The average value of a continuous function $f(x)$ over the interval $[a, b]$ is given by the formula:
$\text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx$

For the function $f(x) = 10 - x$ over $[0, 10]$, we compute:

1. Set up the formula: $\text{Average value} = \frac{1}{10-0} \int_0^{10} (10 - x) \, dx$

2. Simplify: $\text{Average value} = \frac{1}{10} \int_0^{10} (10 - x) \, dx$

3. Compute the integral: $\int (10 - x) \, dx = 10x - \frac{x^2}{2} + C$ Evaluate from 0 to 10: [ \left[ 10(10) - \frac{10^2}{2} \right] - \left[ 10(0) - \frac{0^2}{2} \right] = (100 - 50) - (0 - 0) = 50 ]

4. Divide by the length of the interval: $\text{Average value} = \frac{1}{10} \cdot 50 = 5$

Final Answer:

The average value of $f(x) = 10 - x$ on $[0, 10]$ is $\boxed{5}$.

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Would you like me to explain any specific step in more detail or explore related concepts?

Here are 5 questions to expand on this topic:

1. How do you find the average value of a function over a different interval, such as $[a, b]$?
2. How does the average value of a function compare to its maximum and minimum values?
3. Can we apply this method to find the average value of trigonometric or exponential functions?
4. How does the geometric interpretation of integration relate to the average value?
5. How can the average value formula be extended to higher dimensions?

Tip: When working with definite integrals, always ensure to evaluate carefully over the limits of integration to avoid small calculation errors.