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Question Number 80227 by jagoll last updated on 01/Feb/20

$$howtoprove$$$$\underset{0}{\stackrel{1}{\int }}{x}^n{(1-x)}^{m}dx=\frac{m!\times n!}{(m+n)!}$$$$viaGammafunction$$

Commented by john santu last updated on 01/Feb/20

$$Betafunction$$

Commented by Tony Lin last updated on 01/Feb/20

$${\int }_0^1{x}^n{(1-x)}^{m}dx$$$$=B(n+1,m+1)$$$$=\frac{\mathrm{\Gamma }(n+1)\mathrm{\Gamma }(m+1)}{\mathrm{\Gamma }(n+m+2)}$$$$=\frac{m!n!}{(m+n+1)!}$$

Commented by mathmax by abdo last updated on 01/Feb/20

$$let{A}_{n,m}={\int }_0^1{x}^n{(1-x)}^{m}dxbypsrts{u}^{{}^{\prime }}=x^{n}andv={(1-x)}^m$$$$A_{n,m}={\left[\frac{1}{n+1}{x}^{n+1}{(1-x)}^m\right]}_0^1-{\int }_0^1\frac{x^{n+1}}{n+1}(-m){(1-x)}^{m-1}dx$$$$=\frac{m}{n+1}{A}_{n+1,m-1}=\frac{m(m-1)}{(n+1)(n+2)}{A}_{n+2,m-2}=$$$$=\frac{m(m-1)(m-2)....(m-k+1)}{(n+1)(n+2)...(n+k)}{A}_{n+k,m-k}$$$$k=m\Rightarrow A_{n,m}=\frac{m!}{(n+1)(n+2)...(n+m)}{A}_{n+m,0}$$$$A_{n+m,0}={\int }_0^1{x}^{n+m}dx=\frac{1}{n+m+1}\Rightarrow A_{n,m}=\frac{m!}{(n+1)(n+2)....(n+m+1)}$$$$=\frac{m!\times n!}{(m+n+1)!}$$

Commented by mathmax by abdo last updated on 01/Feb/20

$$errorintheQuestion...$$

Commented by mind is power last updated on 03/Feb/20

$$niceSir$$

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