| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Derive reduction formula by integration by parts |
| Difficulty | Challenging +1.8 This is a Further Maths reduction formula question requiring integration by parts with definite integrals, algebraic manipulation to derive the given formula, and recursive application. While technically demanding with multiple steps and careful bookkeeping of terms, it follows a standard pattern for this topic type with clear guidance on what to prove. |
| Spec | 1.08i Integration by parts4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_{n+2} = \left[\frac{x^{-n-1}}{-n-1}\sin\pi x\right]_{\frac{1}{2}}^{1} - \pi\int_{\frac{1}{2}}^{1}\frac{x^{-n-1}}{-n-1}\cos\pi x\, dx\) | M1 A1 | Integrates by parts |
| \(= \frac{2^{n+1}}{n+1} + \frac{\pi}{n+1}\left(\left[\frac{x^{-n}}{-n}\cos\pi x\right]_{\frac{1}{2}}^{1} + \pi\int_{\frac{1}{2}}^{1}\frac{x^{-n}}{-n}\sin\pi x\, dx\right)\) | M1 | Integrates by parts again |
| \(= \frac{2^{n+1}}{n+1} + \frac{\pi}{n+1}\left(\frac{1}{n} - \frac{\pi}{n}I_n\right)\), so \((n+1)I_{n+2} = 2^{n+1} + \pi\left(\frac{1}{n} - \frac{\pi}{n}I_n\right)\) | M1 | Uses \(I_n\) |
| \(\Rightarrow n(n+1)I_{n+2} = 2^{n+1}n + \pi - \pi^2 I_n\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2I_3 = 4 + \pi - \pi^2 I_1\); \(12I_5 = 48 + \pi - \frac{\pi^2}{2}(4 + \pi - \pi^2 I_1)\) | M1 | Substitutes \(I_3\) into reduction formula |
| \(\Rightarrow I_5 = 4 + \frac{1}{24}(2\pi - 4\pi^2 - \pi^3 + \pi^4 I_1)\) | A1 | AEF, must be exact with fractions simplified |
## Question 3:
### Part 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_{n+2} = \left[\frac{x^{-n-1}}{-n-1}\sin\pi x\right]_{\frac{1}{2}}^{1} - \pi\int_{\frac{1}{2}}^{1}\frac{x^{-n-1}}{-n-1}\cos\pi x\, dx$ | M1 A1 | Integrates by parts |
| $= \frac{2^{n+1}}{n+1} + \frac{\pi}{n+1}\left(\left[\frac{x^{-n}}{-n}\cos\pi x\right]_{\frac{1}{2}}^{1} + \pi\int_{\frac{1}{2}}^{1}\frac{x^{-n}}{-n}\sin\pi x\, dx\right)$ | M1 | Integrates by parts again |
| $= \frac{2^{n+1}}{n+1} + \frac{\pi}{n+1}\left(\frac{1}{n} - \frac{\pi}{n}I_n\right)$, so $(n+1)I_{n+2} = 2^{n+1} + \pi\left(\frac{1}{n} - \frac{\pi}{n}I_n\right)$ | M1 | Uses $I_n$ |
| $\Rightarrow n(n+1)I_{n+2} = 2^{n+1}n + \pi - \pi^2 I_n$ | A1 | AG |
**Total: 5 marks**
### Part 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2I_3 = 4 + \pi - \pi^2 I_1$; $12I_5 = 48 + \pi - \frac{\pi^2}{2}(4 + \pi - \pi^2 I_1)$ | M1 | Substitutes $I_3$ into reduction formula |
| $\Rightarrow I_5 = 4 + \frac{1}{24}(2\pi - 4\pi^2 - \pi^3 + \pi^4 I_1)$ | A1 | AEF, must be exact with fractions simplified |
**Total: 2 marks**
---3 The integral $I _ { n }$, where $n$ is a positive integer, is defined by
$$I _ { n } = \int _ { \frac { 1 } { 2 } } ^ { 1 } x ^ { - n } \sin \pi x \mathrm {~d} x$$
(i) Show that
$$n ( n + 1 ) I _ { n + 2 } = 2 ^ { n + 1 } n + \pi - \pi ^ { 2 } I _ { n }$$
(ii) Find $I _ { 5 }$ in terms of $\pi$ and $I _ { 1 }$.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q3 [7]}}