consider the curve $$ f(x)=\frac{1}{3} x^{3}+\frac{1}{4 x} $$ for $$ 1 \leq x \leq 2 $$. Find the arc length of the curve.

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To find the arc length of the curve $$ f(x) = \frac{1}{3}x^3 + \frac{1}{4x} $$ for $$ 1 \leq x \leq 2 $$, we use the arc length formula:

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

**Step 1: Compute the derivative $$ f'(x) $$**

$$
f(x) = \frac{1}{3}x^3 + \frac{1}{4x}
$$
$$
f'(x) = x^2 - \frac{1}{4x^2}
$$

**Step 2: Compute $$ 1 + \left( f'(x) \right)^2 $$**

$$
\left( f'(x) \right)^2 = \left( x^2 - \frac{1}{4x^2} \right)^2 = x^4 - \frac{1}{2} + \frac{1}{16x^4}
$$
$$
1 + \left( f'(x) \right)^2 = x^4 + \frac{1}{2} + \frac{1}{16x^4}
$$

Notice that this expression can be rewritten as a perfect square:
$$
x^4 + \frac{1}{2} + \frac{1}{16x^4} = \left( x^2 + \frac{1}{4x^2} \right)^2
$$

**Step 3: Simplify the integral**

$$
\sqrt{1 + \left( f'(x) \right)^2} = x^2 + \frac{1}{4x^2}
$$
$$
\text{Arc Length} = \int_{1}^{2} \left( x^2 + \frac{1}{4x^2} \right) dx
$$

**Step 4: Evaluate the integral**

$$
\int x^2 \, dx = \frac{1}{3}x^3
$$
$$
\int \frac{1}{4x^2} \, dx = -\frac{1}{4x}
$$
$$
\text{Arc Length} = \left[ \frac{1}{3}x^3 - \frac{1}{4x} \right]_{1}^{2}
$$

Substituting the limits:

At $$ x = 2 $$:
$$
\frac{1}{3}(2)^3 - \frac{1}{4(2)} = \frac{8}{3} - \frac{1}{8} = \frac{64}{24} - \frac{3}{24} = \frac{61}{24}
$$

At $$ x = 1 $$:
$$
\frac{1}{3}(1)^3 - \frac{1}{4(1)} = \frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12}
$$

Therefore, the arc length is:

$$
\frac{61}{24} - \frac{1}{12} = \frac{59}{24}
$$

**Final Answer:**

The arc length is 59⁄24 units.
The arc length of the curve $$ f(x) = \frac{1}{3}x^3 + \frac{1}{4x} $$ from $$ x = 1 $$ to $$ x = 2 $$ is $$ \frac{59}{24} $$ units.

consider the curve for . Find the arc length of the curve.

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