. Average Values. Find the average value of the function f x x x ( ) = − − (1 1 3 )( ) on the interval 1, . n Your answer will depend on n. Show all your work and simplify your answer. Then find the average when n = 7. AVERAGE ON 1, : n AVERAGE ON 1, 7 :

### Problem Statement:
We are tasked with finding the average value of the function \( f(x) = x - (1 + 1/x^3) \) on the interval \([1, n]\). The solution needs to be expressed as a function of \(n\), and then we will compute the average when \(n = 7\).

### 1. Formula for the Average Value of a Function:
The average value \( \text{Avg}(f) \) of a function \( f(x) \) over an interval \([a, b]\) is given by:

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

For this problem, the interval is \([1, n]\), so \( a = 1 \) and \( b = n \). The function to integrate is \( f(x) = x - \left(1 + \frac{1}{x^3}\right) \).

### 2. Set up the integral for the average value:
\[
\text{Avg}(f) = \frac{1}{n - 1} \int_1^n \left( x - \left( 1 + \frac{1}{x^3} \right) \right) \, dx
\]

Simplifying the function inside the integral:
\[
f(x) = x - 1 - \frac{1}{x^3}
\]

So, the integral becomes:
\[
\text{Avg}(f) = \frac{1}{n - 1} \int_1^n \left( x - 1 - \frac{1}{x^3} \right) \, dx
\]

### 3. Split the integral:
We can break this into three simpler integrals:
\[
\text{Avg}(f) = \frac{1}{n - 1} \left( \int_1^n x \, dx - \int_1^n 1 \, dx - \int_1^n \frac{1}{x^3} \, dx \right)
\]

Now, solve each of these integrals.

### 4. Solve each integral:
- The first integral is straightforward:
  \[
  \int_1^n x \, dx = \left[\frac{x^2}{2}\right]_1^n = \frac{n^2}{2} - \frac{1^2}{2} = \frac{n^2}{2} - \frac{1}{2}
  \]
  
- The second integral:
  \[
  \int_1^n 1 \, dx = [x]_1^n = n - 1
  \]

- The third integral:
  \[
  \int_1^n \frac{1}{x^3} \, dx = \left[-\frac{1}{2x^2}\right]_1^n = -\frac{1}{2n^2} + \frac{1}{2}
  \]

### 5. Combine the results:
Substitute the results back into the formula:
\[
\text{Avg}(f) = \frac{1}{n - 1} \left( \left( \frac{n^2}{2} - \frac{1}{2} \right) - (n - 1) - \left( -\frac{1}{2n^2} + \frac{1}{2} \right) \right)
\]

Simplify the expression:
\[
\text{Avg}(f) = \frac{1}{n - 1} \left( \frac{n^2}{2} - \frac{1}{2} - n + 1 + \frac{1}{2n^2} - \frac{1}{2} \right)
\]
\[
\text{Avg}(f) = \frac{1}{n - 1} \left( \frac{n^2}{2} - n + \frac{1}{2n^2} \right)
\]

### 6. Find the average value when \( n = 7 \):
Now, substitute \( n = 7 \) into the formula:

\[
\text{Avg}(f) = \frac{1}{7 - 1} \left( \frac{7^2}{2} - 7 + \frac{1}{2 \times 7^2} \right)
\]
\[
\text{Avg}(f) = \frac{1}{6} \left( \frac{49}{2} - 7 + \frac{1}{98} \right)
\]
\[
\text{Avg}(f) = \frac{1}{6} \left( 24.5 - 7 + 0.0102 \right)
\]
\[
\text{Avg}(f) = \frac{1}{6} \times 17.5102 = 2.9184
\]

### Final Answer:
The average value of the function \( f(x) \) on the interval \([1, n]\) is:
\[
\text{Avg}(f) = \frac{1}{n - 1} \left( \frac{n^2}{2} - n + \frac{1}{2n^2} \right)
\]
When \( n = 7 \), the average value is approximately \( 2.9184 \).

---

Would you like more details or clarifications? 

Here are five related questions for you to explore:
1. How does the average value change as \( n \) increases?
2. What is the geometric interpretation of the average value of a function?
3. How would the process change if \( f(x) \) involved higher powers of \( x \)?
4. Can the average value formula be applied to periodic functions?
5. What happens if the interval of integration is shifted (e.g., \([2, n]\))?

**Tip:** When integrating complex functions, breaking them into simpler parts often makes solving much easier.

Find the average value of the function f(x) = x - (1 + 1/x^3) on the interval [1, n]. Your answer will depend on n. Show all your work and simplify your answer. Then find the average when n = 7.

Explore how to find the average value of the function f(x) = x - (1 + 1/x^3) over the interval [1, n]. This solution involves using the average value formula and solving integrals of x, constant terms, and x^(-3). The final average value when n = 7 is approximately 2.9184.

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