1. **Problem Statement:** We want to find the surface area of a solid formed by revolving a curve $y=f(x)$ around the x-axis from $x=a$ to $x=b$. 2. **Formula and Explanation:** The surface area $S$ of a frustum (a truncated cone) formed by revolving a small segment of the curve is given by: $$S = 2\pi R_{av} L$$ where $R_{av} = \frac{r_1 + r_2}{2}$ is the average radius of the frustum, and $L$ is the slant height. 3. **Relating to the curve:** For a small change in $x$, the radii are $r_1 = f(x)$ and $r_2 = f(x + \Delta x)$, so the average radius approximates to $f(x)$ for very small $\Delta x$. 4. **Slant height $L$ calculation:** Using the Pythagorean theorem: $$L = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2} \Delta x$$ In the limit as $\Delta x \to 0$, this becomes: $$L = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$ 5. **Surface area of a small segment:** $$dS = 2\pi f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$ 6. **Total surface area:** Summing all small segments from $a$ to $b$ and taking the limit leads to the integral: $$S = 2\pi \int_a^b f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$ 7. **Conclusion:** This formula gives the total surface area of the solid formed by revolving the curve $y=f(x)$ about the x-axis between $x=a$ and $x=b$. **Final formula:** $$\boxed{S = 2\pi \int_a^b f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx}$$ This completes the proof and derivation of the surface area formula for revolution around the x-axis.