| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with logarithmic functions |
| Difficulty | Standard +0.3 This is a standard C4 volumes of revolution question with straightforward integration by parts (part i is given as a hint), routine application of the volume formula about the x-axis, and a slightly more conceptual part (ii)(b) requiring understanding of the cylindrical shell method or subtraction approach. The integration techniques are standard and the question provides significant scaffolding through part (i). |
| Spec | 1.08i Integration by parts4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Treat \(x \ln x\) as a product | M1 | If \(\int \ln x\), use parts \(u = \ln x\), \(dv = 1\) |
| Obtain \(x \cdot \frac{1}{x} + \ln x\) | A1 | \(x \ln x - \int 1\, dx = x \ln x - x\) |
| Show \(x \cdot \frac{1}{x} + \ln x - 1 = \ln x\) WWW AG | A1 3 | And state given result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Part (a) is mainly based on the indefinite integral \(\int (\ln x)^2\, dx\). A candidate stating e.g. \(\int(\ln x)^2\, dx = \int 2\ln x\, dx\) or \(= \int(\ln x - x)^2\, dx\) is awarded 0 for (ii)(a) | — | — |
| Correct use of \(\int \ln x\, dx = x\ln x - x\) anywhere in this part | B1 | Quoted from (i) or derived |
| Use integration by parts on \(\int(\ln x)^2\, dx\) with \(u = \ln x\), \(dv = \ln x\) | M1 | or \(u = (\ln x)^2\), \(dv = 1\) |
| [For 'integration by parts', candidates must get to a 1st stage with format \(f(x) +/- \int g(x)\, dx\)] | — | — |
| \(1^{\text{st}}\) stage \(= \ln x(x\ln x - x) - \int \frac{1}{x}(x\ln x - x)\, dx\) soi | A1 | \(x(\ln x)^2 - \int x \cdot \frac{2}{x} \ln x\, dx\) |
| \(2^{\text{nd}}\) stage \(= x(\ln x)^2 - 2x\ln x + 2x\) AEF (unsimplified) | A1 | — |
| \(\therefore\) Value of definite integral between \(1\) & \(e\) \(= e - 2\) cao | A1 | Use limits on \(2^{\text{nd}}\) stage & produce cao |
| Volume \(= \pi(e-2)\) ISW | A1 6 | Answer as decimal value (only) \(\to\) A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Attempt to connect \(dx\) and \(du\) | M1 | — |
| Becomes \(\int u^2 e^u\, du\) | A1 | — |
| First stage \(u^2 e^u - \int 2u\, e^u\, du\) | A1 | — |
| Third stage \((u^2 - 2u + 2)e^u\) | A1 | — |
| Final A1 A1 available as before | — | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Indication that required vol \(=\) vol cylinder \(-\) vol inner solid | M1 | — |
| Clear demonstration of either vol of cylinder being \(\pi e^2\) (including reason for height \(= \ln e\)) or rotation of \(x = e\) about the \(y\)-axis (including upper limit of \(y = \ln e\)) | A1 | Could appear as \(\pi \int_0^1 e^2\, dy\) |
| \((\pi)\int x^2\, dy = (\pi)\int e^{2y}\, dy\) | B1 | — |
| \(\dfrac{\pi(e^2 + 1)}{2}\) or \(13.2\) or \(13.18\) or better | B1 4 | May be from graphical calculator |
## Question 9:
### Part (i):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Treat $x \ln x$ as a product | M1 | If $\int \ln x$, use parts $u = \ln x$, $dv = 1$ |
| Obtain $x \cdot \frac{1}{x} + \ln x$ | A1 | $x \ln x - \int 1\, dx = x \ln x - x$ |
| Show $x \cdot \frac{1}{x} + \ln x - 1 = \ln x$ WWW **AG** | A1 **3** | And state given result |
---
### Part (ii)(a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Part (a) is mainly based on the indefinite integral $\int (\ln x)^2\, dx$. A candidate stating e.g. $\int(\ln x)^2\, dx = \int 2\ln x\, dx$ or $= \int(\ln x - x)^2\, dx$ is awarded **0** for (ii)(a) | — | — |
| Correct use of $\int \ln x\, dx = x\ln x - x$ anywhere in this part | B1 | Quoted from (i) or derived |
| Use integration by parts on $\int(\ln x)^2\, dx$ with $u = \ln x$, $dv = \ln x$ | M1 | or $u = (\ln x)^2$, $dv = 1$ |
| [For 'integration by parts', candidates must get to a 1st stage with format $f(x) +/- \int g(x)\, dx$] | — | — |
| $1^{\text{st}}$ stage $= \ln x(x\ln x - x) - \int \frac{1}{x}(x\ln x - x)\, dx$ soi | A1 | $x(\ln x)^2 - \int x \cdot \frac{2}{x} \ln x\, dx$ |
| $2^{\text{nd}}$ stage $= x(\ln x)^2 - 2x\ln x + 2x$ AEF (unsimplified) | A1 | — |
| $\therefore$ Value of definite integral between $1$ & $e$ $= e - 2$ cao | A1 | Use limits on $2^{\text{nd}}$ stage & produce cao |
| Volume $= \pi(e-2)$ ISW | A1 **6** | Answer as decimal value (only) $\to$ A0 |
**Alternative method when substitution $u = \ln x$ used:**
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Attempt to connect $dx$ and $du$ | M1 | — |
| Becomes $\int u^2 e^u\, du$ | A1 | — |
| First stage $u^2 e^u - \int 2u\, e^u\, du$ | A1 | — |
| Third stage $(u^2 - 2u + 2)e^u$ | A1 | — |
| Final A1 A1 available as before | — | — |
---
### Part (ii)(b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Indication that required vol $=$ vol cylinder $-$ vol inner solid | M1 | — |
| Clear demonstration of either vol of cylinder being $\pi e^2$ (including reason for height $= \ln e$) or rotation of $x = e$ about the $y$-axis (including upper limit of $y = \ln e$) | A1 | Could appear as $\pi \int_0^1 e^2\, dy$ |
| $(\pi)\int x^2\, dy = (\pi)\int e^{2y}\, dy$ | B1 | — |
| $\dfrac{\pi(e^2 + 1)}{2}$ or $13.2$ or $13.18$ or better | B1 **4** | May be from graphical calculator |
**Total: 13 marks**9 (i) Show that $\frac { \mathrm { d } } { \mathrm { d } x } ( x \ln x - x ) = \ln x$.\\
(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{8492b214-aaac-4354-8649-e317bf7b3535-3_481_725_1064_751}
In the diagram, $C$ is the curve $y = \ln x$. The region $R$ is bounded by $C$, the $x$-axis and the line $x = \mathrm { e }$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact volume of the solid of revolution formed by rotating $R$ completely about the $x$-axis.
\item The region $R$ is rotated completely about the $y$-axis. Explain why the volume of the solid of revolution formed is given by
$$\pi \mathrm { e } ^ { 2 } - \pi \int _ { 0 } ^ { 1 } \mathrm { e } ^ { 2 y } \mathrm {~d} y ,$$
and find this volume.
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2011 Q9 [13]}}