If A is the area bonded by ` x= |y^(2) -1| and y= x -5 ` then A is equal to

Answer

To find the area \( A \) bounded by the curves \( x = |y^2 - 1| \) and \( y = x - 5 \), we will follow these steps:

### Step 1: Understand the curves
The first curve is given by \( x = |y^2 - 1| \). This means we need to consider two cases based on the definition of the absolute value:
1. \( x = y^2 - 1 \) when \( y^2 - 1 \geq 0 \) (i.e., \( |y| \geq 1 \))
2. \( x = -(y^2 - 1) = -y^2 + 1 \) when \( y^2 - 1 < 0 \) (i.e., \( |y| < 1 \))

The second curve is a straight line given by \( y = x - 5 \).

### Step 2: Find points of intersection
To find the area, we need to determine the points where these curves intersect. We will set \( y = x - 5 \) in both cases of the first curve.

**Case 1: \( x = y^2 - 1 \)**
Substituting \( y = x - 5 \):
\[ x = (x - 5)^2 - 1 \]
Expanding and rearranging gives:
\[ x = x^2 - 10x + 25 - 1 \]
\[ 0 = x^2 - 11x + 24 \]
Factoring:
\[ 0 = (x - 3)(x - 8) \]
Thus, \( x = 3 \) and \( x = 8 \).

**Case 2: \( x = -y^2 + 1 \)**
Substituting \( y = x - 5 \):
\[ x = -(x - 5)^2 + 1 \]
Expanding and rearranging gives:
\[ x = - (x^2 - 10x + 25) + 1 \]
\[ 0 = x^2 - 11x + 24 \]
This gives the same solutions \( x = 3 \) and \( x = 8 \).

### Step 3: Determine the corresponding \( y \) values
For \( x = 3 \):
\[ y = 3 - 5 = -2 \]
For \( x = 8 \):
\[ y = 8 - 5 = 3 \]
Thus, the points of intersection are \( (3, -2) \) and \( (8, 3) \).

### Step 4: Set up the area integral
We will compute the area between these curves from \( y = -2 \) to \( y = 3 \).

The area \( A \) can be expressed as:
\[ A = \int_{y_1}^{y_2} (x_{\text{line}} - x_{\text{parabola}}) \, dy \]
Where:
- \( x_{\text{line}} = y + 5 \)
- \( x_{\text{parabola}} = |y^2 - 1| \)

### Step 5: Split the integral
Since the parabola changes at \( y = -1 \):
1. From \( y = -2 \) to \( y = -1 \):
\[ A_1 = \int_{-2}^{-1} ((y + 5) - (-y^2 + 1)) \, dy \]
2. From \( y = -1 \) to \( y = 3 \):
\[ A_2 = \int_{-1}^{3} ((y + 5) - (y^2 - 1)) \, dy \]

### Step 6: Calculate the integrals
**For \( A_1 \):**
\[ A_1 = \int_{-2}^{-1} (y + 5 + y^2 - 1) \, dy = \int_{-2}^{-1} (y^2 + y + 4) \, dy \]
Calculating:
\[ = \left[ \frac{y^3}{3} + \frac{y^2}{2} + 4y \right]_{-2}^{-1} \]
Evaluating this gives:
\[ = \left( \frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 4(-1) \right) - \left( \frac{(-2)^3}{3} + \frac{(-2)^2}{2} + 4(-2) \right) \]
\[ = \left( -\frac{1}{3} + \frac{1}{2} - 4 \right) - \left( -\frac{8}{3} + 2 - 8 \right) \]
Calculating this will yield a value.

**For \( A_2 \):**
\[ A_2 = \int_{-1}^{3} (y + 5 - (y^2 - 1)) \, dy = \int_{-1}^{3} (-y^2 + y + 6) \, dy \]
Calculating:
\[ = \left[ -\frac{y^3}{3} + \frac{y^2}{2} + 6y \right]_{-1}^{3} \]
Evaluating this gives:
\[ = \left( -\frac{27}{3} + \frac{9}{2} + 18 \right) - \left( -\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 6(-1) \right) \]
Calculating this will yield another value.

### Step 7: Add the areas
Finally, add \( A_1 \) and \( A_2 \) to get the total area \( A \).