The area of the region bounded by curves `y=1-x^(2), x+y+1=0 and x-y-1=0` is

Options

A. `(10)/(3)`
B. `(7)/(3)`
C. `(8)/(3)`
D. 3

Answer

To find the area of the region bounded by the curves \( y = 1 - x^2 \), \( x + y + 1 = 0 \), and \( x - y - 1 = 0 \), we will follow these steps:

### Step 1: Identify the curves
The curves are:
1. \( y = 1 - x^2 \) (a downward-opening parabola)
2. \( x + y + 1 = 0 \) (a straight line)
3. \( x - y - 1 = 0 \) (another straight line)

### Step 2: Find the points of intersection
To find the area bounded by these curves, we first need to find the points of intersection.

1. **Intersection of \( y = 1 - x^2 \) and \( x + y + 1 = 0 \)**:
\[ y = -x - 1 \]
Substitute \( y \) in the parabola equation:
\[ 1 - x^2 = -x - 1 \]
Rearranging gives:
\[ x^2 - x - 2 = 0 \]
Factoring:
\[ (x - 2)(x + 1) = 0 \implies x = 2 \text{ or } x = -1 \]
Now substituting back to find \( y \):
- For \( x = 2 \): \( y = -2 - 1 = -3 \) → Point \( (2, -3) \)
- For \( x = -1 \): \( y = 1 - (-1)^2 = 0 \) → Point \( (-1, 0) \)

2. **Intersection of \( y = 1 - x^2 \) and \( x - y - 1 = 0 \)**:
\[ y = x - 1 \]
Substitute \( y \) in the parabola equation:
\[ 1 - x^2 = x - 1 \]
Rearranging gives:
\[ x^2 + x - 2 = 0 \]
Factoring:
\[ (x - 1)(x + 2) = 0 \implies x = 1 \text{ or } x = -2 \]
Now substituting back to find \( y \):
- For \( x = 1 \): \( y = 1 - 1^2 = 0 \) → Point \( (1, 0) \)
- For \( x = -2 \): \( y = -2 - 1 = -3 \) → Point \( (-2, -3) \)

### Step 3: Identify the bounded area
The points of intersection are:
- \( (-1, 0) \)
- \( (1, 0) \)
- \( (2, -3) \)
- \( (-2, -3) \)

The area we need to calculate is between the parabola and the two lines from \( x = -1 \) to \( x = 1 \).

### Step 4: Calculate the area
1. **Area under the parabola from \( x = -1 \) to \( x = 1 \)**:
\[ A_{\text{parabola}} = \int_{-1}^{1} (1 - x^2) \, dx \]
Evaluating the integral:
\[ = \left[ x - \frac{x^3}{3} \right]_{-1}^{1} = \left( 1 - \frac{1}{3} \right) - \left( -1 + \frac{1}{3} \right) \]
\[ = \left( \frac{2}{3} \right) - \left( -\frac{2}{3} \right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \]

2. **Area of the triangle formed by the lines**:
The base of the triangle is from \( (-1, 0) \) to \( (1, 0) \) which is 2 units, and the height is the distance from the x-axis to the line \( y = -3 \) (which is 3 units).
\[ A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 3 = 3 \]

### Step 5: Total area
The total area bounded by the curves is:
\[ \text{Total Area} = A_{\text{parabola}} + A_{\text{triangle}} = \frac{4}{3} + 3 = \frac{4}{3} + \frac{9}{3} = \frac{13}{3} \]

### Final Answer
The area of the region bounded by the curves is \( \frac{13}{3} \) square units.