Set-up the integral to calculate the length of the curve  y=x^3/2 + 1/6x for 3 ≤ x ≤ 5.  Evaluate the integral.

To calculate the length of the curve given by the function \( y = \frac{x^{3/2}}{2} + \frac{1}{6x} \) over the interval \( 3 \leq x \leq 5 \), we use the formula for the length of a curve in Cartesian coordinates:

\[
L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
\]

### Step 1: Find the derivative \( \frac{dy}{dx} \)
The function is \( y = \frac{x^{3/2}}{2} + \frac{1}{6x} \). Differentiating with respect to \( x \):

\[
\frac{dy}{dx} = \frac{3}{4}x^{1/2} - \frac{1}{6x^2}
\]

### Step 2: Set up the integral
The formula for the length of the curve becomes:

\[
L = \int_3^5 \sqrt{1 + \left( \frac{3}{4}x^{1/2} - \frac{1}{6x^2} \right)^2} \, dx
\]

### Step 3: Evaluate the integral

This integral can be evaluated numerically. Let me calculate it.

Set-up the integral to calculate the length of the curve y = x^(3/2)/2 + 1/(6x) for 3 ≤ x ≤ 5. Evaluate the integral.

Learn how to set up and evaluate the integral to calculate the arc length of the curve y = x^(3/2)/2 + 1/(6x) over the interval [3, 5]. This detailed guide covers integral calculus concepts and the formula for curve length, suitable for advanced high school calculus students.

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