Question: Find the arc length of the curve below on the given interval. \[ y = \frac{x^3}{3} +…

Find the arc length of the curve below on the given interval.

\[ y = \frac{x^3}{3} + \frac{1}{4x} \text{ on } [1, 4] \]

Solution

To find the arc length of the curve \( y = \frac{x^3}{3} + \frac{1}{4x} \) on the interval \([1, 4]\), we use the arc length formula: \[ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] First, find the derivative \(\frac{dy}{dx}\). Given: \[ y = \frac{x^3}{3} + \frac{1}{4x} \] Differentiate \(y\) with respect to \(x\): \[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{x^3}{3} \right) + \frac{d}{dx} \left( \frac{1}{4x} \right) \] \[ = x^2 - \frac{1}{4x^2} \] Now, plug \(\frac{dy}{dx}\) into the arc length formula: \[ L = \int_1^4 \sqrt{1 + \left(x^2 - \frac{1}{4x^2} \right)^2} \, dx \] Simplify the expression under the square root: \[ = \int_1^4 \sqrt{1 + x^4 - \frac{1}{2} + \frac{1}{16x^4}} \, dx \] \[ = \int_1^4 \sqrt{x^4 + \frac{1}{16x^4} + \frac{1}{2}} \, dx \] Solve the integral, usually requiring numerical methods for non-trivial integrals. The details of this step depend on integration techniques that might involve substitution or a computer algebra system for approximation. Thus, the exact arc length \(L\) will need to be evaluated using these simplified terms and appropriate methods.