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Question Number 124672 by sdfg last updated on 05/Dec/20

$${\int }_{-3}^{-2}{(y+3)}^6{(y+2)}^{4}dy$$

Commented by mohammad17 last updated on 05/Dec/20

$$put:n=y+3\Rightarrow y=n-3\Rightarrow dy=dn$$$$$$$$y=-2\Rightarrow n=1,y=-3\Rightarrow n=0$$$$$$$$\therefore {\int }_{-3}^{-2}{(y+3)}^6{(y+2)}^{4}dy=$$$$$$$${\int }_0^1{n}^6{(n-1)}^{4}dn={\int }_0^1(n^{10}-4n^9+6n^8-4n^7+n^6)dn$$$$$$$$={(\frac{n^{11}}{11}-\frac{2n^{10}}{5}+\frac{2n^9}{3}-\frac{n^8}{2}+\frac{n^7}{7})}_0^1=\frac{1}{11}-\frac{2}{5}+\frac{2}{3}-\frac{1}{2}+\frac{1}{7}=\frac{1}{2310}$$$$$$$$\left(mohammadAldolaimy\right)$$

Answered by Dwaipayan Shikari last updated on 05/Dec/20

$${\int }_{-3}^{-2}{(y+3)}^6{(y+2)}^{4}y+3=x$$$$={\int }_0^1{x}^6{(x-1)}^{4}dx={\int }_0^1{x}^6{(1-x)}^{4}dx=\frac{\mathrm{\Gamma }\left(7\right)\mathrm{\Gamma }\left(5\right)}{\mathrm{\Gamma }\left(12\right)}=\frac{6!4!}{11!}=\frac{1}{2310}$$

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