1. **Problem statement:** Find the integral expression for the volume of the solid formed by rotating the region bounded by the curve $f(x) = 2x - 1$, the x-axis, and the vertical lines $x=1$ and $x=3$ about the vertical line $x=4$. 2. **Formula used:** When rotating around a vertical line $x = a$, the volume $V$ can be found using the shell method: $$V = 2\pi \int_{x_1}^{x_2} (\text{radius})(\text{height}) \, dx$$ where: - Radius = distance from the shell to the axis of rotation = $|a - x|$ - Height = value of the function = $f(x)$ 3. **Apply to this problem:** - Axis of rotation: $x=4$ - Radius = $4 - x$ - Height = $2x - 1$ - Limits: $x=1$ to $x=3$ 4. **Integral expression:** $$V = 2\pi \int_1^3 (4 - x)(2x - 1) \, dx$$ 5. **Check options:** - Option 1: $\pi \int_1^3 (4 - x)(2x - 1) \, dx$ (missing factor 2) - Option 2: $2\pi \int_1^3 (4 - x)(x^2 - x) \, dx$ (wrong height function) - Option 3: $\pi \int_1^3 (4 - x)^2 (2x - 1) \, dx$ (wrong radius squared, not shell method) - Option 4: $2\pi \int_1^3 (4 - x)(2x - 1) \, dx$ (correct) **Final answer:** $$V = 2\pi \int_1^3 (4 - x)(2x - 1) \, dx$$