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

$$\underset{0}{\stackrel{1}{\int }}x{(1-x)}^{n}dx=$$

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

$$B(p,q)={\int }_0^1{x}^{p-1}{(1-x)}^{q-1}dx,p>0,q>0$$$$\Rightarrow {\int }_0^{1}x{(1-x)}^{n}dx$$$$=B(2,n+1)$$$$=\frac{\mathrm{\Gamma }\left(2\right)\mathrm{\Gamma }(n+1)}{\mathrm{\Gamma }(n+3)}$$$$=\frac{1\times n!}{(n+2)!}$$$$=\frac{1}{(n+1)(n+2)}$$

Commented by abdomathmax last updated on 14/Feb/20

$$bypartsu=xand{v}^{{}^{\prime }}={(1-x)}^n$$$${\int }_0^{1}x{(1-x)}^{n}dx={[-\frac{1}{n+1}x{(1-x)}^{n+1}]}_0^1$$$$+{\int }_0^1\frac{1}{n+1}{(1-x)}^{n+1}dx=\frac{1}{n+1}{\int }_0^1{(1-x)}^{n+1}dx$$$$=\frac{1}{n+1}\times \frac{-1}{n+2}{\left[{(1-x)}^{n+2}\right]}_0^1=\frac{1}{(n+1)(n+2)}$$$$$$

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