Find the area of the region `{(x,y): x^(2)+y^(2) le 1 le x + y}`

Answer

To find the area of the region defined by the inequalities \( x^2 + y^2 \leq 1 \) and \( x + y \geq 1 \), we can follow these steps:

### Step 1: Understand the inequalities
The inequality \( x^2 + y^2 \leq 1 \) represents a circle centered at the origin (0,0) with a radius of 1. The inequality \( x + y \geq 1 \) represents the area above the line \( x + y = 1 \).

### Step 2: Plot the region
We will sketch the circle and the line:
- The circle \( x^2 + y^2 = 1 \) intersects the axes at points (1,0), (0,1), (-1,0), and (0,-1).
- The line \( x + y = 1 \) intersects the axes at points (1,0) and (0,1).

### Step 3: Identify the area of interest
We need to find the area that lies inside the circle and above the line. The area of interest is bounded by the arc of the circle and the line segment from (1,0) to (0,1).

### Step 4: Set up the integration
To find the area, we will set up an integral. We can express the area as:
\[ \text{Area} = \int_0^1 \left( \text{upper curve} - \text{lower curve} \right) \, dx \]
Where:
- The upper curve is given by the circle: \( y = \sqrt{1 - x^2} \)
- The lower curve is given by the line: \( y = 1 - x \)

Thus, we have:
\[ \text{Area} = \int_0^1 \left( \sqrt{1 - x^2} - (1 - x) \right) \, dx \]

### Step 5: Solve the integral
Now we will compute the integral:
\[ \text{Area} = \int_0^1 \left( \sqrt{1 - x^2} - 1 + x \right) \, dx \]

This can be split into two separate integrals:
\[ \text{Area} = \int_0^1 \sqrt{1 - x^2} \, dx - \int_0^1 1 \, dx + \int_0^1 x \, dx \]

Calculating each integral:
1. The integral \( \int_0^1 \sqrt{1 - x^2} \, dx \) is known to equal \( \frac{\pi}{4} \).
2. The integral \( \int_0^1 1 \, dx = 1 \).
3. The integral \( \int_0^1 x \, dx = \frac{1}{2} \).

Putting it all together:
\[ \text{Area} = \frac{\pi}{4} - 1 + \frac{1}{2} \]
\[ \text{Area} = \frac{\pi}{4} - \frac{2}{2} + \frac{1}{2} = \frac{\pi}{4} - \frac{1}{2} \]

### Final Result
Thus, the area of the region is:
\[ \text{Area} = \frac{\pi}{4} - \frac{1}{2} \text{ square units.} \]