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Question Number 163402 by Ahmed777hamouda last updated on 06/Jan/22

$${\int }_0^1{(\mathit{x}-1)}^{10}{(\mathit{x}-3)}^3\mathit{d}\mathit{x}$$

Answered by Ar Brandon last updated on 06/Jan/22

$$I={\int }_0^1{(x-1)}^{10}{(x-3)}^{3}dx,t=x-3$$$$={\int }_{-3}^{-2}{(t+2)}^{10}{t}^{3}dt={\left[\frac{t^3{(t+2)}^{11}}{11}\right]}_{-3}^{-2}-\frac{3}{11}{\int }_{-3}^{-2}{t}^2{(t+2)}^{11}dt$$$$=-\frac{27}{11}-\frac{3}{11}{\left[\frac{t^2{(t+2)}^{12}}{12}\right]}_{-3}^{-2}+\frac{3}{11}\cdot \frac{2}{12}{\int }_{-3}^{-2}t{(t+2)}^{12}dt$$$$=-\frac{27}{11}+\frac{27}{11\times 12}+\frac{1}{22}{\left[\frac{t{(t+2)}^{13}}{13}\right]}_{-3}^{-2}-\frac{1}{22}\cdot \frac{1}{13}{\int }_{-3}^{-2}{(t+2)}^{13}dt$$$$=-\frac{27}{11}+\frac{27}{132}-\frac{1}{22}\cdot \frac{3}{13}-\frac{1}{22\times 13}\cdot {\left[\frac{{(t+2)}^{14}}{14}\right]}_{-3}^{-2}$$$$=-\frac{27}{11}+\frac{27}{132}-\frac{3}{286}-\frac{1}{286\times 14}=-\frac{4525}{2002}$$

Commented by peter frank last updated on 07/Jan/22

$$\mathrm{great}$$

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