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

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