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Question Number 81663 by zainal tanjung last updated on 14/Feb/20

$$\:\underset{\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\:{x}\:\left(\mathrm{1}−{x}\right)^{{n}} \:{dx}\:= \\$$

Commented by Tony Lin last updated on 14/Feb/20

$${B}\left({p},\:{q}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{p}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{q}−\mathrm{1}} {dx},\:{p}>\mathrm{0},{q}>\mathrm{0} \\$$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left(\mathrm{1}−{x}\right)^{{n}} {dx} \\$$$$={B}\left(\mathrm{2},\:{n}+\mathrm{1}\right) \\$$$$=\frac{\Gamma\left(\mathrm{2}\right)\Gamma\left({n}+\mathrm{1}\right)}{\Gamma\left({n}+\mathrm{3}\right)} \\$$$$=\frac{\mathrm{1}×{n}!}{\left({n}+\mathrm{2}\right)!} \\$$$$=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\$$

Commented by abdomathmax last updated on 14/Feb/20

$${by}\:{parts}\:{u}={x}\:{and}\:{v}^{'\:} =\left(\mathrm{1}−{x}\right)^{{n}} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left(\mathrm{1}−{x}\right)^{{n}} \:{dx}\:=\left[−\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}\left(\mathrm{1}−{x}\right)^{{n}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\$$$$+\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{n}+\mathrm{1}}\left(\mathrm{1}−{x}\right)^{{n}+\mathrm{1}} \:{dx}\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{n}+\mathrm{1}} \:{dx} \\$$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}}×\frac{−\mathrm{1}}{{n}+\mathrm{2}}\left[\left(\mathrm{1}−{x}\right)^{{n}+\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\$$$$\\$$

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