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

I_n =∫_0 ^1 (1−u)^n (√(ud(u))) Demonstrate that ∀n∈N, I_n ≥0

In=∫01(1−u)nud(u)Demonstratethat∀n∈N,In≥0

Answered by thfchristopher last updated on 07/Jun/22

I_n =∫_0 ^1 (√u)(1−u)^n du =∫_0 ^1 (√u)(1−u)(1−u)^(n−1) du =∫_0 ^1 (√u)(1−u)^(n−1) du−∫_0 ^1 u^(3/2) (1−u)^(n−1) du =I_(n−1) +(1/n)∫_0 ^1 u^(3/2) d[(1−u)^n ] =I_(n−1) +[(1/n)u^(3/2) (1−u)^n ]_0 ^1 −(1/n)∫_0 ^1 (1−u)^n d(u^(3/2) ) =I_(n−1) −(3/(2n))∫_0 ^1 (√u)(1−u)^n du =I_(n−1) −(3/(2n))I_n ∴ (1+(3/(2n)))I_n =I_(n−1) ⇒I_n =((2n)/(2n+3))I_(n−1) =(((2n)(2n−2)...(2))/((2n+3)(2n+1)...(5)))I_0 I_0 =∫_0 ^1 (√u)du =(2/3)[u^(3/2) ]_0 ^1 =(2/3) ∴ I_n =(((2n)(2n−2)...(2)^2 )/((2n+3)(2n+1)...(5)(3)))

In=∫01u(1−u)ndu=∫01u(1−u)(1−u)n−1du=∫01u(1−u)n−1du−∫01u32(1−u)n−1du=In−1+1n∫01u32d[(1−u)n]=In−1+[1nu32(1−u)n]01−1n∫01(1−u)nd(u32)=In−1−32n∫01u(1−u)ndu=In−1−32nIn∴(1+32n)In=In−1⇒In=2n2n+3In−1=(2n)(2n−2)...(2)(2n+3)(2n+1)...(5)I0I0=∫01udu=23[u32]01=23Missing \left or extra \right

Commented by Kodjo last updated on 07/Jun/22

thanks

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