Find the area bounded by the curves `x = |y^(2)-1|` and `y = x- 5`

Answer

To find the area 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 \( x = |y^2 - 1| \) consists of two parts:
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 \( y = x - 5 \).

### Step 2: Find intersection points
To find the area, we first need to find the points where these two curves intersect.

1. **For \( y^2 - 1 \)**:
\[ y = x - 5 \implies x = y + 5 \]
Substitute into \( x = y^2 - 1 \):
\[ y + 5 = y^2 - 1 \implies y^2 - y - 6 = 0 \]
Factoring gives:
\[ (y - 3)(y + 2) = 0 \implies y = 3 \text{ or } y = -2 \]
Substituting back to find \( x \):
- For \( y = 3 \): \( x = 3 + 5 = 8 \)
- For \( y = -2 \): \( x = -2 + 5 = 3 \)

2. **For \( -y^2 + 1 \)**:
\[ y + 5 = -y^2 + 1 \implies y^2 + y + 4 = 0 \]
This quadratic has no real solutions (discriminant \( < 0 \)).

Thus, the intersection points are \( (3, -2) \) and \( (8, 3) \).

### Step 3: Set up the integral
The area \( A \) between the curves from \( y = -2 \) to \( y = 3 \) can be found by integrating the difference between the right curve and the left curve:
\[ A = \int_{-2}^{3} \left( (y^2 - 1) - (y + 5) \right) \, dy \]

### Step 4: Simplify the integrand
\[ A = \int_{-2}^{3} \left( y^2 - 1 - y - 5 \right) \, dy = \int_{-2}^{3} \left( y^2 - y - 6 \right) \, dy \]

### Step 5: Calculate the integral
Now, we compute the integral:
\[ A = \int_{-2}^{3} (y^2 - y - 6) \, dy \]
Calculating the antiderivative:
\[ \frac{y^3}{3} - \frac{y^2}{2} - 6y \]
Evaluating from \( -2 \) to \( 3 \):
\[ A = \left[ \frac{3^3}{3} - \frac{3^2}{2} - 6 \cdot 3 \right] - \left[ \frac{(-2)^3}{3} - \frac{(-2)^2}{2} - 6 \cdot (-2) \right] \]
Calculating each part:
- Upper limit:
\[ = \left[ 9 - \frac{9}{2} - 18 \right] = 9 - 4.5 - 18 = -13.5 \]
- Lower limit:
\[ = \left[ -\frac{-8}{3} - 2 + 12 \right] = \left[ \frac{8}{3} - 2 + 12 \right] = \frac{8}{3} + 10 = \frac{38}{3} \]

### Step 6: Combine results
\[ A = -13.5 - \left( \frac{38}{3} \right) = -13.5 - 12.67 \approx -26.17 \]
Taking the absolute value for area:
\[ A \approx 26.17 \]

### Final Answer
The area bounded by the curves is approximately \( 26.17 \) square units.