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


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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
	$I_(n+1) =∫_0 ^1 (1−u)^(n+1) u^(1/2) du= =∫_0 ^1 (1−u)^(n+1) d((2/3)u^(3/2) )= ={(2/3)u^(3/2) (1−u)^(n+1) }_0 ^1 −((2(n+1))/3)∫_1 ^0 (1−u)^n u^(3/2) du= =−((2(n+1))/3)(I_(n+1) −I_n ) ⇒((2n+5)/3)I_(n+1) =((2n+2)/3)I_n ⇒I_(n+1) =((2n+2)/(2n+5))I_n ⇒I_n =((2n)/(2n+3)) ((2n−2)/(2n+1)) ... ((2.0+2)/(2.0+5))I_0 I_0 =∫_0 ^1 (√u)du=(2/3)u^(3/2) ]_0 ^1 =(2/3) ⇒I_n =((2.2.4.6. ... . 2n)/(3.5.7. ... .(2n+3)))=((2.2^n (1.2. ... .n))/((2n+3)!!)) I_n =((2^(n+1) n!)/((2n+3)!!)) but (2n+1)!!=1.3. ... .(2n−1)(2n+1)= =((1.2.3. ... .(2n−1)(2n)(2n+1))/(2.4. ... .(2n−2)(2n)))= =(((2n+1)!)/(2^n n!)) ⇒(2n+3)!!=(2(n+1)+1)!!=(((2n+3)!)/(2^(n+1) (n+1)!)) ⇒I_n =((4^(n+1) (n+1)!(n!))/((2n+3)!))$
	$I_{n + 1} = \int_{0}^{1} {(1 - u)}^{n + 1} u^{1 / 2} d u =$ $= \int_{0}^{1} {(1 - u)}^{n + 1} d (\frac{2}{3} u^{3 / 2}) =$ $= {\frac{2}{3} u^{3 / 2} {(1 - u)}^{n + 1}}_{0}^{1} - \frac{2 (n + 1)}{3} \int_{1}^{0} {(1 - u)}^{n} u^{3 / 2} d u =$ $= - \frac{2 (n + 1)}{3} (I_{n + 1} - I_{n})$ $\Rightarrow \frac{2 n + 5}{3} I_{n + 1} = \frac{2 n + 2}{3} I_{n}$ $\Rightarrow I_{n + 1} = \frac{2 n + 2}{2 n + 5} I_{n}$ $\Rightarrow I_{n} = \frac{2 n}{2 n + 3} \frac{2 n - 2}{2 n + 1} . . . \frac{2.0 + 2}{2.0 + 5} I_{0}$ ${I_{0} = \int_{0}^{1} \sqrt{u} d u = \frac{2}{3} u^{3 / 2}]}_{0}^{1} = \frac{2}{3}$ $\Rightarrow I_{n} = \frac{2.2.4.6 . . . . . 2 n}{3.5.7 . . . . . (2 n + 3)} = \frac{2 . 2^{n} (1.2 . . . . . n)}{(2 n + 3)!!}$ $I_{n} = \frac{2^{n + 1} n!}{(2 n + 3)!!}$ $b u t$ $(2 n + 1)!! = 1.3 . . . . . (2 n - 1) (2 n + 1) =$ $= \frac{1.2.3 . . . . . (2 n - 1) (2 n) (2 n + 1)}{2.4 . . . . . (2 n - 2) (2 n)} =$ $= \frac{(2 n + 1)!}{2^{n} n!}$ $\Rightarrow (2 n + 3)!! = (2 (n + 1) + 1)!! = \frac{(2 n + 3)!}{2^{n + 1} (n + 1)!}$ $\Rightarrow I_{n} = \frac{4^{n + 1} (n + 1)! (n!)}{(2 n + 3)!}$
	Commented by Kodjo last updated on 09/Jun/22

	$T h a n k s$
	Commented by ilhamQ last updated on 11/Jun/22

	$]]]]]]]$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum