| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Exponential times polynomial |
| Difficulty | Standard +0.8 This is a standard reduction formula question requiring integration by parts to establish the recurrence relation, then iterative application to find I₃. While methodical, it demands careful algebraic manipulation across multiple steps and is from Further Maths FP2, placing it moderately above average difficulty. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_n = \int_0^1 x^n e^{2x}\,dx\); set \(u=x^n\), \(du=nx^{n-1}dx\), \(dv=e^{2x}dx\), \(v=\frac{1}{2}e^{2x}\) | M1 | Integration by parts |
| \(\Rightarrow I_n = \left[\frac{1}{2}x^n e^{2x}\right]_0^1 - \frac{1}{2}n\int_0^1 x^{n-1}e^{2x}\,dx\) | A1, A1 | Correct way round and correct differentiation; indefinite form acceptable |
| \(I_n = \frac{1}{2}e^2 - \frac{1}{2}nI_{n-1}\) | A1 | Using limits |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_0 = \int_0^1 e^{2x}\,dx = \frac{1}{2}\left[e^{2x}\right]_0^1 = \frac{1}{2}(e^2-1)\) | M1, A1 | Attempt to find \(I_0\) or \(I_1\) |
| \(I_1 = \frac{1}{2}e^2 - \frac{1}{2}I_0 = \frac{1}{2}e^2 - \frac{1}{2}\left(\frac{1}{2}(e^2-1)\right) = \frac{1}{4}e^2 + \frac{1}{4}\) | M1 | Using this to progress, dep |
| \(I_2 = \frac{1}{2}e^2 - I_1 = \frac{1}{2}e^2 - \left(\frac{1}{4}e^2+\frac{1}{4}\right) = \frac{1}{4}e^2 - \frac{1}{4}\) | ||
| \(I_3 = \frac{1}{2}e^2 - \frac{3}{2}I_2 = \frac{1}{2}e^2 - \frac{3}{2}\left(\frac{1}{4}e^2-\frac{1}{4}\right) = \frac{1}{8}e^2 + \frac{3}{8}\) | A1 | |
| [4] |
## Question 4:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_n = \int_0^1 x^n e^{2x}\,dx$; set $u=x^n$, $du=nx^{n-1}dx$, $dv=e^{2x}dx$, $v=\frac{1}{2}e^{2x}$ | M1 | Integration by parts |
| $\Rightarrow I_n = \left[\frac{1}{2}x^n e^{2x}\right]_0^1 - \frac{1}{2}n\int_0^1 x^{n-1}e^{2x}\,dx$ | A1, A1 | Correct way round and correct differentiation; indefinite form acceptable |
| $I_n = \frac{1}{2}e^2 - \frac{1}{2}nI_{n-1}$ | A1 | Using limits |
| **[4]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_0 = \int_0^1 e^{2x}\,dx = \frac{1}{2}\left[e^{2x}\right]_0^1 = \frac{1}{2}(e^2-1)$ | M1, A1 | Attempt to find $I_0$ or $I_1$ |
| $I_1 = \frac{1}{2}e^2 - \frac{1}{2}I_0 = \frac{1}{2}e^2 - \frac{1}{2}\left(\frac{1}{2}(e^2-1)\right) = \frac{1}{4}e^2 + \frac{1}{4}$ | M1 | Using this to progress, dep |
| $I_2 = \frac{1}{2}e^2 - I_1 = \frac{1}{2}e^2 - \left(\frac{1}{4}e^2+\frac{1}{4}\right) = \frac{1}{4}e^2 - \frac{1}{4}$ | | |
| $I_3 = \frac{1}{2}e^2 - \frac{3}{2}I_2 = \frac{1}{2}e^2 - \frac{3}{2}\left(\frac{1}{4}e^2-\frac{1}{4}\right) = \frac{1}{8}e^2 + \frac{3}{8}$ | A1 | |
| **[4]** | | |
---4 You are given that $I _ { n } = \int _ { 0 } ^ { 1 } x ^ { n } \mathrm { e } ^ { 2 x } \mathrm {~d} x$ for $n \geqslant 0$.\\
(i) Show that $I _ { n } = \frac { 1 } { 2 } \mathrm { e } ^ { 2 } - \frac { 1 } { 2 } n I _ { n - 1 }$ for $n \geqslant 1$.\\
(ii) Find $I _ { 3 }$ in terms of e.
\hfill \mbox{\textit{OCR FP2 2013 Q4 [8]}}