1. **State the problem:** Find the surface area generated by revolving the curve $y = x^2 + 4x$ for $0 \leq x \leq 1$ about the x-axis. 2. **Formula for surface area of revolution about x-axis:** $$ S = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$ 3. **Calculate the derivative:** $$ \frac{dy}{dx} = 2x + 4 $$ 4. **Substitute into the formula:** $$ S = 2\pi \int_0^1 (x^2 + 4x) \sqrt{1 + (2x + 4)^2} \, dx $$ 5. **Simplify inside the square root:** $$ 1 + (2x + 4)^2 = 1 + (4x^2 + 16x + 16) = 4x^2 + 16x + 17 $$ 6. **Final integral expression:** $$ S = 2\pi \int_0^1 (x^2 + 4x) \sqrt{4x^2 + 16x + 17} \, dx $$ 7. **Explanation:** This integral represents the surface area of the solid formed by revolving the curve about the x-axis. It may require substitution or numerical methods to evaluate. **Final answer:** $$ S = 2\pi \int_0^1 (x^2 + 4x) \sqrt{4x^2 + 16x + 17} \, dx $$