2. (10 points) Consider the curve $y=\frac{x^{4}}{16}+\frac{1}{2 x^{2}}$ (a) (5 points) Find the length of the curve for $1 \leq x \leq 2$

Step 1 ：Given the curve \(y=\frac{x^{4}}{16}+\frac{1}{2 x^{2}}\), we are asked to find the length of the curve for \(1 \leq x \leq 2\).

Step 2 ：To find the length of the curve, we use the formula for the length of a curve in the plane, which is given by: \(L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx\), where \(f(x)\) is the function defining the curve, and \(f'(x)\) is its derivative.

Step 3 ：First, we need to find the derivative of the function \(f(x) = \frac{x^{4}}{16} + \frac{1}{2x^{2}}\). The derivative is \(f'(x) = \frac{x^{3}}{4} - \frac{1}{x^{3}}\).

Step 4 ：Substitute this derivative into the formula and integrate from \(x = 1\) to \(x = 2\). The integral expression for the length of the curve is \(\int_{1}^{2} \sqrt{1 + \left(\frac{x^{3}}{4} - \frac{1}{x^{3}}\right)^2} dx\).

Step 5 ：Evaluating this integral gives the numerical value of the length, which is approximately 1.3125.

Step 6 ：Final Answer: The length of the curve \(y=\frac{x^{4}}{16}+\frac{1}{2 x^{2}}\) for \(1 \leq x \leq 2\) is \(\boxed{1.3125}\).