Question Number 124675 by sdfg last updated on 05/Dec/20
$$\int_{\mathrm{1}} ^{\mathrm{3}} \left({x}−\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{3}−{x}\right)^{\mathrm{2}} {dx} \\ $$
Commented by mohammad17 last updated on 05/Dec/20
$${put}:{m}={x}−\mathrm{1}\Rightarrow{x}={m}+\mathrm{1}\Rightarrow{dx}={dm} \\ $$$$ \\ $$$${x}=\mathrm{3}\Rightarrow{m}=\mathrm{2}\:\:\:\:,\:\:\:\:{x}=\mathrm{1}\Rightarrow{m}=\mathrm{0} \\ $$$$ \\ $$$$\therefore\int_{\mathrm{1}} ^{\:\mathrm{3}} \left({x}−\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{3}−{x}\right)^{\mathrm{2}} {dx}=\int_{\mathrm{0}} ^{\:\mathrm{2}} {m}^{\mathrm{3}} \left(\mathrm{2}−{m}\right)^{\mathrm{2}} {dm} \\ $$$$ \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{2}} \left(\mathrm{4}{m}^{\mathrm{3}} −\mathrm{4}{m}^{\mathrm{4}} +{m}^{\mathrm{5}} \right){dm} \\ $$$$ \\ $$$$=\left({m}^{\mathrm{4}} −\frac{\mathrm{4}{m}^{\mathrm{5}} }{\mathrm{5}}+\frac{{m}^{\mathrm{6}} }{\mathrm{6}}\right)_{\mathrm{0}} ^{\mathrm{2}} =\mathrm{16}−\frac{\mathrm{128}}{\mathrm{5}}+\frac{\mathrm{64}}{\mathrm{6}}=\frac{\mathrm{16}}{\mathrm{15}} \\ $$$$ \\ $$$$\left({mohammad}\:{Aldolaimy}\right) \\ $$