| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Improper integrals with infinite upper limit (power/logarithm functions) |
| Difficulty | Challenging +1.2 This is a structured multi-part question on improper integrals requiring substitution and integration by parts. Part (a) is trivial recall, part (b)(i) is guided algebraic manipulation, part (b)(ii) requires standard integration by parts with limits, and part (b)(iii) is immediate from the substitution. While it involves multiple techniques and careful handling of limits, the question provides substantial scaffolding and uses standard A-level methods without requiring novel insight—making it moderately above average difficulty for Further Maths. |
| Spec | 1.08i Integration by parts4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks | Guidance |
|---|---|---|
| The interval of integration is infinite | E1 | OE; 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \int -y \ln y^{-2}dy = \int 2y \ln y dy\) | M1, A1 | CSO AG; 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\ldots\ldots = y^2 \ln y - \frac{1}{2}y^2 + c\) | M1, A1, A1 | \(\ldots = ky^2 \ln y \pm \int f(y)dy\) with \(f(y)\) not involving the 'original' ln y; Condone absence of '+ c'; 5 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(= -\frac{1}{2}\) since \(\lim_{a \to 0}a^2 \ln a = 0\) | M1, A1 | CSO Must see clear indication that cand has correctly considered \(\lim_{a \to 0}a^4 \ln a = 0\); 5 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| So \(\int_1^{\infty}\frac{\ln x^2}{x^3}dx = \frac{1}{2}\) | B1F | ft on minus c's value as answer to (b)(ii); 1 mark total |
**(a)**
The interval of integration is infinite | E1 | OE; 1 mark total
**(b)(i)**
$x = \frac{1}{y} \Rightarrow 'dx = -y^{-2}dy'$
$\int\frac{\ln x^2}{x^3}dx \Rightarrow \int (y^3 \ln y^{-2})(-y^{-2})dy$
$= \int -y \ln y^{-2}dy = \int 2y \ln y dy$ | M1, A1 | CSO AG; 2 marks total
**(ii)**
$\int 2y \ln y dy = y^2 \ln y - \int y^2\left(\frac{1}{y}\right)dy$
$\ldots\ldots = y^2 \ln y - \frac{1}{2}y^2 + c$ | M1, A1, A1 | $\ldots = ky^2 \ln y \pm \int f(y)dy$ with $f(y)$ not involving the 'original' ln y; Condone absence of '+ c'; 5 marks total
$\int_0^1 2y \ln y dy = \lim_{a \to 0}\int_a^1 2y \ln y dy$
$= \left(0 - \frac{1}{2}\right) - \lim_{a \to 0}\left[a^2 \ln a - \frac{a^2}{2}\right]$
$= -\frac{1}{2}$ since $\lim_{a \to 0}a^2 \ln a = 0$ | M1, A1 | CSO Must see clear indication that cand has correctly considered $\lim_{a \to 0}a^4 \ln a = 0$; 5 marks total
**(iii)**
So $\int_1^{\infty}\frac{\ln x^2}{x^3}dx = \frac{1}{2}$ | B1F | ft on minus c's value as answer to (b)(ii); 1 mark total
**Total: 9 marks**
---6
\begin{enumerate}[label=(\alph*)]
\item Explain why $\int _ { 1 } ^ { \infty } \frac { \ln x ^ { 2 } } { x ^ { 3 } } \mathrm {~d} x$ is an improper integral.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the substitution $y = \frac { 1 } { x }$ transforms $\int \frac { \ln x ^ { 2 } } { x ^ { 3 } } \mathrm {~d} x$ into $\int 2 y \ln y \mathrm {~d} y$.
\item Evaluate $\int _ { 0 } ^ { 1 } 2 y \ln y \mathrm {~d} y$, showing the limiting process used.
\item Hence write down the value of $\int _ { 1 } ^ { \infty } \frac { \ln x ^ { 2 } } { x ^ { 3 } } \mathrm {~d} x$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2010 Q6 [9]}}