| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Sequential multi-part (building on previous) |
| Difficulty | Standard +0.3 Part (a) involves standard integration by parts applications with trigonometric functions—routine C3 technique requiring careful execution but no novel insight. Part (b) adds a volume of revolution calculation that connects to part (a)(ii), requiring recognition that the integral simplifies to x² sin 4x after squaring. This is slightly above average difficulty due to the multi-step nature and the need to connect parts, but remains a fairly standard exam question testing core techniques. |
| Spec | 1.08i Integration by parts4.08d Volumes of revolution: about x and y axes |
| Answer | Marks |
|---|---|
| \(u = x\), \(\frac{dv}{dx} = \cos 4x \Rightarrow v = \frac{\sin 4x}{4}\) | M1 |
| \(\int x\cos 4x\,dx = \frac{x\sin 4x}{4} - \int \frac{\sin 4x}{4}dx\) | A1 |
| \(= \frac{x\sin 4x}{4} + \frac{\cos 4x}{16} + c\) | A1 A1 |
| Answer | Marks |
|---|---|
| \(u = x^2\), \(\frac{dv}{dx} = \sin 4x \Rightarrow v = -\frac{\cos 4x}{4}\) | M1 |
| \(= -\frac{x^2\cos 4x}{4} + \int \frac{2x\cos 4x}{4}dx\) | A1 |
| Use result from (i): \(= -\frac{x^2\cos 4x}{4} + \frac{x\sin 4x}{8} + \frac{\cos 4x}{32} + c\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = \pi\int_0^{0.2} y^2\,dx = \pi\int_0^{0.2} 64x^2\sin 4x\,dx\) | M1 | |
| Use result from (ii) with limits | M1 | |
| \(V \approx 0.0816\) | A1 | 3 significant figures |
# Question 7:
## Part (a)(i)
| $u = x$, $\frac{dv}{dx} = \cos 4x \Rightarrow v = \frac{\sin 4x}{4}$ | M1 | |
| $\int x\cos 4x\,dx = \frac{x\sin 4x}{4} - \int \frac{\sin 4x}{4}dx$ | A1 | |
| $= \frac{x\sin 4x}{4} + \frac{\cos 4x}{16} + c$ | A1 A1 | |
## Part (a)(ii)
| $u = x^2$, $\frac{dv}{dx} = \sin 4x \Rightarrow v = -\frac{\cos 4x}{4}$ | M1 | |
| $= -\frac{x^2\cos 4x}{4} + \int \frac{2x\cos 4x}{4}dx$ | A1 | |
| Use result from (i): $= -\frac{x^2\cos 4x}{4} + \frac{x\sin 4x}{8} + \frac{\cos 4x}{32} + c$ | M1 A1 | |
## Part (b)
| $V = \pi\int_0^{0.2} y^2\,dx = \pi\int_0^{0.2} 64x^2\sin 4x\,dx$ | M1 | |
| Use result from (ii) with limits | M1 | |
| $V \approx 0.0816$ | A1 | 3 significant figures |
---7
\begin{enumerate}[label=(\alph*)]
\item Use integration by parts to find:
\begin{enumerate}[label=(\roman*)]
\item $\quad \int x \cos 4 x \mathrm {~d} x$;\\
(4 marks)
\item $\int x ^ { 2 } \sin 4 x d x$.\\
(4 marks)
\end{enumerate}\item The region bounded by the curve $y = 8 x \sqrt { ( \sin 4 x ) }$ and the lines $x = 0$ and $x = 0.2$ is rotated through $2 \pi$ radians about the $x$-axis. Find the value of the volume of the solid generated, giving your answer to three significant figures.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2010 Q7 [11]}}