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The average value of a function over a given interval provides useful insight into the behavior of the function across that interval. Imagine you have a wavy function that dips and peaks. The average gives you a single value that represents the function's "typical" value over the range. To find this for a function \(f(t)\) on an interval \([a, b]\), you can use a finite sum. Here's how:
This involves partitioning the interval into equal-length subintervals and then finding the average of function values at selected points within these subintervals. It gives us an approximation of what the function "feels like" over the stretch of interest.
The average value, in our case from the exercise, is computed using the formula: \[\text{Average value} = \frac{1}{n} \sum_{i=1}^{n} f(t_i)\]Where \( f(t_i) \) are evaluations of the function at specific points (midpoints in this case), and \( n \) is the number of partitions. This formula is akin to "mean" from statistics, only applied functionally.

Subinterval midpoints are crucial in numerical methods for estimating integrals or average values. When you partition an interval (e.g., [0, 4]) into smaller segments and calculate their midpoints, these points effectively summarize each segment. Using the midpoint of each subinterval for function evaluation offers a balance and minimizes error in estimating total area or average.
In our exercise, the midpoints \( t_1, t_2, t_3, \) and \( t_4 \) (0.5, 1.5, 2.5, and 3.5) capture central tendencies of their respective subintervals \([0, 1], [1, 2], [2, 3], [3, 4]\). Evaluating the function at these midpoints provides samples representing the function's behavior in that segment.
By using midpoints, we tap into a common numerical integration method called the Midpoint Rule, which often yields better approximations in estimating the function's average value.

Partitioning an interval divides it into segments of equal length. It's a foundational step in numerical approximation techniques. The entire interval length is divided by the number of subintervals desired, giving the width of each subinterval.
For our case, the full interval from 0 to 4 is partitioned into four equal parts, each having a length \( \Delta t = 1 \). This means each subinterval is precisely 1 unit long, i.e., \([0, 1], [1, 2], [2, 3], [3, 4]\).
This method ensures systematic coverage of the entire interval, allowing for fair sampling when evaluating the function across the interval. Equal partitioning simplifies computation and facilitates easier summation when calculating averages.

Function evaluation at specific points involves substituting those points into the given mathematical expression of the function. This provides an output - the function's value at that point.
For the given function \( f(t) = 1 - (\cos(\frac{\pi t}{4}))^4 \), evaluating at the midpoints defined earlier (0.5, 1.5, 2.5, and 3.5) provides the necessary values to compute the estimated average. Performing this step requires calculation skills, often aided by a calculator for precision.
Here's how it plays out: you plug each value of \( t \) into the function. For example:

• At \( t = 0.5 \), compute \( f(0.5) = 1 - (\cos(\frac{\pi \times 0.5}{4}))^4 \).
• Similarly, calculate \( f(t) \) for \( t = 1.5, 2.5, \) and \( 3.5 \).

These computed outputs from each specified midpoint ultimately feed into the formula for finding the average value of the function over the interval.