Mathematics - Area Under Curves Question with Solution | TestHub

$R=\left\{\left(x,y\right):\max \left(0,{\log }_{e}x\right)\le y\le 2^x,\frac{1}{2}\le x\le 2\right\}$

So, the area of the region is

${\int }_{\frac{1}{2}}^2{2}^{x}dx-{\int }_1^2\ell nxdx$

$={\left[\frac{2^x}{\ln 2}\right]}_{1/2}^2-{\left[x\ln x-x\right]}_1^2$

$=\frac{\left(2^2\right)-2^{1/2}}{{\log }_{\mathrm{e}}2}-\left(2\ln 2-1\right)$

$=\frac{\left(2^2-\sqrt{2}\right)}{{\log }_{\mathrm{e}}2}-2\ln 2+1$

$\therefore \alpha =2^2-\sqrt{2},\beta =-2,\gamma =1$

So, ${\left(\alpha +\beta -2\gamma \right)}^2$

$={\left(2^2-\sqrt{2}-2-2\right)}^2$

$={\left(-\sqrt{2}\right)}^2=2$