Question Number 67059 by mhmd last updated on 22/Aug/19
$${find}\:{the}\:{area}\:{abounded}\:{y}=\sqrt{{x}−\mathrm{2}} \\ $$$${and}\:{y}={x}−\mathrm{2}\:? \\ $$
Commented by Tony Lin last updated on 22/Aug/19
$${when}\:{y}=\sqrt{{x}−\mathrm{2}}={x}−\mathrm{2} \\ $$$${x}−\mathrm{2}=\left({x}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$${x}−\mathrm{2}={x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)=\mathrm{0} \\ $$$${x}=\mathrm{2}\:{or}\:{x}=\mathrm{3} \\ $$$${when}\:{x}\in\left[\mathrm{2},\:\mathrm{3}\right] \\ $$$$\Rightarrow\sqrt{{x}−\mathrm{2}}\geqslant{x}−\mathrm{2} \\ $$$${Area} \\ $$$$=\int_{\mathrm{2}} ^{\mathrm{3}} \sqrt{{x}−\mathrm{2}}{dx}−\int_{\mathrm{2}} ^{\mathrm{3}} \left({x}−\mathrm{2}\right){dx} \\ $$$${let}\:{x}−\mathrm{2}={u},\:\frac{{du}}{{dx}}=\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\frac{\mathrm{1}}{\mathrm{2}}} {du}−\int_{\mathrm{0}} ^{\mathrm{1}} {udu} \\ $$$$=\left[\frac{\mathrm{2}}{\mathrm{3}}{u}^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{\mathrm{1}} −\left[\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}} \\ $$