The area bounded by the curve `y^(2)=1-x` and the lines `y=([x])/(x),x=-1, and x=(1)/(2)` is

Options

A. `(3)/(sqrt(2))-(11)/(6)` sq. units
B. `3sqrt(2)-(11)/(4)` sq. units
C. `(6)/(sqrt(2))-(11)/(5)` sq. units
D. None of these

Explanation

To find the area bounded by the curve \( y^2 = 1 - x \) and the lines \( y = \frac{|x|}{x} \), \( x = -1 \), and \( x = \frac{1}{2} \), we can follow these steps:

### Step 1: Understand the equations
The equation \( y^2 = 1 - x \) represents a parabola that opens to the left. The equation \( y = \frac{|x|}{x} \) represents two horizontal lines:
- \( y = 1 \) for \( x > 0 \)
- \( y = -1 \) for \( x < 0 \)

### Step 2: Determine the points of intersection
To find the points of intersection between the parabola and the lines, we will set \( y = 1 \) and \( y = -1 \) in the parabola's equation.

1. For \( y = 1 \):
\[ 1^2 = 1 - x \implies 1 = 1 - x \implies x = 0 \]

2. For \( y = -1 \):
\[ (-1)^2 = 1 - x \implies 1 = 1 - x \implies x = 0 \]

Thus, the only intersection point is at \( (0, 1) \) and \( (0, -1) \).

### Step 3: Set up the integral for the area
We need to compute the area between the curves from \( x = -1 \) to \( x = \frac{1}{2} \).

The area \( A \) can be calculated as:
\[ A = \int_{-1}^{0} (1 - (-\sqrt{1-x})) \, dx + \int_{0}^{\frac{1}{2}} (1 - \sqrt{1-x}) \, dx \]

### Step 4: Calculate the integrals
1. For the first integral \( \int_{-1}^{0} (1 + \sqrt{1-x}) \, dx \):
\[ = \int_{-1}^{0} 1 \, dx + \int_{-1}^{0} \sqrt{1-x} \, dx \]
The first part is:
\[ \int_{-1}^{0} 1 \, dx = [x]_{-1}^{0} = 0 - (-1) = 1 \]

For the second part, we use the substitution \( u = 1 - x \):
\[ dx = -du \quad \text{and when } x = -1, u = 2; \quad \text{when } x = 0, u = 1 \]
Thus:
\[ \int_{-1}^{0} \sqrt{1-x} \, dx = -\int_{2}^{1} \sqrt{u} \, du = \int_{1}^{2} \sqrt{u} \, du = \left[\frac{2}{3} u^{3/2}\right]_{1}^{2} = \frac{2}{3} (2\sqrt{2} - 1) \]

Therefore, the total area for the first integral is:
\[ A_1 = 1 + \frac{2}{3} (2\sqrt{2} - 1) \]

2. For the second integral \( \int_{0}^{\frac{1}{2}} (1 - \sqrt{1-x}) \, dx \):
\[ = \int_{0}^{\frac{1}{2}} 1 \, dx - \int_{0}^{\frac{1}{2}} \sqrt{1-x} \, dx \]
The first part is:
\[ \int_{0}^{\frac{1}{2}} 1 \, dx = \left[x\right]_{0}^{\frac{1}{2}} = \frac{1}{2} \]

For the second part, using the same substitution \( u = 1 - x \):
\[ dx = -du \quad \text{and when } x = 0, u = 1; \quad \text{when } x = \frac{1}{2}, u = \frac{1}{2} \]
Thus:
\[ \int_{0}^{\frac{1}{2}} \sqrt{1-x} \, dx = -\int_{1}^{\frac{1}{2}} \sqrt{u} \, du = \int_{\frac{1}{2}}^{1} \sqrt{u} \, du = \left[\frac{2}{3} u^{3/2}\right]_{\frac{1}{2}}^{1} = \frac{2}{3} (1 - \frac{1}{\sqrt{8}}) = \frac{2}{3} (1 - \frac{1}{2\sqrt{2}}) \]

Therefore, the total area for the second integral is:
\[ A_2 = \frac{1}{2} - \frac{2}{3} (1 - \frac{1}{2\sqrt{2}}) \]

### Step 5: Combine the areas
The total area \( A \) is:
\[ A = A_1 + A_2 \]

### Final Result
After calculating \( A_1 \) and \( A_2 \), we can combine them to get the final area.