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.