6 Show that $$\frac { \mathrm { d } } { \mathrm {~d} x } \left[ x ^ { n - 1 } \sqrt { } \left( 4 - x ^ { 2 } \right) \right] = \frac { 4 ( n - 1 ) x ^ { n - 2 } } { \sqrt { } \left( 4 - x ^ { 2 } \right) } - \frac { n x ^ { n } } { \sqrt { } \left( 4 - x ^ { 2 } \right) }$$ Let $$I _ { n } = \int _ { 0 } ^ { 1 } \frac { x ^ { n } } { \sqrt { } \left( 4 - x ^ { 2 } \right) } \mathrm { d } x$$ where \(n \geqslant 0\). Prove that $$n I _ { n } = 4 ( n - 1 ) I _ { n - 2 } - \sqrt { } 3$$ for \(n \geq 2\). Given that \(I _ { 0 } = \frac { 1 } { 6 } \pi\), find \(I _ { 4 }\), leaving your answer in an exact form.

$\frac{d}{dx}[x^{n-1}\sqrt{4-x^2}] = (n-1)x^{n-2}\sqrt{4-x^2} - x^n/\sqrt{4-x^2}$ | B1 |
Shows that this preliminary result implies first displayed result (AG) | M1A1 |
$[x^{n-1}\sqrt{4-x^2}]_0 = 4(n-1)I_{n-2} - nI_n$ | M1 |
$\Rightarrow ... \Rightarrow$ second displayed result (AG) | A1 |
$I_0 = \pi/6, I_2 = \pi/3 - \sqrt{3}/2, 4I_4 = 12I_2 - \sqrt{3}$ (all) | B1B1B1 |
$I_4 = \pi - 7\sqrt{3}/4$ | M1A1 |

**OR** for last 4 marks: | |
$2I_2 = 4.1.\pi/6 - \sqrt{3}$ | M1 |
$\Rightarrow I_2 = \pi/3 - \sqrt{3}/2$ | A1 |
$\Rightarrow 4I_4 = 4.3(\pi/3 - \sqrt{3}/2) - \sqrt{3}$ | A1 |
$\Rightarrow I_4 = \pi - 7\sqrt{3}/4$ | A1 |

6 Show that

$$\frac { \mathrm { d } } { \mathrm {~d} x } \left[ x ^ { n - 1 } \sqrt { } \left( 4 - x ^ { 2 } \right) \right] = \frac { 4 ( n - 1 ) x ^ { n - 2 } } { \sqrt { } \left( 4 - x ^ { 2 } \right) } - \frac { n x ^ { n } } { \sqrt { } \left( 4 - x ^ { 2 } \right) }$$

Let

$$I _ { n } = \int _ { 0 } ^ { 1 } \frac { x ^ { n } } { \sqrt { } \left( 4 - x ^ { 2 } \right) } \mathrm { d } x$$

where $n \geqslant 0$. Prove that

$$n I _ { n } = 4 ( n - 1 ) I _ { n - 2 } - \sqrt { } 3$$

for $n \geq 2$.

Given that $I _ { 0 } = \frac { 1 } { 6 } \pi$, find $I _ { 4 }$, leaving your answer in an exact form.

\hfill \mbox{\textit{CAIE FP1 2009 Q6 [9]}}