| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Reduction formula or recurrence |
| Difficulty | Challenging +1.2 This is a standard Further Maths reduction formula question requiring integration by parts twice (once to establish the recurrence, once more to find I_0), then recursive substitution. While it involves multiple steps and careful algebraic manipulation, the technique is well-practiced in FP3 and follows a predictable pattern. The definite integral bounds and the presence of ln x make it moderately challenging but not exceptional for Further Maths students. |
| Spec | 1.08i Integration by parts4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore I_n = \frac{e^3}{3} - \frac{n}{3}I_{n-1}\) | M1 A1, DM1, A1cso | Using integration by parts, integrating \(x^2\), differentiating \((\ln x)^n\). Correctly using limits. CSO answer given. |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_1 = \frac{e^3}{3} - \frac{1}{3}I_0\), \(I_2 = \frac{e^3}{3} - \frac{2}{3}I_1\) and \(I_3 = \frac{e^3}{3} - \frac{3}{3}I_2\) so \(I_3 = \frac{4e^3}{27} + \frac{2}{27}\) | M1 A1, M1 A1, M1 A1 | Evaluating \(I_0\) or \(I_1\) by an attempt to integrate something. Finding \(I_3\) (also probably \(I_1\) and \(I_2\)). If 'n' is left in M0. \(I_3\) CAO. |
## Part (a)
**Answer/Working:** $I_n = \left[\frac{x^3}{3}(\ln x)^n\right] - \int \frac{x^3}{3} \cdot \frac{n(\ln x)^{n-1}}{x}dx$
$= \left[\frac{x^3}{3}(\ln x)^n\right] - \int \frac{nx^2(\ln x)^{n-1}}{3}dx$
$\therefore I_n = \frac{e^3}{3} - \frac{n}{3}I_{n-1}$ | **M1 A1, DM1, A1cso** | Using integration by parts, integrating $x^2$, differentiating $(\ln x)^n$. Correctly using limits. CSO answer given.
(4 marks)
## Part (b)
**Answer/Working:** $I_0 = \int x^2 dx = \left[\frac{x^3}{3}\right]_1 = \frac{e^3}{3} - \frac{1}{3}$ or $I_1 = \frac{e^3}{3} - \frac{1}{3}\left(\frac{e^3}{3} - \frac{1}{3}\right) = \frac{2e^3}{9} - \frac{1}{9}$
$I_1 = \frac{e^3}{3} - \frac{1}{3}I_0$, $I_2 = \frac{e^3}{3} - \frac{2}{3}I_1$ and $I_3 = \frac{e^3}{3} - \frac{3}{3}I_2$ so $I_3 = \frac{4e^3}{27} + \frac{2}{27}$ | **M1 A1, M1 A1, M1 A1** | Evaluating $I_0$ or $I_1$ by an attempt to integrate something. Finding $I_3$ (also probably $I_1$ and $I_2$). If 'n' is left in M0. $I_3$ CAO.
(8 marks)
---$$I_n = \int_1^e x^2 (\ln x)^n dx, \quad n \geq 0$$
\begin{enumerate}[label=(\alph*)]
\item Prove that, for $n \geq 1$,
$$I_n = \frac{e^3}{3} - \frac{n}{3} I_{n-1}$$
[4]
\item Find the exact value of $I_3$.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2011 Q4 [8]}}