To find the arc length of the graph of the function $f(x) = \frac{x^3}{6} + \frac{x}{2}$ between $x = 3$ and $x = 4$, we use the formula for the arc length of a function:

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

where $f'(x)$ is the derivative of the function.

Step 1: Find the derivative of the function $f(x)$

Given $f(x) = \frac{x^3}{6} + \frac{x}{2}$, the derivative $f'(x)$ is:

$f'(x) = \frac{d}{dx} \left( \frac{x^3}{6} + \frac{x}{2} \right)$ $f'(x) = \frac{1}{6} \cdot 3x^2 + \frac{1}{2}$ $f'(x) = \frac{x^2}{2} + \frac{1}{2}$

Step 2: Set up the arc length formula

Now we plug $f'(x) = \frac{x^2}{2} + \frac{1}{2}$ into the arc length formula:

$L = \int_{3}^{4} \sqrt{1 + \left( \frac{x^2}{2} + \frac{1}{2} \right)^2} \, dx$

Step 3: Simplify the expression

We now simplify the expression inside the square root:

$\left( \frac{x^2}{2} + \frac{1}{2} \right)^2 = \left( \frac{x^2 + 1}{2} \right)^2 = \frac{(x^2 + 1)^2}{4}$ So,

$L = \int_{3}^{4} \sqrt{1 + \frac{(x^2 + 1)^2}{4}} \, dx$

Now let's evaluate this integral.

The arc length expression simplifies to:

$L = \frac{1}{2} \int_3^4 \sqrt{x^4 + 2x^2 + 5} \, dx$

This integral does not have a simple closed-form solution and is best evaluated numerically. Let's compute the approximate value of this integral.The approximate arc length of the graph of the function $f(x) = \frac{x^3}{6} + \frac{x}{2}$ between $x = 3$ and $x = 4$ is about 6.743 units.

Would you like further details on the calculation or have any questions?

Here are 5 related questions for further exploration:

1. How does changing the range of integration affect the arc length?
2. How do you compute arc length for parametric curves?
3. How does the arc length formula change if the function is in terms of $y$ instead of $x$?
4. What happens to the arc length if the function has a vertical asymptote?
5. Can we estimate arc lengths using numerical methods like trapezoidal or Simpson's rule?

Tip: If the integral for arc length cannot be solved symbolically, numerical methods provide an efficient way to get an approximation.