Key Concepts and Formulas

• Area Between Curves: The area between two curves $f(x)$ and $g(x)$ from $a$ to $b$, where $f(x) \ge g(x)$ on $[a,b]$, is given by $\int_a^b [f(x) - g(x)] dx$.
• Integral of Exponential Functions: $\int a^x dx = \frac{a^x}{\ln a} + C$.
• Integration by Parts: $\int u \, dv = uv - \int v \, du$.

Step-by-Step Solution

Step 1: Define the Region and Identify the Curves

The problem defines the region $R$ as: $R = \left\{ {(x,y):\max \{ 0,{{\log }_e}x\} \le y \le {2^x},{1 \over 2} \le x \le 2} \right\}$ This tells us:

• Upper curve: $y_U = 2^x$
• Lower curve: $y_L = \max\{0, \ln x\}$
• Interval: $x \in \left[\frac{1}{2}, 2\right]$

Step 2: Analyze the Lower Curve

The lower curve $y_L = \max\{0, \ln x\}$ is a piecewise function. We need to consider when $\ln x$ is positive or negative. $\ln x = 0$ when $x = 1$.

• For $\frac{1}{2} \le x \le 1$, $\ln x \le 0$, so $y_L = \max\{0, \ln x\} = 0$.
• For $1 \le x \le 2$, $\ln x \ge 0$, so $y_L = \max\{0, \ln x\} = \ln x$.

Step 3: Set Up the Integral for the Area

Since the lower curve changes definition at $x = 1$, we split the integral into two parts:

$\text{Area} = \int_{1/2}^1 (2^x - 0) \, dx + \int_1^2 (2^x - \ln x) \, dx$ $\text{Area} = \int_{1/2}^1 2^x \, dx + \int_1^2 2^x \, dx - \int_1^2 \ln x \, dx$ $\text{Area} = \int_{1/2}^2 2^x \, dx - \int_1^2 \ln x \, dx$ Let $I_1 = \int_{1/2}^2 2^x \, dx$ and $I_2 = \int_1^2 \ln x \, dx$. Then, $\text{Area} = I_1 - I_2$.

Step 4: Evaluate $I_1 = \int_{1/2}^2 2^x \, dx$

Using the formula $\int a^x dx = \frac{a^x}{\ln a} + C$: $I_1 = \left[ \frac{2^x}{\ln 2} \right]_{1/2}^2 = \frac{2^2}{\ln 2} - \frac{2^{1/2}}{\ln 2} = \frac{4 - \sqrt{2}}{\ln 2}$

Step 5: Evaluate $I_2 = \int_1^2 \ln x \, dx$

We use integration by parts: $u = \ln x$, $dv = dx$. Then $du = \frac{1}{x} dx$, $v = x$. $\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C$ $I_2 = [x \ln x - x]_1^2 = (2 \ln 2 - 2) - (1 \ln 1 - 1) = 2 \ln 2 - 2 - (0 - 1) = 2 \ln 2 - 1$

Step 6: Calculate the Area

$\text{Area} = I_1 - I_2 = \frac{4 - \sqrt{2}}{\ln 2} - (2 \ln 2 - 1) = \frac{4 - \sqrt{2}}{\ln 2} - 2 \ln 2 + 1$

Step 7: Match the Area to the Given Form

The area is given as $\alpha (\ln 2)^{-1} + \beta (\ln 2) + \gamma$. Comparing this with our result: $\text{Area} = (4 - \sqrt{2}) (\ln 2)^{-1} + (-2) (\ln 2) + 1$ So, $\alpha = 4 - \sqrt{2}$, $\beta = -2$, and $\gamma = 1$.

Step 8: Calculate $(\alpha + \beta - 2\gamma)^2$

$\alpha + \beta - 2\gamma = (4 - \sqrt{2}) + (-2) - 2(1) = 4 - \sqrt{2} - 2 - 2 = -\sqrt{2}$ $(\alpha + \beta - 2\gamma)^2 = (-\sqrt{2})^2 = 2$

Common Mistakes & Tips

• Remember to split the integral when the lower bound function changes.
• Be careful with signs during integration by parts.
• Double-check the limits of integration after applying integration by parts.

Summary

We calculated the area of the region by splitting the integral based on the piecewise definition of the lower bounding curve. We then used integration by parts to evaluate the integral of $\ln x$. Finally, we matched the calculated area to the given form and found the required expression's value to be 2.

Final Answer

The final answer is \boxed{2}, which corresponds to option (B).