Area enclosed = ∫01x.dx−∫01x.dx\int^1_0 \sqrt{x}. dx-\int^1_0 x.dx\\∫01x.dx−∫01x.dx
=[x3232−x22]01=[23−12]−[0−0]=16=[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^2}{2}]^1_0\\ =[\frac{2}{3}-\frac{1}{2}]-[0-0]\\ =\frac{1}{6}\\=[23x23−2x2]01=[32−21]−[0−0]=61
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