Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
y =

x − 1
,  y = 0,  x = 5;    about the x-axis
V =

Incorrect: Your answer is incorrect.

To find the volume \( V \) of the solid obtained by rotating the region bounded by the given curves about the x-axis, we use the **disk method**. The curves provided are:

\[
y = x - 1, \quad y = 0, \quad \text{and} \quad x = 5
\]

### Step 1: Determine the bounds of integration
The region is bounded horizontally by \( y = 0 \) (the x-axis) and \( y = x - 1 \), and vertically by \( x = 1 \) (where \( y = 0 \)) and \( x = 5 \).

### Step 2: Set up the disk method integral
When rotating around the x-axis, the volume is given by:

\[
V = \pi \int_{a}^{b} [f(x)]^2 \, dx
\]

Here, \( f(x) = x - 1 \), and the limits of integration are from \( x = 1 \) to \( x = 5 \).

### Step 3: Write the integral
\[
V = \pi \int_{1}^{5} (x - 1)^2 \, dx
\]

### Step 4: Expand the integrand
\[
(x - 1)^2 = x^2 - 2x + 1
\]

### Step 5: Integrate
Now, integrate term by term:
\[
V = \pi \int_{1}^{5} (x^2 - 2x + 1) \, dx
\]

\[
V = \pi \left[ \frac{x^3}{3} - x^2 + x \right]_{1}^{5}
\]

### Step 6: Evaluate the integral at the bounds
Evaluate at \( x = 5 \):
\[
\left( \frac{5^3}{3} - 5^2 + 5 \right) = \left( \frac{125}{3} - 25 + 5 \right) = \frac{125}{3} - 20
\]

Evaluate at \( x = 1 \):
\[
\left( \frac{1^3}{3} - 1^2 + 1 \right) = \left( \frac{1}{3} - 1 + 1 \right) = \frac{1}{3}
\]

Now, subtract the two values:
\[
V = \pi \left[ \left( \frac{125}{3} - 20 \right) - \frac{1}{3} \right] = \pi \left( \frac{125}{3} - \frac{60}{3} - \frac{1}{3} \right) = \pi \left( \frac{125 - 60 - 1}{3} \right) = \pi \left( \frac{64}{3} \right)
\]

### Final Answer
\[
V = \frac{64\pi}{3} \text{ cubic units}
\]

This is the correct volume of the solid.

Do you need further details or have any questions? 

Here are 5 related questions for you:

1. How would the integral change if the region was rotated about the y-axis instead?
2. What if the curve was rotated around the line \( y = 2 \)? How would that affect the volume?
3. Can you find the volume if the region was between \( x = 2 \) and \( x = 5 \)?
4. How does the method change if the curves were more complex, such as a quadratic function?
5. What if the problem involved finding the surface area instead of volume?

**Tip:** For rotating a region around a horizontal or vertical line other than the axes, always adjust the radius to account for the distance from the axis of rotation.

Learn how to calculate the volume of a solid obtained by rotating the region bounded by y = x - 1, y = 0, and x = 5 about the x-axis. Detailed step-by-step solution using the disk method integration. Suitable for advanced high school students and anyone learning calculus.

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