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 $2.9184$.

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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.