I-n-0-1-1-u-ud-u-Demonstrate-that-n-N-I-n-1-I-n-1-u-n-u-3-2-d-u-and-deduce-the-meaning-of-variations-of-I-n-N-

Question Number 171090 by Kodjo last updated on 07/Jun/22

I_{n} = \int_{0}^{1} (1 - u) \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 + 1} - I_{n} = {(1 - u)}^{n} u^{\frac{3}{2}} d (u) a n d d e d u c e t h e m e a n i n g o f v a r i a t i o n s o f (I_{n}) \in N

Commented by Kodjo last updated on 07/Jun/22

Excuse me it's a mistake here is the real subject

Commented by Kodjo last updated on 07/Jun/22

Commented by Kodjo last updated on 07/Jun/22

Answered by aleks041103 last updated on 09/Jun/22

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

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

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

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

\Rightarrow I_{n + 1} - I_{n} = \int_{1}^{0} {(1 - u)}^{n} u^{3 / 2} d u

Commented by aleks041103 last updated on 09/Jun/22

I_{n + 1} = \int_{0}^{1} {(1 - u)}^{n + 1} u^{1 / 2} d u =

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

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

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

\Rightarrow \frac{2 n + 5}{3} I_{n + 1} = \frac{2 n + 2}{3} I_{n}

\Rightarrow I_{n + 1} = \frac{2 n + 2}{2 n + 5} I_{n}

\Rightarrow I_{n} = \frac{2 n}{2 n + 3} \frac{2 n - 2}{2 n + 1} \dots \frac{2.0 + 2}{2.0 + 5} I_{0}

{I_{0} = \int_{0}^{1} \sqrt{u} d u = \frac{2}{3} u^{3 / 2}]}_{0}^{1} = \frac{2}{3}

\Rightarrow I_{n} = \frac{2.2.4.6 . \dots . 2 n}{3.5.7 . \dots . (2 n + 3)} = \frac{2 . 2^{n} (1.2 . \dots . n)}{(2 n + 3)!!}

I_{n} = \frac{2^{n + 1} n!}{(2 n + 3)!!}

b u t

(2 n + 1)!! = 1.3 . \dots . (2 n - 1) (2 n + 1) =

= \frac{1.2.3 . \dots . (2 n - 1) (2 n) (2 n + 1)}{2.4 . \dots . (2 n - 2) (2 n)} =

= \frac{(2 n + 1)!}{2^{n} n!}

\Rightarrow (2 n + 3)!! = (2 (n + 1) + 1)!! = \frac{(2 n + 3)!}{2^{n + 1} (n + 1)!}

\Rightarrow I_{n} = \frac{4^{n + 1} (n + 1)! (n!)}{(2 n + 3)!}

Commented by Kodjo last updated on 09/Jun/22

T h a n k s

Commented by ilhamQ last updated on 11/Jun/22

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