Problem 74 - Thomas Calculus

Question

It would be nice if average values of integrable functions obeyed the following rules on an interval \([a, b] .\) a. \(\operatorname{av}(f+g)=\operatorname{av}(f)+\operatorname{av}(g)\) b. \(\operatorname{av}(k f)=k \operatorname{av}(f) \quad\) (any number \(k )\) c. \(\mathrm{av}(f) \leq \operatorname{av}(g) \quad\) if \(\quad f(x) \leq g(x) \quad\) on \(\quad[a, b]\) Do these rules everhold? Give reasons for vour answers.

Step-by-Step Solution

Answer

Yes, all rules hold as they align with properties of integrals.

Step 1: Understanding Average Values

The average value av(f) of a function f over an interval \([a, b]\) is defined as the integral of f over the interval divided by the length of the interval. Mathematically, this is expressed as: \[ \operatorname{av}(f) = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]

Step 2: Rule A: Linear Combination of Functions

The first rule is av(f + g) = av(f) + av(g). This can be checked by integrating the sum: \[ \operatorname{av}(f + g) = \frac{1}{b-a} \int_{a}^{b} (f(x) + g(x)) \, dx = \frac{1}{b-a} \left( \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \right) \]Using the property of integrals, \(\frac{1}{b-a} \int_{a}^{b} f(x) \, dx + \frac{1}{b-a} \int_{a}^{b} g(x) \, dx = \operatorname{av}(f) + \operatorname{av}(g)\).Thus, the rule holds.

Step 3: Rule B: Scalar Multiplication of Functions

The second rule is av(k f) = k av(f). Using the definition, \[ \operatorname{av}(k f) = \frac{1}{b-a} \int_{a}^{b} k f(x) \, dx = \frac{k}{b-a} \int_{a}^{b} f(x) \, dx = k \cdot \operatorname{av}(f) \]This confirms that the rule holds since constant factors can be factored out of integrals.

Step 4: Rule C: Comparison of Functions

The third rule states av(f) ≤ av(g) if f(x) ≤ g(x) on [a, b]. Since the integral of f over [a, b] is less than or equal to that of g whenever f(x) ≤ g(x),\[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \leq \frac{1}{b-a} \int_{a}^{b} g(x) \, dx \]Thus, the rule holds because integration preserves inequalities.

Key Concepts

Linear Combination of Functions

When we talk about the linear combination of functions, we are essentially combining two or more functions by adding them together. Consider two functions, \( f(x) \) and \( g(x) \). A linear combination of these functions would be \( f(x) + g(x) \). The average value over an interval \([a, b]\) for this combination can be found by integrating the sum of these two functions over that interval, and then dividing by the length of the interval. This gives us:

• \( \operatorname{av}(f+g) = \frac{1}{b-a} \int_{a}^{b} (f(x) + g(x)) \, dx \)

Utilizing the properties of integrals, we see that we can separate the integral of a sum into the sum of integrals:

• \( \operatorname{av}(f+g) = \operatorname{av}(f) + \operatorname{av}(g) \)

This shows that the average value of the sum is the sum of the average values. This is an essential concept because it demonstrates how the interaction of functions through addition affects their overall behavior over the selected interval.
It underlines the foundational property that the average value respects the additive nature of functions within calculus.

Scalar Multiplication of Functions

Scalar multiplication involves multiplying a function by a constant value, or scalar. This operation is straightforward: if you have a function \( f(x) \) and a constant \( k \), then scalar multiplication refers to \( k \times f(x) \). For the average value, this is expressed as:

• \( \operatorname{av}(k f) = \frac{1}{b-a} \int_{a}^{b} k f(x) \, dx \)

We take advantage of the rules of integrals, which allow us to factor the constant \( k \) out of the integral:

• \( \operatorname{av}(k f) = k \cdot \operatorname{av}(f) \)

This demonstrates a property that scalar multiplication of a function inside an integral maintains the scalar outside the average value computed.
This principle is vital since it offers a way to see how scaling a function impacts its average value which is linearly related to the original function. It's much like seeing how the height of a graph is stretched or compressed along the y-axis by scaling.

Comparison of Functions

In terms of comparison, if we have two functions, \( f(x) \) and \( g(x) \), where for all values within a given interval \([a,b]\), \( f(x) \leq g(x) \), then it follows logically and mathematically that the average value of \( f \) over that interval will also be less than or equal to that of \( g \).

• If \( f(x) \leq g(x) \) then \( \operatorname{av}(f) \leq \operatorname{av}(g) \)

This is supported by the idea that integration is a linear operation that preserves inequality. When you integrate functions that are bound by an inequality over a closed interval, the inequality is sustained within their averages.Understanding this concept of comparison is critical as it offers a method to predict relationships between different functions by comparing their behavior over a specified domain.
This forms the basis of reasoning in numerous applied mathematics and analysis fields, where function comparison leads to deduction of relationships and constraints in real-world scenarios.