| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 7 |
| 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 C4 volumes of revolution question where part (i) guides students through the integration by parts needed for part (ii). The setup is straightforward (π∫y² dx), and the substitution from x^(1/2)e^(-x) to the given integral is direct. While it requires competent execution of integration by parts and careful algebra, the structure is typical and the hint in part (i) removes the problem-solving element. |
| Spec | 1.08i Integration by parts4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int xe^{-2x}\,dx\), let \(u = x\), \(dv/dx = e^{-2x}\) | M1 | Integration by parts with \(u = x\), \(dv/dx = e^{-2x}\) |
| \(\Rightarrow v = -\frac{1}{2}e^{-2x}\) | A1 | \(= -\frac{1}{2}xe^{-2x} + \int\frac{1}{2}e^{-2x}\,dx\) |
| \(= -\frac{1}{2}xe^{-2x} + \int\frac{1}{2}e^{-2x}\,dx\) | ||
| \(= -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + c\) | ||
| \(= -\frac{1}{4}e^{-2x}(1 + 2x) + c\)* | E1 | condone omission of \(c\) |
| or \(\frac{d}{dx}\left[-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + c\right] = -\frac{1}{2}e^{-2x} + xe^{-2x} + \frac{1}{2}e^{-2x} = xe^{-2x}\) | M1, A1, E1 | product rule |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = \int_0^2 \pi y^2\,dx\) | M1 | Using formula, condone omission of limits |
| \(= \int_0^2 \pi(x^{1/2}e^{-x})^2\,dx\) | ||
| \(= \pi\int_0^2 xe^{-2x}\,dx\) | A1 | \(y^2 = xe^{-2x}\), condone omission of limits and \(\pi\) |
| \(= \pi\left[-\frac{1}{4}e^{-2x}(1+2x)\right]_0^2\) | DM1 | condone omission of \(\pi\) (need limits) |
| \(= \pi(-\frac{1}{4}e^{-4} \cdot 5 + \frac{1}{4})\) | ||
| \(= \frac{1}{4}\pi(1 - \frac{5}{e^4})\)* | E1 | |
| [4] |
## Question 7:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int xe^{-2x}\,dx$, let $u = x$, $dv/dx = e^{-2x}$ | M1 | Integration by parts with $u = x$, $dv/dx = e^{-2x}$ |
| $\Rightarrow v = -\frac{1}{2}e^{-2x}$ | A1 | $= -\frac{1}{2}xe^{-2x} + \int\frac{1}{2}e^{-2x}\,dx$ |
| $= -\frac{1}{2}xe^{-2x} + \int\frac{1}{2}e^{-2x}\,dx$ | | |
| $= -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + c$ | | |
| $= -\frac{1}{4}e^{-2x}(1 + 2x) + c$* | E1 | condone omission of $c$ |
| or $\frac{d}{dx}\left[-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + c\right] = -\frac{1}{2}e^{-2x} + xe^{-2x} + \frac{1}{2}e^{-2x} = xe^{-2x}$ | M1, A1, E1 | product rule |
| **[3]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = \int_0^2 \pi y^2\,dx$ | M1 | Using formula, condone omission of limits |
| $= \int_0^2 \pi(x^{1/2}e^{-x})^2\,dx$ | | |
| $= \pi\int_0^2 xe^{-2x}\,dx$ | A1 | $y^2 = xe^{-2x}$, condone omission of limits and $\pi$ |
| $= \pi\left[-\frac{1}{4}e^{-2x}(1+2x)\right]_0^2$ | DM1 | condone omission of $\pi$ (need limits) |
| $= \pi(-\frac{1}{4}e^{-4} \cdot 5 + \frac{1}{4})$ | | |
| $= \frac{1}{4}\pi(1 - \frac{5}{e^4})$* | E1 | |
| **[4]** | | |
---7 (i) Show that $\int x \mathrm { e } ^ { - 2 x } \mathrm {~d} x = - \frac { 1 } { 4 } \mathrm { e } ^ { - 2 x } ( 1 + 2 x ) + c$.
A vase is made in the shape of the volume of revolution of the curve $y = x ^ { 1 / 2 } \mathrm { e } ^ { - x }$ about the $x$-axis between $x = 0$ and $x = 2$ (see Fig. 5).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-5_718_751_638_654}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
(ii) Show that this volume of revolution is $\frac { 1 } { 4 } \pi \left( 1 \frac { 5 } { \mathrm { e } ^ { 4 } } \right)$.
\hfill \mbox{\textit{OCR MEI C4 Q7 [7]}}