| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Improper integrals with infinite upper limit (power/logarithm functions) |
| Difficulty | Standard +0.8 This is a Further Maths question requiring understanding of limits and improper integrals with logarithmic functions. Part (a) involves a substitution to establish a general limit result (ln p / p^k → 0), then part (b) applies this to evaluate an improper integral using integration by parts and the limiting process. While the techniques are standard for FP3, it requires careful handling of limits at infinity and combining multiple concepts (substitution, integration by parts, improper integrals), making it moderately challenging but within expected Further Maths scope. |
| Spec | 4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lim_{p\to\infty}\frac{\ln p}{p^k} = \lim_{a\to 0^+}\frac{\ln(1/a)}{(1/a)^k}\) | M1 | Changing \(\lim_{p\to\infty}\) to \(\lim_{a\to 0^+}\) |
| \(= \lim_{a\to 0^+} -a^k \ln a\) | M1 | Changing \(\frac{\ln p}{p^k}\) to \(-a^k \ln a\) |
| \(= 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(u=\ln x\), \(\frac{dv}{dx}=x^{-7}\), \(\frac{du}{dx}=\frac{1}{x}\), \(v=\frac{x^{-6}}{-6}\) | M1 | \(\frac{du}{dx}=\frac{1}{x}\), \(v=\frac{x^{-6}}{k}\) with \(k=-6\) or \(6\) |
| \(= -\frac{1}{6}x^{-6}\ln x - \frac{1}{36}x^{-6} \; (+c)\) | A1 | ACF |
| \(\int_1^\infty \frac{\ln x}{x^7}dx = \lim_{p\to\infty}\left(-\frac{1}{6}\frac{\ln p}{p^6}-\frac{1}{36p^6}\right)-\left(0-\frac{1}{36}\right)\) | M1 | \(\lim_{p\to\infty}[F(p)-F(1)]\) OE; consistent use of same letter; \(F\) follows from integration by parts attempt |
| \(\int_1^\infty \frac{\ln x}{x^7}dx = 0-0-0+\frac{1}{36} = \frac{1}{36}\) | A1 | \(\frac{1}{36}\) together with split into two relevant limit expressions with indication of \(0\) evaluations or reference to (a) with general \(k\) or \(k=6\) |
# Question 6:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lim_{p\to\infty}\frac{\ln p}{p^k} = \lim_{a\to 0^+}\frac{\ln(1/a)}{(1/a)^k}$ | M1 | Changing $\lim_{p\to\infty}$ to $\lim_{a\to 0^+}$ |
| $= \lim_{a\to 0^+} -a^k \ln a$ | M1 | Changing $\frac{\ln p}{p^k}$ to $-a^k \ln a$ |
| $= 0$ | A1 | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $u=\ln x$, $\frac{dv}{dx}=x^{-7}$, $\frac{du}{dx}=\frac{1}{x}$, $v=\frac{x^{-6}}{-6}$ | M1 | $\frac{du}{dx}=\frac{1}{x}$, $v=\frac{x^{-6}}{k}$ with $k=-6$ or $6$ |
| $= -\frac{1}{6}x^{-6}\ln x - \frac{1}{36}x^{-6} \; (+c)$ | A1 | ACF |
| $\int_1^\infty \frac{\ln x}{x^7}dx = \lim_{p\to\infty}\left(-\frac{1}{6}\frac{\ln p}{p^6}-\frac{1}{36p^6}\right)-\left(0-\frac{1}{36}\right)$ | M1 | $\lim_{p\to\infty}[F(p)-F(1)]$ OE; consistent use of same letter; $F$ follows from integration by parts attempt |
| $\int_1^\infty \frac{\ln x}{x^7}dx = 0-0-0+\frac{1}{36} = \frac{1}{36}$ | A1 | $\frac{1}{36}$ together with split into two relevant limit expressions with indication of $0$ evaluations or reference to (a) with general $k$ or $k=6$ |
---6
\begin{enumerate}[label=(\alph*)]
\item Use the substitution $a = \frac { 1 } { p }$ to find $\lim _ { p \rightarrow \infty } \left[ \frac { \ln p } { p ^ { k } } \right]$, where $k > 0$.
\item Evaluate the improper integral $\int _ { 1 } ^ { \infty } \frac { \ln x } { x ^ { 7 } } \mathrm {~d} x$, showing the limiting process used.\\[0pt]
[4 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{0b9b947d-824b-4d3a-b66d-4bfd8d49be17-16_2039_1719_671_148}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2016 Q6 [7]}}