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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

$${find}\:{the}\:{value}\:{of}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \sqrt{\mathrm{1}−{x}}{dx} \\ $$

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)) .

$${chsngement}\:\sqrt{\mathrm{1}−{x}}={t}\:{give} \\ $$$$\mathrm{1}−{x}={t}^{\mathrm{2}} \:\:\Rightarrow{x}=\mathrm{1}−{t}^{\mathrm{2}} \:\:{and}\: \\ $$$${I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} \:{t}\:\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}} \:\left(\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(−{t}^{\mathrm{2}} \right)^{{k}} \right){dt} \\ $$$$=\mathrm{2}\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{k}+\mathrm{2}} \:{dt} \\ $$$$=\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{3}}\:{changement}\:{of}\:{indice} \\ $$$${k}+\mathrm{1}={p}\:{give} \\ $$$${I}_{{n}} \:=\:\mathrm{2}\:\sum_{{p}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} }{\mathrm{2}{p}+\mathrm{1}}\:\Rightarrow \\ $$$${I}_{{n}} \:=\mathrm{2}\:\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{\mathrm{2}{k}+\mathrm{1}}\:\:. \\ $$

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