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Question Number 17604 by tawa tawa last updated on 08/Jul/17

$${\int }_0^{\mathrm{n}}{\mathrm{x}}^2{(\mathrm{n}-\mathrm{x})}^{\mathrm{p}}\mathrm{dx}\mathrm{for}\mathrm{p}>0$$

Answered by sma3l2996 last updated on 08/Jul/17

$$t=n-x\Rightarrow dt=-dx$$$$I={\int }_0^n{x}^2{(n-x)}^{p}dx={\int }_0^n{(n-t)}^2{t}^{p}dt={\int }_0^n(n^2{t}^p-2{nt}^{p+1}+t^{p+2})dt$$$$={[\frac{n^2}{p+1}{t}^{p+1}-\frac{2n}{p+2}{t}^{p+2}+\frac{t^{p+3}}{p+3}]}_0^n$$$$=\frac{n^2}{p+1}{n}^{p+1}-\frac{2n}{p+2}{n}^{p+2}+\frac{n^{p+3}}{p+3}$$$$=n^{p+3}(\frac{1}{p+1}-\frac{2}{p+2}+\frac{1}{p+3})=n^{p+3}\left(\frac{(p+3)(p+2)-2(p+1)(p+3)+(p+1)(p+2)}{(p+1)(p+2)(p+3)}\right)$$$$=n^{p+3}\left(\frac{p^2+5p+6-2p^2-8p-6+p^2+3p+2}{(p+1)(p+2)(p+3)}\right)$$$$I=\frac{2n^{p+3}}{(p+1)(p+2)(p+3)}$$

Commented by tawa tawa last updated on 08/Jul/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by alex041103 last updated on 08/Jul/17

$$\mathrm{We}\mathrm{use}\mathrm{the}\mathrm{substitution}$$$$u=n-x\Rightarrow -du=dx\mathrm{and}{x}^2={(n-u)}^2$$$$\mathrm{And}\mathrm{we}\mathrm{change}\mathrm{the}\mathrm{limits}\mathrm{of}\mathrm{integration}$$$$x=0x=n$$$$u=nu=0$$$$\Rightarrow \underset{0}{\stackrel{n}{\int }}{x}^2{(n-x)}^{p}dx=-\underset{n}{\stackrel{0}{\int }}{(n-u)}^2{u}^{p}du$$$$=\underset{0}{\stackrel{n}{\int }}(n^2-2nu+u^2)u^{p}du$$$$=n^2\underset{0}{\stackrel{n}{\int }}{u}^{p}du-2n\underset{0}{\stackrel{n}{\int }}{u}^{p+1}du+\underset{0}{\stackrel{n}{\int }}{u}^{p+2}du$$$$=n^2\frac{n^{p+1}-0}{p+1}-2n\frac{n^{p+2}-0}{p+2}+\frac{n^{p+3}-0}{p+3}$$$$=\frac{n^{p+3}}{p+1}-2\frac{n^{p+3}}{p+2}+\frac{n^{p+3}}{p+3}$$$$=n^{p+3}(\frac{1}{p+1}-\frac{2}{p+2}+\frac{1}{p+3})$$$$\mathrm{This}\mathrm{can}\mathrm{be}\mathrm{expressed}\mathrm{as}$$$$\underset{0}{\stackrel{n}{\int }}{x}^2{(n-x)}^{p}dx=\frac{2n^{p+3}}{(p+1)(p+2)(p+3)}$$

Commented by tawa tawa last updated on 08/Jul/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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