Problem 53 - Calculus Early Transcendentals: Pearson New International Edition

Question

Assuming that \(u\) and \(v\) can be integrated over the interval \([a, b]\) and that the average values over the interval are denoted by \(\bar{u}\) and \(\bar{v}\), prove or disprove that (a) \(\bar{u}+\bar{v}=\overline{u+v}\), (b) \(k \bar{u}=\overline{k u}\), where \(k\) is any constant; (c) if \(u \leq v\) then \(\bar{u} \leq \bar{v}\).

Step-by-Step Solution

Answer

(a), (b), and (c) are all true.

Step 1: Define Average Value of a Function

The average value of a function \( f \) over an interval \([a, b]\) is defined as: \[ \bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx. \] We will use this definition to prove each part of the problem.

Step 2a: Prove (a) \( \bar{u} + \bar{v} = \overline{u + v} \)

To prove \( \bar{u} + \bar{v} = \overline{u+v} \), we first express each side:- \( \bar{u} = \frac{1}{b-a} \int_a^b u(x) \, dx \)- \( \bar{v} = \frac{1}{b-a} \int_a^b v(x) \, dx \)- \( \overline{u+v} = \frac{1}{b-a} \int_a^b (u(x) + v(x)) \, dx \)We need to show that:\[ \bar{u} + \bar{v} = \frac{1}{b-a} \left( \int_a^b u(x) \, dx + \int_a^b v(x) \, dx \right) = \frac{1}{b-a} \int_a^b (u(x) + v(x)) \, dx = \overline{u+v}. \]Since the integrals are additive, the equality holds. Thus, (a) is **true**.

Step 2b: Prove (b) \( k \bar{u} = \overline{k u} \)

To prove \( k \bar{u} = \overline{k u} \), we first express each term:- \( k \bar{u} = k \cdot \frac{1}{b-a} \int_a^b u(x) \, dx \)- \( \overline{k u} = \frac{1}{b-a} \int_a^b k u(x) \, dx \)Then, we need to show that:\[ k \cdot \frac{1}{b-a} \int_a^b u(x) \, dx = \frac{1}{b-a} \int_a^b ku(x) \, dx \]Since a constant can be factored out of an integral:\[ \frac{1}{b-a} \cdot k \int_a^b u(x) \, dx = \frac{1}{b-a} \int_a^b ku(x) \, dx \]This equality holds, so (b) is **true**.

Step 2c: Prove (c) If \( u \leq v \) then \( \bar{u} \leq \bar{v} \)

To prove \(\bar{u} \leq \bar{v}\) given \( u \leq v \), we express:- \( \bar{u} = \frac{1}{b-a} \int_a^b u(x) \, dx \)- \( \bar{v} = \frac{1}{b-a} \int_a^b v(x) \, dx \)If \( u(x) \leq v(x) \) for all \( x \) in \([a, b]\), then:\[ \int_a^b u(x) \, dx \leq \int_a^b v(x) \, dx. \]Thus:\[ \frac{1}{b-a} \int_a^b u(x) \, dx \leq \frac{1}{b-a} \int_a^b v(x) \, dx \]Hence, \( \bar{u} \leq \bar{v} \), making (c) **true**.

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that helps calculate areas, volumes, and other quantities that arise by summing infinitesimal contributions. In particular, it deals with integrals, which can be thought of as generalized sums. Integrals play a crucial role in finding the average value of a function over an interval.

When you have a function defined across an interval \([a, b]\), the average value of the function is essentially the sum of its values over this interval, divided by its length. This is calculated using the integral of the function over \([a, b]\). The formula to find the average value \(\bar{f}\) is:

• \[ \bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx. \]

Understanding how to apply integration in this way can offer insights into many real-world phenomena, such as calculating average speed, average environmental pollutant levels, or even average economic indicators over a period of time.

Properties of Integrals

Integrals have several important properties that are fundamental for proving complex mathematical relationships, like those involved in average values. Two of these key properties involve additivity and the constant rule.

• Additivity: This property states that the integral of a sum of functions is equal to the sum of their integrals. Mathematically, if you have two functions \(u(x)\) and \(v(x)\), it holds that: \[ \int_a^b (u(x) + v(x)) \, dx = \int_a^b u(x) \, dx + \int_a^b v(x) \, dx. \]
• Constant Rule: This states that you can factor a constant out of an integral, which is essential when scaling a function by a constant. If \(k\) is a constant, then: \[ \int_a^b k \, u(x) \, dx = k \int_a^b u(x) \, dx. \]

These properties are utilized to prove fundamental relationships, such as the average of the sum of functions equals the sum of their averages and the average of a constant multiple of a function.

Function Inequalities

Function inequalities are important tools in integral calculus for comparing functions over an interval. Particularly, they help establish order relations among functions by comparing their integrals.

For instance, if you have two functions where \(u(x) \leq v(x)\) for all \( x \) in the interval \([a, b]\), this implies an order relationship for their integrals:

• \[ \int_a^b u(x) \, dx \leq \int_a^b v(x) \, dx. \]

Translating this integral inequality into average values, you have:

• \[ \frac{1}{b-a} \int_a^b u(x) \, dx \leq \frac{1}{b-a} \int_a^b v(x) \, dx. \]

This gives us that \( \bar{u} \leq \bar{v}\) if \( u \leq v \) across the interval, supporting the logical consistency in comparing average values of functions. Function inequalities thus provide a robust framework for analyzing and proving such mathematical propositions.