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Step 1: Understand the Region
The region R is defined by the set of points
$ \displaystyle R = \left\{(x,y) : \max \{ 0,\ln(x)\} \leq y \leq 2^x \text{ and } \tfrac{1}{2} \leq x \leq 2 \right\}.$
This means:
• For $x$ between $1/2$ and $1$, $\ln(x)$ is negative, so $\max\{0,\ln(x)\} = 0.$
• For $x$ between $1$ and $2$, $\ln(x)$ is non-negative, so $\max\{0,\ln(x)\} = \ln(x).$
Hence, the region can be split into two parts:
1) From $x = 1/2$ to $x = 1$, the region lies between $y=0$ and $y=2^x.$
2) From $x = 1$ to $x = 2$, the region lies between $y=\ln(x)$ and $y=2^x.$
Step 2: Express the Area as Integrals
The total area of R can be computed by summing the areas of the two parts:
$ \displaystyle \text{Area}(R)
= \int_{1/2}^1 \bigl[2^x - 0\bigr]\,dx \;+\; \int_{1}^2 \bigl[2^x - \ln(x)\bigr]\,dx.$
Another equivalent way (as given in the solution) is to combine the integrals:
$ \displaystyle \int_{1/2}^2 2^x \,dx \;-\; \int_{1}^2 \ln(x)\,dx,$
because from $x=1/2$ to $1$, the lower boundary is $0,$ and from $x=1$ to $2,$ the lower boundary is $\ln(x).$
Step 3: Evaluate the Integral of $2^x$ from $x=\tfrac{1}{2}$ to $x=2$
$ \displaystyle \int_{1/2}^2 2^x\,dx \;=\; \left[\frac{2^x}{\ln(2)}\right]_{1/2}^2
= \frac{2^2 - 2^{1/2}}{\ln(2)}
= \frac{4 - \sqrt{2}}{\ln(2)}.
$
Step 4: Evaluate the Integral of $\ln(x)$ from $x=1$ to $x=2$
$ \displaystyle \int_{1}^2 \ln(x)\,dx
= \left[x\,\ln(x) - x\right]_{1}^{2}
= \bigl(2 \ln(2) - 2\bigr) - \bigl(1 \cdot \ln(1) - 1\bigr)
= 2\ln(2) - 2 - (0 - 1)
= 2\ln(2) - 2 + 1
= 2\ln(2) - 1.
$
Step 5: Combine the Results to Find the Area
$ \displaystyle \text{Area}(R)
= \int_{1/2}^2 2^x\,dx \;-\; \int_{1}^2 \ln(x)\,dx
= \frac{4 - \sqrt{2}}{\ln(2)} \;-\; \bigl(2\ln(2) - 1\bigr).
$
Rewriting:
$ \displaystyle \text{Area}(R)
= \frac{4 - \sqrt{2}}{\ln(2)} \;-\; 2\ln(2) \;+\; 1.
$
This expression matches the form
$ \displaystyle \alpha\, \bigl(\ln(2)\bigr)^{-1} \;+\; \beta\, \ln(2) \;+\; \gamma,$
so we identify:
$ \alpha = 4 - \sqrt{2},\,$
$ \beta = -\,2,\,$
$ \gamma = 1.\,$
Step 6: Compute the Required Combination $(\alpha + \beta - 2\gamma)^2$
From the problem statement, we need
$ \displaystyle (\alpha + \beta - 2\lambda)^2
\text{ or equivalently } (\alpha + \beta - 2\gamma)^2
\text{ (depending on the variable naming).}$
Given our identified values:
$ \displaystyle \alpha + \beta - 2\gamma
= (4 - \sqrt{2}) + (-2) - 2(1)
= 4 - \sqrt{2} - 2 - 2
= 0 - \sqrt{2}
= -\,\sqrt{2}.
$
Hence,
$ \displaystyle (\alpha + \beta - 2\gamma)^2
= (-\,\sqrt{2})^2
= 2.
$
Step 7: Final Answer
The value of $ \displaystyle (\alpha + \beta - 2\gamma)^2 $ is
2.
