Find the area of the region `{(x, y) : x^(2) + y^(2) le 4, x + y ge 2}`, using the method of integration.

Answer

To find the area of the region defined by the inequalities \( x^2 + y^2 \leq 4 \) and \( x + y \geq 2 \) using integration, we will follow these steps:

### Step 1: Understand the Region
The equation \( x^2 + y^2 = 4 \) describes a circle with a radius of 2 centered at the origin (0,0). The line \( x + y = 2 \) can be rewritten as \( y = 2 - x \). We need to find the area of the region that lies inside the circle and above the line.

### Step 2: Find Points of Intersection
To find the points where the line intersects the circle, we substitute \( y = 2 - x \) into the circle's equation:
\[ x^2 + (2 - x)^2 = 4 \]
Expanding this:
\[ x^2 + (4 - 4x + x^2) = 4 \]
\[ 2x^2 - 4x + 4 - 4 = 0 \]
\[ 2x^2 - 4x = 0 \]
Factoring out \( 2x \):
\[ 2x(x - 2) = 0 \]
This gives us \( x = 0 \) and \( x = 2 \). We can find the corresponding \( y \) values:
- For \( x = 0 \): \( y = 2 \)
- For \( x = 2 \): \( y = 0 \)

Thus, the points of intersection are \( (0, 2) \) and \( (2, 0) \).

### Step 3: Set Up the Integral
We will integrate the area between the line and the circle from \( x = 0 \) to \( x = 2 \). The upper curve is given by the circle, and the lower curve is given by the line.

The equation of the circle can be expressed as:
\[ y = \sqrt{4 - x^2} \]
The equation of the line is:
\[ y = 2 - x \]

### Step 4: Write the Area Integral
The area \( A \) can be calculated using the integral:
\[ A = \int_{0}^{2} \left( \sqrt{4 - x^2} - (2 - x) \right) \, dx \]

### Step 5: Solve the Integral
Now we will compute the integral:
\[ A = \int_{0}^{2} \left( \sqrt{4 - x^2} - 2 + x \right) \, dx \]

We can break this into two separate integrals:
\[ A = \int_{0}^{2} \sqrt{4 - x^2} \, dx - \int_{0}^{2} 2 \, dx + \int_{0}^{2} x \, dx \]

1. **Calculate \( \int_{0}^{2} \sqrt{4 - x^2} \, dx \)**:
This is the area of a quarter-circle with radius 2:
\[ \int_{0}^{2} \sqrt{4 - x^2} \, dx = \frac{1}{4} \pi (2^2) = \frac{1}{4} \pi \cdot 4 = \pi \]

2. **Calculate \( \int_{0}^{2} 2 \, dx \)**:
\[ \int_{0}^{2} 2 \, dx = 2x \bigg|_{0}^{2} = 2(2) - 2(0) = 4 \]

3. **Calculate \( \int_{0}^{2} x \, dx \)**:
\[ \int_{0}^{2} x \, dx = \frac{x^2}{2} \bigg|_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = 2 \]

### Step 6: Combine the Results
Now substituting back into the area formula:
\[ A = \pi - 4 + 2 = \pi - 2 \]

### Final Answer
Thus, the area of the region is:
\[ \boxed{\pi - 2} \]