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

Answer

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

### Step 1: Find the intersection points of the curves
We start by substituting \( y = x - 5 \) into the equation \( x = |y^2 - 1| \).

1. Substitute \( y = x - 5 \) into \( x = |y^2 - 1| \):
\[ x = |(x - 5)^2 - 1| \]
This gives us two cases to consider based on the definition of absolute value.

### Step 2: Solve for both cases
**Case 1:** \( (x - 5)^2 - 1 \geq 0 \)
\[ x = (x - 5)^2 - 1 \]
Expanding and rearranging:
\[ x = x^2 - 10x + 25 - 1 \implies x^2 - 11x + 24 = 0 \]
Factoring:
\[ (x - 3)(x - 8) = 0 \implies x = 3 \text{ or } x = 8 \]

**Case 2:** \( (x - 5)^2 - 1 < 0 \)
\[ x = -((x - 5)^2 - 1) \implies x = -((x^2 - 10x + 25) - 1) \]
Rearranging gives:
\[ x = -x^2 + 10x - 24 \implies x^2 - 9x + 24 = 0 \]
This quadratic does not yield real solutions (discriminant is negative).

### Step 3: Find corresponding y-values
Now, we substitute \( x = 3 \) and \( x = 8 \) back into \( y = x - 5 \):
- For \( x = 3 \):
\[ y = 3 - 5 = -2 \]
- For \( x = 8 \):
\[ y = 8 - 5 = 3 \]

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = 3 \) to \( x = 8 \) can be expressed as:
\[ A = \int_{3}^{8} (y_{\text{top}} - y_{\text{bottom}}) \, dx \]
Where:
- \( y_{\text{top}} = x - 5 \)
- \( y_{\text{bottom}} = \sqrt{x + 1} \) (from \( x = |y^2 - 1| \))

### Step 5: Calculate the area using integration
Thus, we have:
\[ A = \int_{3}^{8} \left((x - 5) - \sqrt{x + 1}\right) \, dx \]

Calculating the integral:
1. Integrate \( x - 5 \):
\[ \int (x - 5) \, dx = \frac{x^2}{2} - 5x \]
2. Integrate \( \sqrt{x + 1} \):
\[ \int \sqrt{x + 1} \, dx = \frac{2}{3}(x + 1)^{3/2} \]

Putting it all together:
\[ A = \left[ \left(\frac{x^2}{2} - 5x\right) - \frac{2}{3}(x + 1)^{3/2} \right]_{3}^{8} \]

### Step 6: Evaluate the definite integral
Calculating the upper limit \( x = 8 \):
\[ \left(\frac{8^2}{2} - 5 \cdot 8\right) - \frac{2}{3}(8 + 1)^{3/2} = \left(32 - 40\right) - \frac{2}{3}(9)^{3/2} = -8 - \frac{2}{3}(27) = -8 - 18 = -26 \]

Calculating the lower limit \( x = 3 \):
\[ \left(\frac{3^2}{2} - 5 \cdot 3\right) - \frac{2}{3}(3 + 1)^{3/2} = \left(\frac{9}{2} - 15\right) - \frac{2}{3}(4)^{3/2} = \left(\frac{9}{2} - 15\right) - \frac{2}{3}(8) = \left(\frac{9 - 30}{2}\right) - \frac{16}{3} = -\frac{21}{2} - \frac{16}{3} \]

Finding a common denominator and simplifying gives the final area.

### Final Area Calculation
Combining the results from the upper and lower limits, we find:
\[ A = \text{(Upper Limit)} - \text{(Lower Limit)} = -26 - \left(-\frac{21}{2} - \frac{16}{3}\right) \]

### Conclusion
The area of the region bounded by the curves is:
\[ A = \frac{61}{6} \text{ square units} \]