Question

In the following exercise, find the average value f_ave of f between a and b, and determine a point c where f(c) equals the average value.

Answer 1

The average value
`\( f_{\text{ave}} \) of \( f \) between \( a \) and \( b \) is \( f_{\text{ave}} = (1)/(b-a) \int_(a)^(b) f(x) \, dx \). A point \( c \) where \( f(c) \) equals the average value is \( c \) in the interval \([a, b]\) such that \( f(c) = f_{\text{ave}} \).`

Step-by-step explanation:

To find the average value
`\( f_{\text{ave}} \) of \( f \) between \( a \) and \( b \), we use the formula \( f_{\text{ave}} = (1)/(b-a) \int_(a)^(b) f(x) \, dx \).`This formula involves finding the definite integral of \( f \) over the interval \([a, b]\) and then dividing by the length of the interval.

The integral represents the accumulated area under the curve of \( f \) over the given interval. Dividing by the length of the interval (\( b - a \)) normalizes this accumulated area, providing the average value of \( f \) over the interval.

Now, to determine a point \( c \) where \( f(c) \) equals the average value, we look for a specific \( c \) in the interval \([a, b]\). This point \( c \) satisfies
`\( f(c) = f_{\text{ave}} \)`, making the function value at \( c \) equal to the average value of the function over the interval. This can be interpreted as a balance point where the function is equally distributed above and below its average value in the given interval.