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Question Number 122166 by physicstutes last updated on 14/Nov/20

$$\mathrm{Given}\mathrm{that}{I}_n=\underset{0}{\stackrel{1}{\int }}x{(1-x)}^{n}dx$$$$\mathrm{obtain}\mathrm{a}\mathrm{reduction}\mathrm{formulae}\mathrm{for}{I}_n\mathrm{in}\mathrm{terms}$$$$\mathrm{of}{I}_{n-2}\mathrm{Hence}\mathrm{evaluate}\underset{0}{\stackrel{1}{\int }}x{(1-x)}^{5}dx.$$

Answered by liberty last updated on 14/Nov/20

$${\underset{0}{\stackrel{1}{\int }}\mathrm{x}{(1-\mathrm{x})}^5\mathrm{dx}=-\frac{\mathrm{x}{(1-\mathrm{x})}^6}{6}]}_0^1+\underset{0}{\stackrel{1}{\int }}\frac{{(1-\mathrm{x})}^6}{6}\mathrm{dx}$$$$=-\frac{1}{42}{\left[(1-\mathrm{x})\right]}_0^1=\frac{1}{42}$$

Commented by physicstutes last updated on 14/Nov/20

$$\mathrm{from}\mathrm{your}\mathrm{method}\mathrm{i}\mathrm{deduce}\mathrm{that}:$$$$I_n=-{\left[\frac{x{(1-x)}^{n+1}}{n+1}\right]}_0^1+\underset{0}{\stackrel{1}{\int }}\frac{{(1-x)}^{n+1}}{n+1}dx$$$$I_n=\frac{1}{n+1}\underset{0}{\stackrel{1}{\int }}{(1-x)}^{n+1}dx$$$$\mathrm{Thanks}$$

Commented by Dwaipayan Shikari last updated on 14/Nov/20

$${\int }_0^{1}x{(1-x)}^{n}dx={\int }_0^1(1-x)x^{n}dx=\frac{1}{n+1}-\frac{1}{n+2}$$$$heren=5so,\frac{1}{42}$$

Answered by Dwaipayan Shikari last updated on 14/Nov/20

$${\int }_0^{1}x{(1-x)}^{n}dx$$$$=\beta (2,n+1)=\frac{\mathrm{\Gamma }\left(2\right)\mathrm{\Gamma }(n+1)}{\mathrm{\Gamma }(n+3)}=\frac{\mathrm{\Gamma }(n+1)}{\mathrm{\Gamma }(n+3)}$$$$So$$$${\int }_0^{1}x{(1-x)}^{5}dt=\frac{\mathrm{\Gamma }(5+1)}{\mathrm{\Gamma }(5+3)}=\frac{5!}{7!}=\frac{1}{42}$$

Commented by physicstutes last updated on 14/Nov/20

$$\mathrm{Magnifcent}!$$

Answered by mathmax by abdo last updated on 14/Nov/20

$${\mathrm{I}}_{\mathrm{n}}={\int }_0^1\mathrm{x}{(1-\mathrm{x})}^{\mathrm{n}}\mathrm{dx}\mathrm{we}\mathrm{know}\mathrm{B}(\mathrm{p},\mathrm{q})={\int }_0^1{\mathrm{x}}^{\mathrm{p}-1}{(1-\mathrm{x})}^{\mathrm{q}-1}\mathrm{dx}\Rightarrow$$$$\mathrm{B}(\mathrm{p},\mathrm{q})=\frac{\mathrm{\Gamma }\left(\mathrm{p}\right).\mathrm{\Gamma }\left(\mathrm{q}\right)}{\mathrm{\Gamma }(\mathrm{p}+\mathrm{q})}\Rightarrow {\mathrm{I}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{2-1}{(1-\mathrm{x})}^{\mathrm{n}+1-1}\mathrm{dx}=\mathrm{B}(2,\mathrm{n}+1)$$$$=\frac{\mathrm{\Gamma }\left(2\right).\mathrm{\Gamma }(\mathrm{n}+1)}{\mathrm{\Gamma }(2+\mathrm{n}+1)}=\frac{1!\mathrm{n}!}{(\mathrm{n}+2)!}=\frac{\mathrm{n}!}{(\mathrm{n}+2)(\mathrm{n}+1)\mathrm{n}!}=\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}$$$$\frac{{\mathrm{I}}_{\mathrm{n}}}{{\mathrm{I}}_{\mathrm{n}-2}}=\frac{1}{(\mathrm{n}+1)(\mathrm{n}+2)}.(\mathrm{n}-1)\mathrm{n}=\frac{\mathrm{n}(\mathrm{n}-1)}{(\mathrm{n}+1)(\mathrm{n}+2)}\Rightarrow$$$${\mathrm{I}}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}-1)}{(\mathrm{n}+1)(\mathrm{n}+2)}{\mathrm{I}}_{\mathrm{n}-2}(\mathrm{n}⩾2)$$$${\int }_0^1\mathrm{x}{(1-\mathrm{x})}^5\mathrm{dx}={\mathrm{I}}_5=\frac{1}{6.7}=\frac{1}{42}$$

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