1. **State the problem:** Find the arc length of the curve given by
$$y = \frac{x^3}{6} + \frac{1}{2x}$$
on the interval $$\left[\frac{1}{2}, 2\right]$$.
2. **Formula for arc length:** The arc length $$L$$ of a curve $$y=f(x)$$ from $$x=a$$ to $$x=b$$ is given by
$$
L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
$$
3. **Find the derivative $$\frac{dy}{dx}$$:**
$$
y = \frac{x^3}{6} + \frac{1}{2x} = \frac{x^3}{6} + \frac{1}{2}x^{-1}
$$
Differentiate term-by-term:
$$
\frac{dy}{dx} = \frac{3x^2}{6} - \frac{1}{2}x^{-2} = \frac{x^2}{2} - \frac{1}{2x^2}
$$
4. **Square the derivative:**
$$
\left(\frac{dy}{dx}\right)^2 = \left(\frac{x^2}{2} - \frac{1}{2x^2}\right)^2 = \left(\frac{x^2}{2}\right)^2 - 2 \cdot \frac{x^2}{2} \cdot \frac{1}{2x^2} + \left(\frac{1}{2x^2}\right)^2
$$
$$
= \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4}
$$
5. **Add 1 inside the square root:**
$$
1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4} = \frac{1}{2} + \frac{x^4}{4} + \frac{1}{4x^4}
$$
6. **Rewrite the expression inside the root:**
$$
\frac{1}{2} + \frac{x^4}{4} + \frac{1}{4x^4} = \frac{2}{4} + \frac{x^4}{4} + \frac{1}{4x^4} = \frac{x^4 + 2 + \frac{1}{x^4}}{4}
$$
7. **Recognize a perfect square:**
$$
x^4 + 2 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2
$$
8. **So the integrand simplifies to:**
$$
\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{\frac{\left(x^2 + \frac{1}{x^2}\right)^2}{4}} = \frac{1}{2} \left| x^2 + \frac{1}{x^2} \right|
$$
Since $$x$$ is in $$\left[\frac{1}{2}, 2\right]$$ and positive, absolute value is not needed:
$$
= \frac{1}{2} \left(x^2 + \frac{1}{x^2}\right)
$$
9. **Set up the integral for arc length:**
$$
L = \int_{1/2}^2 \frac{1}{2} \left(x^2 + \frac{1}{x^2}\right) dx = \frac{1}{2} \int_{1/2}^2 \left(x^2 + x^{-2}\right) dx
$$
10. **Integrate term-by-term:**
$$
\int x^2 dx = \frac{x^3}{3}, \quad \int x^{-2} dx = -\frac{1}{x}
$$
11. **Evaluate the definite integral:**
$$
L = \frac{1}{2} \left[ \frac{x^3}{3} - \frac{1}{x} \right]_{1/2}^2 = \frac{1}{2} \left( \left(\frac{2^3}{3} - \frac{1}{2}\right) - \left(\frac{(1/2)^3}{3} - \frac{1}{1/2}\right) \right)
$$
Calculate each term:
$$
\frac{2^3}{3} = \frac{8}{3}, \quad -\frac{1}{2} = -\frac{1}{2}
$$
$$
\frac{(1/2)^3}{3} = \frac{1/8}{3} = \frac{1}{24}, \quad -\frac{1}{1/2} = -2
$$
So:
$$
L = \frac{1}{2} \left( \left(\frac{8}{3} - \frac{1}{2}\right) - \left(\frac{1}{24} - 2\right) \right) = \frac{1}{2} \left( \frac{8}{3} - \frac{1}{2} - \frac{1}{24} + 2 \right)
$$
12. **Find common denominator 24:**
$$
\frac{8}{3} = \frac{64}{24}, \quad \frac{1}{2} = \frac{12}{24}, \quad 2 = \frac{48}{24}
$$
Sum inside parentheses:
$$
\frac{64}{24} - \frac{12}{24} - \frac{1}{24} + \frac{48}{24} = \frac{64 - 12 - 1 + 48}{24} = \frac{99}{24}
$$
13. **Multiply by $$\frac{1}{2}$$:**
$$
L = \frac{1}{2} \times \frac{99}{24} = \frac{99}{48} = 2.0625
$$
14. **Round to nearest thousandth:**
$$
L \approx 2.063
$$
**Final answer:** The arc length of the curve on $$\left[\frac{1}{2}, 2\right]$$ is approximately $$2.063$$.
Arc Length C5E0Bf
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