| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Polynomial times trigonometric |
| Difficulty | Challenging +1.2 This is a standard reduction formula question requiring integration by parts twice to establish the recurrence relation, followed by straightforward substitution. While it involves Further Maths content (reduction formulae), the technique is methodical and well-practiced, with no novel insight required. The definite integral limits are clean (0 to π/2), making evaluation straightforward. Slightly above average difficulty due to the algebraic manipulation and two-stage application, but remains a textbook exercise. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_n = \left[-x^n \cos x\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} nx^{n-1} \cos x \, dx\) | M1A1 | Uses integration by parts with \(u = x^n\) |
| \(= 0 + \left[nx^{n-1} \sin x\right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} n(n-1)x^{n-2} \sin x \, dx\) | A1 | |
| \(= n\left(\frac{1}{2}\pi\right)^{n-1} - n(n-1)I_{n-2} \Rightarrow I_n + n(n-1)I_{n-2} = n\left(\frac{1}{2}\pi\right)^{n-1}\) (AG) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_1 = \int_0^{\frac{\pi}{2}} x\sin x \, dx = \left[-x\cos x\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos x \, dx\) | M1 | |
| \(= \left[\sin x\right]_0^{\frac{\pi}{2}} = 1\) | A1 | |
| \(n=3 \Rightarrow I_3 = 3\left(\frac{\pi}{2}\right)^2 - 3\times2\times1 = \frac{3}{4}\pi^2 - 6\) | B1 FT |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_n = \left[-x^n \cos x\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} nx^{n-1} \cos x \, dx$ | M1A1 | Uses integration by parts with $u = x^n$ |
| $= 0 + \left[nx^{n-1} \sin x\right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} n(n-1)x^{n-2} \sin x \, dx$ | A1 | |
| $= n\left(\frac{1}{2}\pi\right)^{n-1} - n(n-1)I_{n-2} \Rightarrow I_n + n(n-1)I_{n-2} = n\left(\frac{1}{2}\pi\right)^{n-1}$ (AG) | A1 | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_1 = \int_0^{\frac{\pi}{2}} x\sin x \, dx = \left[-x\cos x\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos x \, dx$ | M1 | |
| $= \left[\sin x\right]_0^{\frac{\pi}{2}} = 1$ | A1 | |
| $n=3 \Rightarrow I_3 = 3\left(\frac{\pi}{2}\right)^2 - 3\times2\times1 = \frac{3}{4}\pi^2 - 6$ | B1 FT | |6 Let $I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x ^ { n } \sin x \mathrm {~d} x$.\\
(i) Prove that, for $n \geqslant 2$,
$$I _ { n } + n ( n - 1 ) I _ { n - 2 } = n \left( \frac { 1 } { 2 } \pi \right) ^ { n - 1 } .$$
(ii) Calculate the exact value of $I _ { 1 }$ and deduce the exact value of $I _ { 3 }$.\\
\hfill \mbox{\textit{CAIE FP1 2017 Q6 [7]}}