Math Problem Statement

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

Solution

To solve for the average value of the function f(x)=x23f(x) = x^2 - 3 on the interval [0,9][0, 9], we can apply the formula for the average value of a function on an interval [a,b][a, b]:

Average value of f(x) on [a,b]=1baabf(x)dx\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)=x23f(x) = x^2 - 3
  • The interval is [0,9][0, 9], so a=0a = 0 and b=9b = 9

Now, calculate:

Average value=19009(x23)dx\text{Average value} = \frac{1}{9 - 0} \int_0^9 (x^2 - 3) \, dx =1909(x23)dx= \frac{1}{9} \int_0^9 (x^2 - 3) \, dx

Step 1: Compute the integral

09(x23)dx=09x2dx093dx\int_0^9 (x^2 - 3) \, dx = \int_0^9 x^2 \, dx - \int_0^9 3 \, dx

First part:

09x2dx=[x33]09=933033=7293=243\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:

093dx=3x09=3(9)3(0)=27\int_0^9 3 \, dx = 3x \bigg|_0^9 = 3(9) - 3(0) = 27

Step 2: Combine results

09(x23)dx=24327=216\int_0^9 (x^2 - 3) \, dx = 243 - 27 = 216

Step 3: Find the average value

Average value=19×216=24\text{Average value} = \frac{1}{9} \times 216 = 24

Thus, the average value of the function f(x)=x23f(x) = x^2 - 3 on the interval [0,9][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)=x23f(x) = x^2 - 3 on a different interval, such as [1,4][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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Definite Integral
Average Value of a Function

Formulas

Average value of f(x) on [a, b] = (1 / (b - a)) * ∫[a to b] f(x) dx

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12 or Early College (Calculus)