I-n-0-1-1-u-n-ud-u-Demonstrate-that-n-N-I-n-0-

Question Number 171039 by Kodjo last updated on 06/Jun/22

I_{n} = \int_{0}^{1} {(1 - u)}^{n} \sqrt{u d (u)}

D e m o n s t r a t e t h a t \forall n \in N, I_{n} \geq 0

Answered by thfchristopher last updated on 07/Jun/22

I_{n} = \int_{0}^{1} \sqrt{u} {(1 - u)}^{n} d u

= \int_{0}^{1} \sqrt{u} (1 - u) {(1 - u)}^{n - 1} d u

= \int_{0}^{1} \sqrt{u} {(1 - u)}^{n - 1} d u - \int_{0}^{1} u^{\frac{3}{2}} {(1 - u)}^{n - 1} d u

= I_{n - 1} + \frac{1}{n} \int_{0}^{1} u^{\frac{3}{2}} d [{(1 - u)}^{n}]

= I_{n - 1} + {[\frac{1}{n} u^{\frac{3}{2}} {(1 - u)}^{n}]}_{0}^{1} - \frac{1}{n} \int_{0}^{1} {(1 - u)}^{n} d (u^{\frac{3}{2}})

= I_{n - 1} - \frac{3}{2 n} \int_{0}^{1} \sqrt{u} {(1 - u)}^{n} d u

= I_{n - 1} - \frac{3}{2 n} I_{n}

∴ (1 + \frac{3}{2 n}) I_{n} = I_{n - 1}

\Rightarrow I_{n} = \frac{2 n}{2 n + 3} I_{n - 1}

= \frac{(2 n) (2 n - 2) \dots (2)}{(2 n + 3) (2 n + 1) \dots (5)} I_{0}

I_{0} = \int_{0}^{1} \sqrt{u} d u

= \frac{2}{3} {[u^{\frac{3}{2}}]}_{0}^{1} = \frac{2}{3}

Missing \left or extra \right

Commented by Kodjo last updated on 07/Jun/22

thanks