If `abs(x-1) le y le sqrt(5-x^2)`, then the area of region bounded by the curves is

Options

A. `(5pi)/4-1/2`
B. `(5pi)/4-3/2`
C. `(3pi)/4-1/2`
D. `cos^-1(1/3)-1/2`

Answer

To find the area of the region bounded by the curves defined by the inequalities \( |x-1| \leq y \leq \sqrt{5-x^2} \), we will follow these steps:

### Step 1: Understand the inequalities
The inequalities can be split into two parts:
1. \( y \geq |x-1| \)
2. \( y \leq \sqrt{5 - x^2} \)

### Step 2: Analyze the curves
- The curve \( y = |x-1| \) represents two lines:
- \( y = x - 1 \) for \( x \geq 1 \)
- \( y = -x + 1 \) for \( x < 1 \)

- The curve \( y = \sqrt{5 - x^2} \) represents the upper half of a circle with radius \( \sqrt{5} \) centered at the origin.

### Step 3: Find intersection points
To find the points where these curves intersect, we set \( |x-1| = \sqrt{5 - x^2} \).

#### Case 1: \( x - 1 = \sqrt{5 - x^2} \)
Squaring both sides:
\[ (x - 1)^2 = 5 - x^2 \]
Expanding and rearranging gives:
\[ x^2 - 2x + 1 + x^2 - 5 = 0 \implies 2x^2 - 2x - 4 = 0 \implies x^2 - x - 2 = 0 \]
Factoring:
\[ (x - 2)(x + 1) = 0 \implies x = 2 \text{ or } x = -1 \]

#### Case 2: \( -x + 1 = \sqrt{5 - x^2} \)
Squaring both sides:
\[ (-x + 1)^2 = 5 - x^2 \]
Expanding and rearranging gives:
\[ x^2 - 2x + 1 + x^2 - 5 = 0 \implies 2x^2 - 2x - 4 = 0 \implies x^2 - x - 2 = 0 \]
This case also yields \( x = 2 \) or \( x = -1 \).

### Step 4: Find corresponding y-values
For \( x = 2 \):
\[ y = |2 - 1| = 1 \]
For \( x = -1 \):
\[ y = |-1 - 1| = 2 \]
Thus, the points of intersection are \( (2, 1) \) and \( (-1, 2) \).

### Step 5: Set up the area integral
The area can be found by integrating the difference between the upper curve and the lower curve from \( x = -1 \) to \( x = 2 \):
\[ \text{Area} = \int_{-1}^{2} \left( \sqrt{5 - x^2} - |x - 1| \right) \, dx \]

### Step 6: Break the integral into two parts
From \( x = -1 \) to \( x = 1 \), \( |x - 1| = -x + 1 \):
\[ \text{Area}_1 = \int_{-1}^{1} \left( \sqrt{5 - x^2} - (-x + 1) \right) \, dx = \int_{-1}^{1} \left( \sqrt{5 - x^2} + x - 1 \right) \, dx \]

From \( x = 1 \) to \( x = 2 \), \( |x - 1| = x - 1 \):
\[ \text{Area}_2 = \int_{1}^{2} \left( \sqrt{5 - x^2} - (x - 1) \right) \, dx = \int_{1}^{2} \left( \sqrt{5 - x^2} - x + 1 \right) \, dx \]

### Step 7: Calculate the integrals
1. **For Area_1**:
\[ \int_{-1}^{1} \sqrt{5 - x^2} \, dx + \int_{-1}^{1} x \, dx - \int_{-1}^{1} 1 \, dx \]
The integral \( \int_{-1}^{1} x \, dx = 0 \) and \( \int_{-1}^{1} 1 \, dx = 2 \).
The integral \( \int_{-1}^{1} \sqrt{5 - x^2} \, dx \) can be calculated using the area of a semicircle.

2. **For Area_2**:
\[ \int_{1}^{2} \sqrt{5 - x^2} \, dx - \int_{1}^{2} x \, dx + \int_{1}^{2} 1 \, dx \]
The integral \( \int_{1}^{2} x \, dx \) and \( \int_{1}^{2} 1 \, dx \) can be calculated directly.

### Step 8: Combine the areas
Finally, combine both areas to get the total area.

### Final Answer
Thus, the area of the region bounded by the curves is:
\[ \text{Area} = \text{Area}_1 + \text{Area}_2 \]