nsider the curve \( f(x)=\frac{1}{6} x^{3}+\frac{1}{2 x} \) for \( 1 \leq x \leq 3 \). Find the exact arc length of the curve.

To find the arc length of the curve \( f(x) = \frac{1}{6} x^3 + \frac{1}{2x} \) from \( x = 1 \) to \( x = 3 \), we use the arc length formula: \[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] First, we need to calculate \( \frac{dy}{dx} \): 1. Differentiate \( f(x) \): \[ f'(x) = \frac{d}{dx} \left( \frac{1}{6} x^3 + \frac{1}{2x} \right) = \frac{1}{2} x^2 - \frac{1}{2x^2} \] 2. Now, simplify \( f'(x) \): \[ f'(x) = \frac{x^4 - 1}{2x^2} \] 3. Next, compute \( \left( f'(x) \right)^2 \): \[ \left( f'(x) \right)^2 = \left( \frac{x^4 - 1}{2x^2} \right)^2 = \frac{(x^4 - 1)^2}{4x^4} \] 4. Substitute \( \left( f'(x) \right)^2 \) into the arc length formula: \[ L = \int_1^3 \sqrt{1 + \frac{(x^4 - 1)^2}{4x^4}} \, dx \] 5. Simplify the term under the square root: \[ 1 + \frac{(x^4 - 1)^2}{4x^4} = \frac{4x^4 + (x^4 - 1)^2}{4x^4} = \frac{4x^4 + x^8 - 2x^4 + 1}{4x^4} = \frac{x^8 + 2x^4 + 1}{4x^4} = \frac{(x^4 + 1)^2}{4x^4} \] Therefore, \[ \sqrt{1 + \left( f'(x) \right)^2} = \sqrt{\frac{(x^4 + 1)^2}{4x^4}} = \frac{x^4 + 1}{2x^2} \] 6. Now we can rewrite the arc length: \[ L = \int_1^3 \frac{x^4 + 1}{2x^2} \, dx = \frac{1}{2} \int_1^3 (x^2 + \frac{1}{x^2}) \, dx \] 7. Now, split the integral: \[ L = \frac{1}{2} \left( \int_1^3 x^2 \, dx + \int_1^3 \frac{1}{x^2} \, dx \right) \] 8. Calculate \( \int_1^3 x^2 \, dx \): \[ \int_1^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^3 = \frac{27}{3} - \frac{1}{3} = \frac{26}{3} \] 9. Calculate \( \int_1^3 \frac{1}{x^2} \, dx \): \[ \int_1^3 \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_1^3 = -\frac{1}{3} + 1 = \frac{2}{3} \] 10. Combine the results: \[ L = \frac{1}{2} \left( \frac{26}{3} + \frac{2}{3} \right) = \frac{1}{2} \cdot \frac{28}{3} = \frac{14}{3} \] Thus, the exact arc length of the curve from \( x = 1 \) to \( x = 3 \) is \[ \boxed{\frac{14}{3}}. \]