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The problem in the image asks to calculate the volume of a solid of revolution formed by rotating the region bounded by the lines \( y = 5x - 7 \), \( y = 0 \), \( x = 2 \), and \( x = 8 \) around the **x-axis**.

### Approach:

1. **Set up the boundaries**:
   - The curve \( y = 5x - 7 \) describes the upper boundary.
   - The limits of integration are from \( x = 2 \) to \( x = 8 \).

2. **Volume of a solid of revolution**:
   The formula for the volume of a solid of revolution around the x-axis is given by:
   \[
   V = \pi \int_{a}^{b} [f(x)]^2 \, dx
   \]
   where \( f(x) \) is the function being rotated (in this case, \( y = 5x - 7 \)), and \( a \) and \( b \) are the bounds on the x-axis (here, \( x = 2 \) to \( x = 8 \)).

3. **Substitute the function**:
   Using \( y = 5x - 7 \), the volume integral becomes:
   \[
   V = \pi \int_{2}^{8} (5x - 7)^2 \, dx
   \]

4. **Expand the integrand**:
   \[
   (5x - 7)^2 = 25x^2 - 70x + 49
   \]
   Thus, the volume integral becomes:
   \[
   V = \pi \int_{2}^{8} (25x^2 - 70x + 49) \, dx
   \]

5. **Compute the integral**:
   \[
   V = \pi \left[ \frac{25x^3}{3} - 35x^2 + 49x \right]_{2}^{8}
   \]

6. **Evaluate the definite integral**:
   First, evaluate the expression at \( x = 8 \):
   \[
   \frac{25(8)^3}{3} - 35(8)^2 + 49(8) = \frac{25(512)}{3} - 35(64) + 392 = \frac{12800}{3} - 2240 + 392
   \]
   \[
   = \frac{12800}{3} - 1848
   \]

   Now evaluate at \( x = 2 \):
   \[
   \frac{25(2)^3}{3} - 35(2)^2 + 49(2) = \frac{25(8)}{3} - 35(4) + 98 = \frac{200}{3} - 140 + 98 = \frac{200}{3} - 42
   \]

   Subtracting the results gives:
   \[
   V = \pi \left( \frac{12800}{3} - 1848 - \left( \frac{200}{3} - 42 \right) \right)
   = \pi \left( \frac{12800 - 200}{3} - 1848 + 42 \right)
   = \pi \left( \frac{12600}{3} - 1806 \right)
   = \pi \left( 4200 - 1806 \right)
   = \pi (2394)
   \]

7. **Final answer**:
   \[
   V = 2394\pi \, \text{units}^3
   \]

The volume of the solid of revolution is \( 2394\pi \) cubic units.

---

Would you like more details or have any questions? Here are some related questions:

1. How does the formula for the volume of revolution around the x-axis change if we rotate around the y-axis?
2. What are other methods for finding volumes of solids of revolution besides using integrals?
3. How can we calculate the surface area of the solid generated in this problem?
4. What happens to the volume if we rotate around a different axis, say \( y = 5 \)?
5. How would the volume change if the limits were \( x = 0 \) to \( x = 8 \)?

**Tip**: When computing volumes of revolution, always visualize the shape and boundaries to ensure the correct limits and function.

A região delimitada pelas retas y = 5x - 7, y = 0, x = 2 e x = 8, rotaciona em torno do eixo x. Nesse caso, o volume do sólido de revolução gerado é:

This problem asks to calculate the volume of a solid of revolution formed by rotating the region bounded by y = 5x - 7, y = 0, and vertical lines x = 2 and x = 8 around the x-axis. The solution involves applying the formula for the volume of a solid of revolution using a definite integral.

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