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To find the length of the curve described by \( x = \frac{y^3}{6} + \frac{1}{2y} \) from \( y = 2 \) to \( y = 3 \), we use the formula for the length of a parametric curve:

\[
L = \int_{y_1}^{y_2} \sqrt{\left(\frac{dx}{dy}\right)^2 + 1} \, dy.
\]

### Step 1: Differentiate \( x \) with respect to \( y \)
Given:
\[
x = \frac{y^3}{6} + \frac{1}{2y}.
\]

Differentiating term by term:
\[
\frac{dx}{dy} = \frac{3y^2}{6} - \frac{1}{2y^2} = \frac{y^2}{2} - \frac{1}{2y^2}.
\]

### Step 2: Substitute into the length formula
The length formula becomes:
\[
L = \int_{2}^{3} \sqrt{\left(\frac{y^2}{2} - \frac{1}{2y^2}\right)^2 + 1} \, dy.
\]

Expand \(\left(\frac{y^2}{2} - \frac{1}{2y

Find the length of the curve described by x = (y^3/6) + 1/(2y) from y = 2 to y = 3.

Find the length of the curve defined by x = y^3/6 + 1/(2y) for y ranging from 2 to 3. This problem involves using calculus and the parametric curve length formula, suitable for advanced high school or early college students.

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