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.

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.