| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Improper integrals with discontinuity |
| Difficulty | Challenging +1.3 This is a multi-part Further Maths question requiring integration by parts, limit evaluation, and handling an improper integral. While it involves several techniques, each part follows standard procedures: (a) is routine integration by parts, (b) is a standard limit (L'Hôpital's rule or series), (c) requires recognizing the discontinuity at x=0, and (d) involves careful limit evaluation where terms cancel to give ln 2. The question is methodical rather than requiring novel insight, making it moderately above average difficulty but accessible to well-prepared FP3 students. |
| Spec | 1.08i Integration by parts4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks |
|---|---|
| By parts: \(u = x\), \(\frac{dv}{dx} = \cos 8x\) | M1 |
| \(= \frac{x\sin 8x}{8} - \int \frac{\sin 8x}{8}\, dx\) | A1 |
| \(= \frac{x\sin 8x}{8} + \frac{\cos 8x}{64} + c\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lim_{x \to 0}\left[\frac{1}{x}\sin 2x\right]\) | M1 | Use of L'Hôpital or small angle approximation |
| \(= \lim_{x \to 0}\frac{2\cos 2x}{1} = 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| The integrand contains \(-\frac{1}{x}\) which is undefined (has a singularity/discontinuity) at \(x = 0\) | B1 | Must reference \(x=0\) and the term \(-\frac{1}{x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \lim_{\varepsilon \to 0^+}\int_\varepsilon^{\pi/4}\left(2\cot 2x - \frac{1}{x} + x\cos 8x\right)dx\) | M1 | Correct limiting process shown |
| Integrate: \(\left[\ln\sin 2x - \ln x + \frac{x\sin 8x}{8} + \frac{\cos 8x}{64}\right]_\varepsilon^{\pi/4}\) | A1ft | Using parts (a) and (b) |
| Answer | Marks | Guidance |
|---|---|---|
| As \(\varepsilon \to 0\): \(\ln\sin 2\varepsilon - \ln\varepsilon \to \ln 2\) (using part (b)) | M1 | Correct evaluation of limit |
| \(= -\ln\frac{\pi}{4} + \frac{1}{64} - \ln 2 = \ln\frac{2}{\pi} + \frac{1}{64}\) | A1 | Single term answer |
# Question 5:
## Part (a):
$\int x\cos 8x\, dx$
By parts: $u = x$, $\frac{dv}{dx} = \cos 8x$ | M1 |
$= \frac{x\sin 8x}{8} - \int \frac{\sin 8x}{8}\, dx$ | A1 |
$= \frac{x\sin 8x}{8} + \frac{\cos 8x}{64} + c$ | A1 |
## Part (b):
$\lim_{x \to 0}\left[\frac{1}{x}\sin 2x\right]$ | M1 | Use of L'Hôpital or small angle approximation
$= \lim_{x \to 0}\frac{2\cos 2x}{1} = 2$ | A1 |
## Part (c):
The integrand contains $-\frac{1}{x}$ which is undefined (has a singularity/discontinuity) at $x = 0$ | B1 | Must reference $x=0$ and the term $-\frac{1}{x}$
## Part (d):
$\int_0^{\pi/4}\left(2\cot 2x - \frac{1}{x} + x\cos 8x\right)dx$
$= \lim_{\varepsilon \to 0^+}\int_\varepsilon^{\pi/4}\left(2\cot 2x - \frac{1}{x} + x\cos 8x\right)dx$ | M1 | Correct limiting process shown
Integrate: $\left[\ln\sin 2x - \ln x + \frac{x\sin 8x}{8} + \frac{\cos 8x}{64}\right]_\varepsilon^{\pi/4}$ | A1ft | Using parts (a) and (b)
At $x = \frac{\pi}{4}$: $\ln\sin\frac{\pi}{2} - \ln\frac{\pi}{4} + 0 + \frac{1}{64}= -\ln\frac{\pi}{4} + \frac{1}{64}$
As $\varepsilon \to 0$: $\ln\sin 2\varepsilon - \ln\varepsilon \to \ln 2$ (using part (b)) | M1 | Correct evaluation of limit
$= -\ln\frac{\pi}{4} + \frac{1}{64} - \ln 2 = \ln\frac{2}{\pi} + \frac{1}{64}$ | A1 | Single term answer5
\begin{enumerate}[label=(\alph*)]
\item Find $\int x \cos 8 x \mathrm {~d} x$.
\item Find $\lim _ { x \rightarrow 0 } \left[ \frac { 1 } { x } \sin 2 x \right]$.
\item Explain why $\int _ { 0 } ^ { \frac { \pi } { 4 } } \left( 2 \cot 2 x - \frac { 1 } { x } + x \cos 8 x \right) \mathrm { d } x$ is an improper integral.
\item Evaluate $\int _ { 0 } ^ { \frac { \pi } { 4 } } \left( 2 \cot 2 x - \frac { 1 } { x } + x \cos 8 x \right) \mathrm { d } x$, showing the limiting process used. Give your answer as a single term.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2014 Q5 [4]}}