Find the average value fave of the function f on the given interval.
f(x) = x^2/(x3 + 10)^2
,   [−2, 2]

To find the average value \( f_{\text{ave}} \) of the function \( f(x) = \frac{x^2}{(x^3 + 10)^2} \) on the interval \([−2, 2]\), we use the formula for the average value of a function over a closed interval:

\[
f_{\text{ave}} = \frac{1}{b - a} \int_a^b f(x) \, dx
\]

Here, \( f(x) = \frac{x^2}{(x^3 + 10)^2} \), and the interval is \( [-2, 2] \), so \( a = -2 \) and \( b = 2 \).

The formula becomes:

\[
f_{\text{ave}} = \frac{1}{2 - (-2)} \int_{-2}^{2} \frac{x^2}{(x^3 + 10)^2} \, dx
\]

This simplifies to:

\[
f_{\text{ave}} = \frac{1}{4} \int_{-2}^{2} \frac{x^2}{(x^3 + 10)^2} \, dx
\]

Next, we can solve the integral using appropriate techniques (such as substitution or numerical methods if necessary). Since the function doesn't have an elementary antiderivative, this will likely require numerical integration.

Let me calculate the integral and provide the result.The average value \( f_{\text{ave}} \) of the function \( f(x) = \frac{x^2}{(x^3 + 10)^2} \) over the interval \([-2, 2]\) is approximately \( 0.037 \).

Would you like further details or have any questions? Here are five related questions for deeper understanding:

1. What are the general steps for calculating the average value of a function on an interval?
2. How would the average value change if the interval were different?
3. Can the integral of \( f(x) = \frac{x^2}{(x^3 + 10)^2} \) be computed analytically?
4. How does the behavior of \( f(x) \) influence the result of its average value?
5. What is the significance of average values of functions in practical applications?

**Tip**: The average value of a function gives insight into its overall behavior across an interval, summarizing the function's "height" in terms of an area interpretation.

Find the average value fave of the function f on the given interval. f(x) = x^2/(x^3 + 10)^2, [−2, 2]

Discover how to find the average value of the function f(x) = x^2/(x^3 + 10)^2 on the interval [-2, 2] using calculus techniques. This solution includes the average value formula and application of definite integrals. Suitable for high school or college-level calculus students.

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