Find the area between x-axis and the curve `y=(x-1)^(2)-25`.

Answer

To find the area between the x-axis and the curve \( y = (x-1)^2 - 25 \), we will follow these steps:

### Step 1: Find the points of intersection with the x-axis
To find the points where the curve intersects the x-axis, we set \( y = 0 \):

\[ (x-1)^2 - 25 = 0 \]

### Step 2: Solve the equation
Rearranging the equation gives:

\[ (x-1)^2 = 25 \]

Taking the square root of both sides, we have:

\[ x - 1 = 5 \quad \text{or} \quad x - 1 = -5 \]

This leads to:

\[ x = 6 \quad \text{or} \quad x = -4 \]

### Step 3: Set up the integral for the area
The area \( A \) between the curve and the x-axis from \( x = -4 \) to \( x = 6 \) can be found using the integral:

\[ A = \int_{-4}^{6} |(x-1)^2 - 25| \, dx \]

### Step 4: Determine the function's behavior
Next, we need to determine where the curve is above or below the x-axis. We can evaluate the function at a point in the interval, say \( x = 0 \):

\[ y = (0-1)^2 - 25 = 1 - 25 = -24 \]

Since \( y < 0 \) at \( x = 0 \), the curve is below the x-axis between \( x = -4 \) and \( x = 6 \). Thus, we can drop the absolute value:

\[ A = \int_{-4}^{6} -((x-1)^2 - 25) \, dx \]

### Step 5: Simplify the integral
This simplifies to:

\[ A = \int_{-4}^{6} (25 - (x-1)^2) \, dx \]

### Step 6: Expand the integrand
Expanding \( (x-1)^2 \):

\[ (x-1)^2 = x^2 - 2x + 1 \]

Thus, the integrand becomes:

\[ A = \int_{-4}^{6} (25 - (x^2 - 2x + 1)) \, dx = \int_{-4}^{6} (24 + 2x - x^2) \, dx \]

### Step 7: Integrate
Now we can integrate term by term:

\[ A = \int_{-4}^{6} (24 + 2x - x^2) \, dx = \left[ 24x + x^2 - \frac{x^3}{3} \right]_{-4}^{6} \]

### Step 8: Evaluate the definite integral
Calculating at the bounds:

1. At \( x = 6 \):

\[ 24(6) + (6)^2 - \frac{(6)^3}{3} = 144 + 36 - 72 = 108 \]

2. At \( x = -4 \):

\[ 24(-4) + (-4)^2 - \frac{(-4)^3}{3} = -96 + 16 + \frac{64}{3} = -80 + \frac{64}{3} = -\frac{240}{3} + \frac{64}{3} = -\frac{176}{3} \]

### Step 9: Combine the results
Thus, the area \( A \) is:

\[ A = 108 - \left(-\frac{176}{3}\right) = 108 + \frac{176}{3} = \frac{324}{3} + \frac{176}{3} = \frac{500}{3} \]

### Final Result
The area between the x-axis and the curve \( y = (x-1)^2 - 25 \) is:

\[ \boxed{\frac{500}{3}} \]