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To solve for the average value of the function \( f(x) = x^2 - 3 \) on the interval \([0, 9]\), we can apply the formula for the average value of a function on an interval \([a, b]\):

\[
\text{Average value of } f(x) \text{ on } [a, b] = \frac{1}{b - a} \int_a^b f(x) \, dx
\]

For this specific problem:
- The function is \( f(x) = x^2 - 3 \)
- The interval is \( [0, 9] \), so \( a = 0 \) and \( b = 9 \)

Now, calculate:

\[
\text{Average value} = \frac{1}{9 - 0} \int_0^9 (x^2 - 3) \, dx
\]
\[
= \frac{1}{9} \int_0^9 (x^2 - 3) \, dx
\]

### Step 1: Compute the integral
\[
\int_0^9 (x^2 - 3) \, dx = \int_0^9 x^2 \, dx - \int_0^9 3 \, dx
\]

#### First part:
\[
\int_0^9 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^9 = \frac{9^3}{3} - \frac{0^3}{3} = \frac{729}{3} = 243
\]

#### Second part:
\[
\int_0^9 3 \, dx = 3x \bigg|_0^9 = 3(9) - 3(0) = 27
\]

### Step 2: Combine results
\[
\int_0^9 (x^2 - 3) \, dx = 243 - 27 = 216
\]

### Step 3: Find the average value
\[
\text{Average value} = \frac{1}{9} \times 216 = 24
\]

Thus, the average value of the function \( f(x) = x^2 - 3 \) on the interval \([0, 9]\) is **24**.

Would you like further details or have any questions about this process?

Here are five related questions to explore further:
1. What is the average value of \( f(x) = x^2 - 3 \) on a different interval, such as \([1, 4]\)?
2. How does the average value formula change for different types of functions, like trigonometric or exponential functions?
3. Can the average value of a function ever be negative, and under what conditions?
4. How would you find the average value of a function if given in a piecewise form?
5. What is the physical interpretation of the average value of a function in real-world contexts?

**Tip:** Always carefully compute definite integrals before dividing by the length of the interval to avoid errors in finding the average value of a function.

Find the average value of the function f(x) = x^2 - 3 on [0, 9].

Learn how to find the average value of the function f(x) = x^2 - 3 over the interval [0, 9]. This detailed solution includes key calculus concepts such as definite integrals and the fundamental theorem of calculus. Suitable for advanced high school or early college students.

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