Algebra Questions

Question Number 149516 by mathdanisur last updated on 05/Aug/21

$$\Omega\:=\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{3}} {\int}}\:\left(\sqrt{\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} \:-\:\mathrm{8}{x}}\right)\:{dx}\:=\:? \\$$

Answered by Ar Brandon last updated on 05/Aug/21

$$\:\:=\int_{\mathrm{0}} ^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}}{dx}=\int_{\mathrm{0}} ^{\mathrm{3}} \sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{dx} \\$$$$\:\:=\int_{\mathrm{0}} ^{\mathrm{3}} \mid{x}−\mathrm{2}\mid{dx}=\int_{\mathrm{2}} ^{\mathrm{3}} \left({x}−\mathrm{2}\right){dx}−\int_{\mathrm{0}} ^{\mathrm{2}} \left({x}−\mathrm{2}\right){dx} \\$$$$\:\:=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{x}\right]_{\mathrm{2}} ^{\mathrm{3}} −\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{x}\right]_{\mathrm{0}} ^{\mathrm{2}} =\left(\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{6}\right)−\mathrm{2}\left(\mathrm{2}−\mathrm{4}\right) \\$$$$\:\:=−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{4}=\frac{\mathrm{5}}{\mathrm{2}} \\$$

Commented by mathdanisur last updated on 06/Aug/21

$$\mathrm{Thankyou}\:\boldsymbol{\mathrm{Ser}} \\$$