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

$\frac{1}{b - a} \int_a^b f(x) \, dx$

For this problem, $f(x) = x^3 - 9x$, and the interval is $[-1, 3]$. Let's calculate the average value.

Step 1: Set up the integral

The formula for the average value of the function becomes:

$\text{Average value} = \frac{1}{3 - (-1)} \int_{-1}^{3} (x^3 - 9x) \, dx$

Simplifying the denominator:

$\text{Average value} = \frac{1}{4} \int_{-1}^{3} (x^3 - 9x) \, dx$

Step 2: Compute the integral

We now compute the integral of $x^3 - 9x$.

$\int (x^3 - 9x) \, dx = \int x^3 \, dx - \int 9x \, dx$

Using the power rule for integration, we have:

$\int x^3 \, dx = \frac{x^4}{4}, \quad \int 9x \, dx = \frac{9x^2}{2}$

Thus, the indefinite integral of $x^3 - 9x$ is:

$\frac{x^4}{4} - \frac{9x^2}{2}$

Step 3: Evaluate the definite integral

Now, we evaluate the integral from $x = -1$ to $x = 3$:

$\int_{-1}^{3} \left( x^3 - 9x \right) dx = \left[ \frac{x^4}{4} - \frac{9x^2}{2} \right]_{-1}^{3}$

Substitute $x = 3$:

$\frac{3^4}{4} - \frac{9 \cdot 3^2}{2} = \frac{81}{4} - \frac{9 \cdot 9}{2} = \frac{81}{4} - \frac{81}{2} = \frac{81}{4} - \frac{162}{4} = \frac{-81}{4}$

Substitute $x = -1$:

$\frac{(-1)^4}{4} - \frac{9 \cdot (-1)^2}{2} = \frac{1}{4} - \frac{9}{2} = \frac{1}{4} - \frac{18}{4} = \frac{-17}{4}$

Step 4: Compute the difference

Now, subtract the value at $x = -1$ from the value at $x = 3$:

$\frac{-81}{4} - \frac{-17}{4} = \frac{-81 + 17}{4} = \frac{-64}{4} = -16$

Step 5: Final calculation

Now, multiply the result by $\frac{1}{4}$ to find the average value:

$\text{Average value} = \frac{1}{4} \times (-16) = -4$

Thus, the average value of $x^3 - 9x$ on the interval $[-1, 3]$ is:

$\boxed{-4}$

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Would you like further details or have any questions on the steps? Here are some related questions you might find interesting:

1. How do we find the average value of a different function on a given interval?
2. What happens if the interval is symmetric about zero (e.g., $[-a, a]$)?
3. How do we apply the same method to find the average value of a trigonometric function?
4. What would happen if we had to compute the average value of a function that involves absolute value or piecewise components?
5. How does the concept of the average value of a function relate to its geometric interpretation?

Tip: To simplify integrals like this one, always break them into simpler, individual terms that are easier to integrate.