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The problem asks for the area of the surface generated when the given curve is revolved about the x-axis. The curve provided is:

\[
y = \frac{x^3}{6} + \frac{1}{2x}, \quad \text{for } \frac{1}{2} \leq x \leq 2
\]

### Formula for Surface Area of Revolution
To find the surface area of a curve \(y = f(x)\) revolved around the x-axis, we use the surface area formula:

\[
A = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
\]

### Step-by-Step Process

1. **Differentiate \(y\) with respect to \(x\):**
   We need to find \(\frac{dy}{dx}\).

   \[
   \frac{dy}{dx} = \frac{d}{dx} \left(\frac{x^3}{6} + \frac{1}{2x}\right)
   \]

   This derivative can be computed as:

   \[
   \frac{dy}{dx} = \frac{1}{6} \cdot 3x^2 - \frac{1}{2} \cdot \frac{1}{x^2} = \frac{x^2}{2} - \frac{1}{2x^2}
   \]

2. **Square the derivative:**
   Now, square the result of \(\frac{dy}{dx}\).

   \[
   \left(\frac{dy}{dx}\right)^2 = \left(\frac{x^2}{2} - \frac{1}{2x^2}\right)^2
   \]

3. **Set up the surface area integral:**

   The formula becomes:

   \[
   A = 2\pi \int_{\frac{1}{2}}^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right) \sqrt{1 + \left(\frac{x^2}{2} - \frac{1}{2x^2}\right)^2} \, dx
   \]

4. **Evaluate the integral:**
   This involves solving the integral, which might require symbolic integration tools for the square root term. Let me compute the exact surface area value for you.

Find the area of the surface generated when the given curve y = (x^3)/6 + 1/(2x), for 1/2 ≤ x ≤ 2 is revolved about the x-axis.

Learn how to calculate the surface area generated by revolving the curve y = (x^3)/6 + 1/(2x) about the x-axis. This detailed calculus problem covers differentiation, integration, and the surface area of revolution formula. Suitable for college-level calculus students.

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