The formula for the average value of a function $f(x)$ on the interval $[a, b]$ is:

$\text{Avg}(f) = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx$

For the function $f(x) = \frac{x^2 - 1}{x}$, on the interval $[1, 4]$, we can first simplify the function:

$f(x) = \frac{x^2 - 1}{x} = x - \frac{1}{x}$

Now, applying the formula for the average value, the integral setup becomes:

$\text{Avg}(f) = \frac{1}{4 - 1} \int_{1}^{4} \left( x - \frac{1}{x} \right) dx$

Simplifying further:

$\text{Avg}(f) = \frac{1}{3} \int_{1}^{4} \left( x - \frac{1}{x} \right) dx$

This is the integral that represents the average value of the function on the interval $[1, 4]$.

Would you like me to solve the integral next?

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5 related questions:

1. How do you evaluate the integral $\int (x - \frac{1}{x}) dx$?
2. What are the geometric interpretations of the average value of a function?
3. How do changes in the interval $[a, b]$ affect the average value of a function?
4. Can we apply this average value formula to any continuous function?
5. How does the average value differ for polynomial versus rational functions?

Tip: Always simplify the given function if possible before setting up the integral for the average value.