| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Polynomial times trigonometric |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring integration by parts twice to derive a reduction formula, then applying it cleverly in part (ii) where arc length leads to integrating t^4 sin t after simplification. The derivation is standard but the multi-step nature, algebraic manipulation, and connection between parts elevates it above typical questions. |
| Spec | 1.08i Integration by parts4.08f Integrate using partial fractions |
| Answer | Marks |
|---|---|
| (i) \(\int_0^{t/2} t^n sdt = [-t^n c]_0^{t/2} + n\int_0^{t/2} t^{n-1} cdt\) | M1A1 (LR) |
| \(= [nt^{n-1}s]_0^t - n(n-1)I_{n-2}\) | M1A1 |
| \(\Rightarrow I_n = n(\pi/2)^{n-1} - n(n-1)I_{n-2}\) (AG) | A1 |
| (ii) \(L = \text{length of arc} = \int_0^{\pi/2}\sqrt{x^2 + y^2}dt\) with integrand expressed in terms of \(t\) | M1 |
| \(\sqrt{x^2 + y^2} = 2t^4 s\) | B1 |
| \(L = \int 2t^4 s \, dt = 2I_4\) | A1 |
| \(I_2 = \pi - 2\) | A1 |
| \(L = \pi^3 - 24\pi + 48\) (AEF), e.g., \(L = 8(\pi-2)^2 - 24(\pi-2)\) Accept 3.61 | A1 |
**(i)** $\int_0^{t/2} t^n sdt = [-t^n c]_0^{t/2} + n\int_0^{t/2} t^{n-1} cdt$ | M1A1 (LR) |
$= [nt^{n-1}s]_0^t - n(n-1)I_{n-2}$ | M1A1 |
$\Rightarrow I_n = n(\pi/2)^{n-1} - n(n-1)I_{n-2}$ (AG) | A1 |
**(ii)** $L = \text{length of arc} = \int_0^{\pi/2}\sqrt{x^2 + y^2}dt$ with integrand expressed in terms of $t$ | M1 |
$\sqrt{x^2 + y^2} = 2t^4 s$ | B1 |
$L = \int 2t^4 s \, dt = 2I_4$ | A1 |
$I_2 = \pi - 2$ | A1 |
$L = \pi^3 - 24\pi + 48$ (AEF), e.g., $L = 8(\pi-2)^2 - 24(\pi-2)$ Accept 3.61 | A1 |8 (i) Given that
$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } t ^ { n } \sin t \mathrm {~d} t$$
show that, for $n \geqslant 2$,
$$I _ { n } = n \left( \frac { \pi } { 2 } \right) ^ { n - 1 } - n ( n - 1 ) I _ { n - 2 } .$$
(ii) A curve $C$ in the $x - y$ plane is defined parametrically in terms of $t$. It is given that
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = t ^ { 4 } ( 1 - \cos 2 t ) \quad \text { and } \quad \frac { \mathrm { d } y } { \mathrm {~d} t } = t ^ { 4 } \sin 2 t .$$
Find the length of the arc of $C$ from the point where $t = 0$ to the point where $t = \frac { 1 } { 2 } \pi$.
\hfill \mbox{\textit{CAIE FP1 2008 Q8 [10]}}