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Question Number 36410 by abdo.msup.com last updated on 01/Jun/18

find the value of I_n = ∫_0 ^1 x^n (√(1−x))dx

findthevalueofIn=∫01xn1−xdx

Commented by abdo.msup.com last updated on 04/Jun/18

chsngement (√(1−x))=t give 1−x=t^2 ⇒x=1−t^2 and I_n = ∫_0 ^1 (1−t^2 )^n t (2t)dt =2 ∫_0 ^1 t^2 ( Σ_(k=0) ^n (−t^2 )^k )dt =2Σ_(k=0) ^n (−1)^k ∫_0 ^1 t^(2k+2) dt =2 Σ_(k=0) ^n (((−1)^k )/(2k+3)) changement of indice k+1=p give I_n = 2 Σ_(p=1) ^(n+1) (((−1)^(p−1) )/(2p+1)) ⇒ I_n =2 Σ_(k=1) ^(n+1) (((−1)^(k−1) )/(2k+1)) .

chsngement1−x=tgive1−x=t2⇒x=1−t2andIn=∫01(1−t2)nt(2t)dt=2∫01t2(∑k=0n(−t2)k)dt=2∑k=0n(−1)k∫01t2k+2dt=2∑k=0n(−1)k2k+3changementofindicek+1=pgiveIn=2∑p=1n+1(−1)p−12p+1⇒In=2∑k=1n+1(−1)k−12k+1.

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