Consider the curve $$ f(x)=\frac{1}{16} x^{4}+\frac{1}{2 x^{2}} $$ 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 $$ f(x) = \frac{1}{16} x^4 + \frac{1}{2 x^2} $$ from $$ x = 1 $$ to $$ x = 2 $$, we'll use the formula for the arc length of a function $$ y = f(x) $$:

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

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

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

Differentiate with respect to $$ x $$:

$$
f'(x) = \frac{4}{16} x^3 - \frac{2}{2 x^3} = \frac{1}{4} x^3 - \frac{1}{x^3}
$$

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

$$
\left(f'(x)\right)^2 = \left(\frac{1}{4} x^3 - \frac{1}{x^3}\right)^2 = \frac{1}{16} x^6 - \frac{2}{4} + \frac{1}{x^6} = \frac{1}{16} x^6 - \frac{1}{2} + \frac{1}{x^6}
$$

Adding 1:

$$
1 + \left(f'(x)\right)^2 = \frac{1}{2} + \frac{1}{16} x^6 + \frac{1}{x^6}
$$

Notice that this expression can be written as a perfect square:

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

**Step 3: Compute the integral for arc length**

$$
\text{Arc Length} = \int_{1}^{2} \left(\frac{1}{4} x^3 + \frac{1}{x^3}\right) dx
$$

Integrate term by term:

$$
\int \frac{1}{4} x^3 \, dx = \frac{1}{4} \cdot \frac{x^4}{4} = \frac{x^4}{16}
$$

$$
\int \frac{1}{x^3} \, dx = \int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}
$$

Combine the results:

$$
\text{Arc Length} = \left[\frac{x^4}{16} - \frac{1}{2x^2}\right]_{1}^{2}
$$

Evaluate from $$ x = 1 $$ to $$ x = 2 $$:

At $$ x = 2 $$:

$$
\frac{2^4}{16} - \frac{1}{2 \cdot 2^2} = \frac{16}{16} - \frac{1}{8} = 1 - \frac{1}{8} = \frac{7}{8}
$$

At $$ x = 1 $$:

$$
\frac{1^4}{16} - \frac{1}{2 \cdot 1^2} = \frac{1}{16} - \frac{1}{2} = \frac{1}{16} - \frac{8}{16} = -\frac{7}{16}
$$

Subtract the two results:

$$
\frac{7}{8} - \left(-\frac{7}{16}\right) = \frac{7}{8} + \frac{7}{16} = \frac{14}{16} + \frac{7}{16} = \frac{21}{16}
$$

**Final Answer:**

The arc length of the curve is $$ \frac{21}{16} $$ units.
The arc length of the curve is $$ \frac{21}{16} $$ units.

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

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