| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| 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, then applying it iteratively. While it involves Further Maths content (FP2), the technique is methodical and well-practiced. The algebraic manipulation is straightforward, and finding I₄ requires only systematic application of the formula with evaluation of base cases I₀ and I₂. |
| Spec | 1.08i Integration by parts8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Reasonable attempt at parts; get \(x\sin x - \int \sin x \cdot nx^{n-1}\,dx\) | M1 | Involving second integral |
| Attempt parts again accurately | M1 | |
| Clearly derive AG | A1 | |
| A1 | Indicate \(\left(\frac{1}{2}\pi\right)^n\) and \(0\) from limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Get \(I_4 = \left(\frac{1}{2}\pi\right)^4 - 12I_2\) or \(I_2 = \left(\frac{1}{2}\pi\right)^2 - 2I_0\) | B1 | |
| Show clearly \(I_0 = 1\) | B1 | May use \(I_2\) |
| Replace their values in relation | M1 | |
| Get \(I_4 = \frac{1}{16}\pi^4 - 3\pi^2 + 24\) | A1 | cao |
## Question 5:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Reasonable attempt at parts; get $x\sin x - \int \sin x \cdot nx^{n-1}\,dx$ | M1 | Involving second integral |
| Attempt parts again accurately | M1 | |
| Clearly derive AG | A1 | |
| | A1 | Indicate $\left(\frac{1}{2}\pi\right)^n$ and $0$ from limits |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Get $I_4 = \left(\frac{1}{2}\pi\right)^4 - 12I_2$ or $I_2 = \left(\frac{1}{2}\pi\right)^2 - 2I_0$ | B1 | |
| Show clearly $I_0 = 1$ | B1 | May use $I_2$ |
| Replace their values in relation | M1 | |
| Get $I_4 = \frac{1}{16}\pi^4 - 3\pi^2 + 24$ | A1 | cao |
---5 It is given that, for non-negative integers $n$,
$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } x ^ { n } \cos x \mathrm {~d} x$$
(i) Prove that, for $n \geqslant 2$,
$$I _ { n } = \left( \frac { 1 } { 2 } \pi \right) ^ { n } - n ( n - 1 ) I _ { n - 2 } .$$
(ii) Find $I _ { 4 }$ in terms of $\pi$.
\hfill \mbox{\textit{OCR FP2 2007 Q5 [9]}}