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.

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UpStudy AI · Accepted Answer

To find the exact arc length of the curve $$ f(x) = \frac{1}{6}x^3 + \frac{1}{2x} $$ for $$ 1 \leq x \leq 3 $$, we'll use the arc length formula for a function $$ f(x) $$:

$$
\text{Arc Length} = \int_{a}^{b} \sqrt{1 + \left(f'(x)\right)^2} \, dx
$$

### Step 1: Compute the First Derivative $$ f'(x) $$

Given:
$$
f(x) = \frac{1}{6}x^3 + \frac{1}{2x}
$$

Differentiate with respect to $$ x $$:
$$
f'(x) = \frac{1}{6} \cdot 3x^2 - \frac{1}{2} \cdot x^{-2} = \frac{1}{2}x^2 - \frac{1}{2x^2}
$$

### Step 2: Compute $$ \left(f'(x)\right)^2 $$

$$
\left(f'(x)\right)^2 = \left(\frac{1}{2}x^2 - \frac{1}{2x^2}\right)^2 = \frac{(x^4 - 2 + \frac{1}{x^4})}{4}
$$

### Step 3: Set Up the Arc Length Integral

$$
\sqrt{1 + \left(f'(x)\right)^2} = \sqrt{1 + \frac{x^4 - 2 + \frac{1}{x^4}}{4}} = \sqrt{\frac{x^4 + 2 + \frac{1}{x^4}}{4}} = \frac{x^2 + \frac{1}{x^2}}{2}
$$

Thus, the arc length integral becomes:
$$
\text{Arc Length} = \int_{1}^{3} \frac{x^2 + \frac{1}{x^2}}{2} \, dx = \frac{1}{2} \left( \int_{1}^{3} x^2 \, dx + \int_{1}^{3} \frac{1}{x^2} \, dx \right)
$$

### Step 4: Evaluate the Integrals

Compute $$ \int x^2 \, dx $$ from 1 to 3:
$$
\int_{1}^{3} x^2 \, dx = \left[\frac{x^3}{3}\right]_{1}^{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}
$$

Compute $$ \int \frac{1}{x^2} \, dx $$ from 1 to 3:
$$
\int_{1}^{3} \frac{1}{x^2} \, dx = \left[-\frac{1}{x}\right]_{1}^{3} = -\frac{1}{3} + 1 = \frac{2}{3}
$$

### Step 5: Combine the Results

$$
\text{Arc Length} = \frac{1}{2} \left( \frac{26}{3} + \frac{2}{3} \right) = \frac{1}{2} \cdot \frac{28}{3} = \frac{14}{3}
$$

**Final Answer:**  
The exact arc length is 14⁄3 units.
The exact arc length of the curve is $$ \frac{14}{3} $$ units.

nsider the curve for . Find the exact arc length of the curve.

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