Answer
See the explanation
Work Step by Step
The average value of an integrable function \( f(x) \) over a closed interval \([a, b]\) can be calculated using the definite integral formula:
\[ \text{Average value} = \frac{1}{b - a} \int_a^b f(x) \, dx \]
This formula represents the total area under the curve of \( f(x) \) over the interval \([a, b]\), divided by the length of the interval.
As for whether the function must assume its average value, the answer is not necessarily. The function may or may not assume its average value over the interval. For example, consider a function that oscillates above and below its average value within the interval. In such cases, the function does not necessarily equal its average value at any point within the interval.
However, according to the Mean Value Theorem for Integrals, there exists at least one point \( c \) in the interval \([a, b]\) such that \( f(c) \) is equal to the average value of the function over that interval. This theorem guarantees the existence of at least one point where the function equals its average value, but it doesn't guarantee that the function must assume its average value at every point within the interval.
