| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with trigonometric functions |
| Difficulty | Standard +0.3 This is a standard volumes of revolution question requiring expansion of a squared expression, integration of trigonometric terms using double angle identities, and application of the volume formula. Part (i) scaffolds the main calculation needed for part (ii). While it involves multiple steps and trigonometric manipulation, it follows a routine procedure with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempt to multiply out \((x+\cos 2x)^2\) | M1 | Min of 2 correct terms |
| Finding \(\int 2x\cos 2x\,dx\): Use \(u=2x\), \(dv=\cos 2x\) | M1 | 1st stage \(f(x)+/-\int g(x)\,dx\) |
| 1st stage: \(x\sin 2x - \int \sin 2x\,dx\) | A1 | |
| \(\therefore \int 2x\cos 2x\,dx = x\sin 2x + \frac{1}{2}\cos 2x\) | A1 | |
| Finding \(\int \cos^2 2x\,dx\): Change to \(k\int +/-1+/-\cos 4x\,dx\) | M1 | where \(k=\frac{1}{2}, 2\) or \(1\) |
| Correct version \(\frac{1}{2}\int 1+\cos 4x\,dx\) | A1 | |
| \(\int \cos 4x\,dx = \frac{1}{4}\sin 4x\) | B1 | seen anywhere in this part |
| Result \(= \frac{1}{2}x + \frac{1}{8}\sin 4x\) | A1 | |
| Answer: \(\frac{1}{3}x^3 + x\sin 2x + \frac{1}{2}\cos 2x + \frac{1}{2}x + \frac{1}{8}\sin 4x\,(+c)\) | A1 | 9 marks Fully correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(V = \pi\int_0^{\frac{1}{2}\pi}(x+\cos 2x)^2\,dx\) | M1 | |
| Use limits \(0\) & \(\frac{1}{2}\pi\) correctly on their (i) answer | M1 | |
| (i) correct value \(= \frac{1}{24}\pi^3 - \frac{1}{2} + \frac{1}{4}\pi - \frac{1}{2}\) | A1 | |
| Final answer \(= \pi\left(\frac{1}{24}\pi^3 + \frac{1}{4}\pi - 1\right)\) | A1 | 4 marks c.a.o. No follow-through |
# Question 9(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempt to multiply out $(x+\cos 2x)^2$ | M1 | Min of 2 correct terms |
| **Finding $\int 2x\cos 2x\,dx$:** Use $u=2x$, $dv=\cos 2x$ | M1 | 1st stage $f(x)+/-\int g(x)\,dx$ |
| 1st stage: $x\sin 2x - \int \sin 2x\,dx$ | A1 | |
| $\therefore \int 2x\cos 2x\,dx = x\sin 2x + \frac{1}{2}\cos 2x$ | A1 | |
| **Finding $\int \cos^2 2x\,dx$:** Change to $k\int +/-1+/-\cos 4x\,dx$ | M1 | where $k=\frac{1}{2}, 2$ or $1$ |
| Correct version $\frac{1}{2}\int 1+\cos 4x\,dx$ | A1 | |
| $\int \cos 4x\,dx = \frac{1}{4}\sin 4x$ | B1 | seen anywhere in this part |
| Result $= \frac{1}{2}x + \frac{1}{8}\sin 4x$ | A1 | |
| Answer: $\frac{1}{3}x^3 + x\sin 2x + \frac{1}{2}\cos 2x + \frac{1}{2}x + \frac{1}{8}\sin 4x\,(+c)$ | A1 | **9 marks** Fully correct |
# Question 9(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $V = \pi\int_0^{\frac{1}{2}\pi}(x+\cos 2x)^2\,dx$ | M1 | |
| Use limits $0$ & $\frac{1}{2}\pi$ correctly on their (i) answer | M1 | |
| (i) correct value $= \frac{1}{24}\pi^3 - \frac{1}{2} + \frac{1}{4}\pi - \frac{1}{2}$ | A1 | |
| Final answer $= \pi\left(\frac{1}{24}\pi^3 + \frac{1}{4}\pi - 1\right)$ | A1 | **4 marks** c.a.o. No follow-through |
**Total: 13 marks**9 (i) Find $\int ( x + \cos 2 x ) ^ { 2 } \mathrm {~d} x$.\\
(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{80f94db1-39be-46f5-896e-277c93cbe4b8-3_538_935_383_646}
The diagram shows the part of the curve $y = x + \cos 2 x$ for $0 \leqslant x \leqslant \frac { 1 } { 2 } \pi$. The shaded region bounded by the curve, the axes and the line $x = \frac { 1 } { 2 } \pi$ is rotated completely about the $x$-axis to form a solid of revolution of volume $V$. Find $V$, giving your answer in an exact form.
\hfill \mbox{\textit{OCR C4 2010 Q9 [13]}}