Question Number 145774 by Engr_Jidda last updated on 08/Jul/21
$${find}\:{the}\:{area}\:{bounded}\:{by}\:{y}=\mathrm{2}{x},\:{y}=\frac{{x}}{\mathrm{2}}\:{and}?{xy}=\mathrm{2} \\ $$
Answered by ArielVyny last updated on 08/Jul/21
$$\forall{x}\in\mathbb{R}\:\mathrm{2}{x}\geqslant\frac{{x}}{\mathrm{2}}\:\:{A}=\int\left(\mathrm{2}{x}−\frac{{x}}{\mathrm{2}}\right){dx}=\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \\ $$$${A}={x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}}{x}^{\mathrm{2}} \:\:\:\:{A}=\frac{\mathrm{3}}{\mathrm{4}}{x}^{\mathrm{2}} \\ $$$${y}=\mathrm{2}{x}\rightarrow{yx}=\mathrm{2}{x}^{\mathrm{2}} =\mathrm{2}\rightarrow{x}=\pm\mathrm{1} \\ $$$${y}=\frac{{x}}{\mathrm{2}}\rightarrow{yx}=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} =\mathrm{2}\rightarrow{x}^{\mathrm{2}} =\mathrm{4}\rightarrow{x}=\pm\mathrm{2} \\ $$$${then}\:{xy}=\mathrm{2}\:{solutions}\:{are}\:{incompatibles} \\ $$