}-2e^{-x}\right]_0^{10} = (-(10)^2e^{-10}-2\times10e^{-10}-2e^{-10})-(-2)\) | M1 | Substitutes limits 10 and 0 and subtracts |
| \(= 2-122e^{-10}\ (1.9944...)\) | A1 | Or awrt 1.99 |
| \(\text{Density} = \dfrac{5000}{200\pi\times 1.9944...}\) | dM1 | Dependent on second M; attempts \(\frac{5000}{\text{their Volume}}\) with 5000 in numerator |
| \(\approx 3.99\ \text{g/cm}^3\) | A1 | awrt 3.99 g/cm³; accept \(\frac{5000}{200\pi(2-122e^{-10})}\) |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = \pi\int_0^{10}\left(10xe^{-\frac{1}{2}x}\right)^2 dx = 100\pi\int_0^{10}x^2 e^{-x}\,dx$ | M1, A1 | M1: correct unsimplified volume expression; A1: achieves $100\pi\int_0^{10}x^2e^{-x}dx$; condone missing $dx$ but limits must be present |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int x^2 e^{-x}\,dx = -x^2 e^{-x} + 2\int xe^{-x}\,dx$ | M1 | Integration by parts giving $...x^2e^{-x}\pm...\int xe^{-x}dx$; condone missing $dx$ |
| $= -x^2e^{-x}+2\left\{-xe^{-x}+\int e^{-x}dx\right\}$ | dM1 | Second integration by parts giving $...x^2e^{-x}\pm...xe^{-x}\pm...\int e^{-x}dx$; dependent on previous M |
| $= -x^2e^{-x}-2xe^{-x}-2e^{-x}\ (+c)$ | A1 | With or without constant of integration |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Total volume $= 2\times 100\pi\int_0^{10}x^2e^{-x}\,dx$ | M1 | Correct strategy with their $k$ |
| $\int_0^{10}x^2e^{-x} = \left[-x^2e^{-x}-2xe^{-x}-2e^{-x}\right]_0^{10} = (-(10)^2e^{-10}-2\times10e^{-10}-2e^{-10})-(-2)$ | M1 | Substitutes limits 10 and 0 and subtracts |
| $= 2-122e^{-10}\ (1.9944...)$ | A1 | Or awrt 1.99 |
| $\text{Density} = \dfrac{5000}{200\pi\times 1.9944...}$ | dM1 | Dependent on second M; attempts $\frac{5000}{\text{their Volume}}$ with 5000 in numerator |
| $\approx 3.99\ \text{g/cm}^3$ | A1 | awrt 3.99 g/cm³; accept $\frac{5000}{200\pi(2-122e^{-10})}$ |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2dffe245-b18a-4486-af8e-bad598ceb6e8-26_446_492_434_447}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2dffe245-b18a-4486-af8e-bad598ceb6e8-26_441_495_402_1139}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 2 shows the curve with equation
$$y = 10 x \mathrm { e } ^ { - \frac { 1 } { 2 } x } \quad 0 \leqslant x \leqslant 10$$
The finite region $R$, shown shaded in Figure 2, is bounded by the curve, the $x$-axis and the line with equation $x = 10$
The region $R$ is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.\\
(a) Show that the volume, $V$, of this solid is given by
$$V = k \int _ { 0 } ^ { 10 } x ^ { 2 } \mathrm { e } ^ { - x } \mathrm {~d} x$$
where $k$ is a constant to be found.\\
(b) Find $\int x ^ { 2 } e ^ { - x } d x$
Figure 3 represents an exercise weight formed by joining two of these solids together.\\
The exercise weight has mass 5 kg and is 20 cm long.\\
Given that
$$\text { density } = \frac { \text { mass } } { \text { volume } }$$
and using your answers to part (a) and part (b),\\
(c) find the density of this exercise weight. Give your answer in grams per $\mathrm { cm } ^ { 3 }$ to 3 significant figures.
\hfill \mbox{\textit{Edexcel P4 2022 Q8 [10]}}