Question Number 89228 by cindiaulia last updated on 16/Apr/20
$$\underset{\mathrm{4}} {\overset{\mathrm{5}} {\int}}\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}−\mathrm{4}} \\ $$
Commented by jagoll last updated on 16/Apr/20
$$\sqrt{{x}−\mathrm{4}}\:=\:{t}\:\Rightarrow{x}={t}^{\mathrm{2}} +\mathrm{4} \\ $$$${dx}\:=\:\mathrm{2}{t}\:{dt} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left({t}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} {t}\:\left(\mathrm{2}{t}\right)\:{dt}\:= \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\mathrm{2}{t}^{\mathrm{2}} \left({t}^{\mathrm{4}} +\mathrm{8}{t}^{\mathrm{2}} +\mathrm{16}\right){dt}\:= \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\:\mathrm{2}{t}^{\mathrm{6}} +\mathrm{16}{t}^{\mathrm{4}} +\mathrm{32}{t}^{\mathrm{2}} \:{dt}\:= \\ $$$$\frac{\mathrm{2}}{\mathrm{7}}+\frac{\mathrm{16}}{\mathrm{5}}+\frac{\mathrm{32}}{\mathrm{3}} \\ $$