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Question Number 128373 by BHOOPENDRA last updated on 06/Jan/21

∫_0 ^1  x^(3/2) (1−x)^(1/2)  dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{dx} \\ $$

Answered by Dwaipayan Shikari last updated on 06/Jan/21

∫_0 ^1 x^(3/2) (1−x)^(1/2) dx=β((5/2),(3/2))=((Γ((5/2))Γ((3/2)))/(Γ(4)))=(3/(8.6))Γ^2 ((1/2))  =(((√π).(√π))/(16))=(π/(16))

$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx}=\beta\left(\frac{\mathrm{5}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{4}\right)}=\frac{\mathrm{3}}{\mathrm{8}.\mathrm{6}}\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\sqrt{\pi}.\sqrt{\pi}}{\mathrm{16}}=\frac{\pi}{\mathrm{16}} \\ $$

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