1. **Problem:** Find the volume of the solid obtained by rotating the region bounded by $y = 2 - \frac{1}{2}x$, $y=0$, $x=1$, and $x=2$ about the x-axis. 2. **Formula:** For rotation about the x-axis, the volume $V$ is given by the disk method: $$V = \pi \int_a^b [f(x)]^2 \, dx$$ where $f(x)$ is the function defining the radius of the disk. 3. **Identify bounds and function:** Here, $f(x) = 2 - \frac{1}{2}x$, $a=1$, $b=2$. 4. **Set up the integral:** $$V = \pi \int_1^2 \left(2 - \frac{1}{2}x\right)^2 \, dx$$ 5. **Expand the square:** $$\left(2 - \frac{1}{2}x\right)^2 = 4 - 2x + \frac{1}{4}x^2$$ 6. **Integral becomes:** $$V = \pi \int_1^2 \left(4 - 2x + \frac{1}{4}x^2\right) dx$$ 7. **Integrate term-by-term:** $$\int_1^2 4 \, dx = 4x \Big|_1^2 = 4(2) - 4(1) = 8 - 4 = 4$$ $$\int_1^2 (-2x) \, dx = -2 \cdot \frac{x^2}{2} \Big|_1^2 = -x^2 \Big|_1^2 = -(4 - 1) = -3$$ $$\int_1^2 \frac{1}{4}x^2 \, dx = \frac{1}{4} \cdot \frac{x^3}{3} \Big|_1^2 = \frac{1}{12}(8 - 1) = \frac{7}{12}$$ 8. **Sum the integrals:** $$4 - 3 + \frac{7}{12} = 1 + \frac{7}{12} = \frac{12}{12} + \frac{7}{12} = \frac{19}{12}$$ 9. **Multiply by $\pi$:** $$V = \pi \cdot \frac{19}{12} = \frac{19\pi}{12}$$ **Final answer:** $$\boxed{\frac{19\pi}{12}}$$ This is the volume of the solid formed by rotating the given region about the x-axis.