To find the area enclosed by the curve \( y = 2^x \) and the lines defined by \( \max \{|x|, |y| \} = 1 \), we will follow these steps:
### Step 1: Understand the boundaries defined by \( \max \{|x|, |y|\} = 1 \)
The equation \( \max \{|x|, |y|\} = 1 \) defines a square with vertices at \( (1, 1) \), \( (1, -1) \), \( (-1, 1) \), and \( (-1, -1) \). This means that the area we are interested in is bounded by the lines \( x = 1 \), \( x = -1 \), \( y = 1 \), and \( y = -1 \).
**Hint:** Sketch the square defined by the lines to visualize the area of interest.
### Step 2: Find the intersection points of the curve and the lines
We need to find where the curve \( y = 2^x \) intersects the lines \( y = 1 \) and \( y = -1 \).
1. **For \( y = 1 \):**
\[
2^x = 1 \implies x = 0
\]
Thus, the point of intersection is \( (0, 1) \).
2. **For \( y = -1 \):**
The function \( y = 2^x \) is always positive, so there are no intersection points with \( y = -1 \).
**Hint:** Check the behavior of the function \( y = 2^x \) to see if it can intersect with negative values.
### Step 3: Determine the area segments
The area can be divided into three segments:
- **Area \( A_1 \)**: The rectangle formed by the lines \( y = 1 \) and \( y = -1 \) from \( x = -1 \) to \( x = 1 \).
- **Area \( A_2 \)**: The square formed by the lines \( x = 1 \) and \( x = -1 \) from \( y = 1 \) to \( y = -1 \).
- **Area \( A_3 \)**: The area under the curve \( y = 2^x \) from \( x = -1 \) to \( x = 0 \).
### Step 4: Calculate the areas
1. **Area \( A_1 \)**:
\[
A_1 = \text{Base} \times \text{Height} = 2 \times 1 = 2
\]
2. **Area \( A_2 \)**:
\[
A_2 = 1 \times 1 = 1
\]
3. **Area \( A_3 \)**:
\[
A_3 = \int_{-1}^{0} 2^x \, dx
\]
The integral of \( 2^x \) is:
\[
\int 2^x \, dx = \frac{2^x}{\ln 2}
\]
Evaluating from \( -1 \) to \( 0 \):
\[
A_3 = \left[ \frac{2^x}{\ln 2} \right]_{-1}^{0} = \frac{2^0}{\ln 2} - \frac{2^{-1}}{\ln 2} = \frac{1}{\ln 2} - \frac{1/2}{\ln 2} = \frac{1 - \frac{1}{2}}{\ln 2} = \frac{1/2}{\ln 2} = \frac{1}{2 \ln 2}
\]
### Step 5: Total area
The total area enclosed by the curve and the lines is:
\[
\text{Total Area} = A_1 + A_2 + A_3 = 2 + 1 + \frac{1}{2 \ln 2} = 3 + \frac{1}{2 \ln 2}
\]
### Final Answer:
The area enclosed by the curve \( y = 2^x \) and the lines defined by \( \max \{|x|, |y|\} = 1 \) is:
\[
\boxed{3 + \frac{1}{2 \ln 2}}
\]
---
