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=(√(x−2))=x−2 x−2=(x−2)^2 x−2=x^2 −4x+4 x^2 −5x+6=0 (x−2)(x−3)=0 x=2 or x=3 when x∈[2, 3] ⇒(√(x−2))≥x−2 Area =∫_2 ^3 (√(x−2))dx−∫_2 ^3 (x−2)dx let x−2=u, (du/dx)=1 ⇒∫_0 ^1 u^(1/2) du−∫_0 ^1 udu =[(2/3)u^(3/2) ]_0 ^1 −[(u^2 /2)]_0 ^1 =(2/3)−(1/2) =(1/6)](https://www.tinkutara.com/question/Q67060.png)
$${when}\:{y}=\sqrt{{x}−\mathrm{2}}={x}−\mathrm{2} \\ $$$${x}−\mathrm{2}=\left({x}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$${x}−\mathrm{2}={x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)=\mathrm{0} \\ $$$${x}=\mathrm{2}\:{or}\:{x}=\mathrm{3} \\ $$$${when}\:{x}\in\left[\mathrm{2},\:\mathrm{3}\right] \\ $$$$\Rightarrow\sqrt{{x}−\mathrm{2}}\geqslant{x}−\mathrm{2} \\ $$$${Area} \\ $$$$=\int_{\mathrm{2}} ^{\mathrm{3}} \sqrt{{x}−\mathrm{2}}{dx}−\int_{\mathrm{2}} ^{\mathrm{3}} \left({x}−\mathrm{2}\right){dx} \\ $$$${let}\:{x}−\mathrm{2}={u},\:\frac{{du}}{{dx}}=\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\frac{\mathrm{1}}{\mathrm{2}}} {du}−\int_{\mathrm{0}} ^{\mathrm{1}} {udu} \\ $$$$=\left[\frac{\mathrm{2}}{\mathrm{3}}{u}^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{\mathrm{1}} −\left[\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}} \\ $$