If the area of the bounded region R={(x, y): | vedclass

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(D) The region $R$ is defined by $\frac{1}{2} \leq x \leq 2$ and $0 \leq y \leq 2^x$ for $x \in [\frac{1}{2}, 1]$,and $\log_e x \leq y \leq 2^x$ for $

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$2$

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(D) The region $R$ is defined by $\frac{1}{2} \leq x \leq 2$ and $0 \leq y \leq 2^x$ for $x \in [\frac{1}{2}, 1]$,and $\log_e x \leq y \leq 2^x$ for $x \in [1, 2]$.Area $A = \int_{1/2}^{1} 2^x \, dx + \int_{1}^{2} (2^x - \log_e x) \, dx$$A = \int_{1/2}^{2} 2^x \, dx - \int_{1}^{2} \log_e x \, dx$$A = \left[ \frac{2^x}{\log_e 2} \right]_{1/2}^{2} - [x \log_e x - x]_{1}^{2}$$A = \frac{2^2 - 2^{1/2}}{\log_e 2} - ((2 \log_e 2 - 2) - (1 \log_e 1 - 1))$$A = \frac{4 - \sqrt{2}}{\log_e 2} - (2 \log_e 2 - 2 + 1)$$A = (4 - \sqrt{2})(\log_e 2)^{-1} - 2(\log_e 2) + 1$Comparing with $\alpha(\log_e 2)^{-1} + \beta(\log_e 2) + \gamma$,we get $\alpha = 4 - \sqrt{2}$,$\beta = -2$,$\gamma = 1$.Now,$(\alpha + \beta - 2\gamma)^2 = (4 - \sqrt{2} - 2 - 2(1))^2 = (4 - \sqrt{2} - 2 - 2)^2 = (-\sqrt{2})^2 = 2$.

If the area of the bounded region $R=\{(x, y): \max \{0, \log _{e} x\} \leq y \leq 2^{x}, \frac{1}{2} \leq x \leq 2\}$ is $\alpha(\log _{e} 2)^{-1}+\beta(\log _{e} 2)+\gamma$,then the value of $(\alpha+\beta-2 \gamma)^{2}$ is equal to:

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