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

All the statements (a), (b), and (c) are true.

Step 1: Understand the Problem

We are given functions \(u\) and \(v\), and we need to determine whether certain properties of their average values over an interval hold true. The average value of a function \(f\) over an interval \([a, b]\) is given by \(\frac{1}{b-a}\int_a^b f(x)\,dx\). We will use this formula to determine whether the given properties hold.

Step 2: Prove Part (a): Average of Sum Property

The average value of \(u\) is \(\bar{u} = \frac{1}{b-a}\int_a^b u(x)\,dx\), and the average value of \(v\) is \(\bar{v} = \frac{1}{b-a}\int_a^b v(x)\,dx\). The average value of \(u + v\) is \(\overline{u+v} = \frac{1}{b-a}\int_a^b (u(x) + v(x))\,dx\). By linearity of integration, this simplifies to \(\frac{1}{b-a}(\int_a^b u(x)\,dx + \int_a^b v(x)\,dx) = \bar{u} + \bar{v}\). Thus, \(\bar{u} + \bar{v} = \overline{u+v}\).

Step 3: Prove Part (b): Scalar Multiplication Property

The average value of \(ku\), where \(k\) is a constant, is \(\overline{ku} = \frac{1}{b-a}\int_a^b k u(x)\,dx\). The constant \(k\) can be factored out of the integral, giving \(\frac{k}{b-a}\int_a^b u(x)\,dx = k \bar{u}\). Therefore, \(k \bar{u} = \overline{ku}\).

Step 4: Prove Part (c): Order Preservation Property

If \(u \leq v\) for all \(x\) in \([a, b]\), then by properties of integration \(\int_a^b u(x)\,dx \leq \int_a^b v(x)\,dx\). Dividing both sides by \(b-a\) gives \(\bar{u} = \frac{1}{b-a}\int_a^b u(x)\,dx \leq \frac{1}{b-a}\int_a^b v(x)\,dx = \bar{v}\). Therefore, if \(u \leq v\), it follows that \(\bar{u} \leq \bar{v}\).

Key Concepts

Average Value of a Function

The average value of a function over an interval provides a sense of the "typical" value of the function within that interval. It is calculated using integration, which is a fundamental part of calculus. If you have a function \( f(x) \) and you want to find its average value over an interval \([a, b]\), you use the following formula:

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

This formula essentially computes the area under the function curve over the interval and then divides it by the length of the interval, \(b-a\). This gives you the average height or value of the function over that interval.
Think of it as spreading the "total effect" of the function evenly across the interval, which helps us understand the overall behavior of the function in a quantifiable way.

Linearity of Integration

Linearity of integration is a powerful property that simplifies integration when dealing with a sum of functions or constant multiples of functions. In more practical terms, it means that integration does not disrupt addition or scalar multiplication.

For any functions \( u(x) \) and \( v(x) \), and any constant \( k \), the following hold true:

• \( \int_a^b (u(x) + v(x))\,dx = \int_a^b u(x)\,dx + \int_a^b v(x)\,dx \)
• \( \int_a^b k\cdot u(x)\,dx = k\cdot \int_a^b u(x)\,dx \)

This means when you have a function expressed as a sum, you can split it and integrate each part separately before summing the results.
Moreover, when a constant is involved (like with scalar multiplication), you can "pull" this constant out of the integral, making the problem easier to tackle. This property underpins why the average value of a sum \( \bar{u} + \bar{v} = \overline{u+v} \), and why multiplying by a constant \( k\bar{u} = \overline{ku} \) holds true.

Order Preservation in Integration

Order preservation in integration ensures that if one function is always less than or equal to another over a given interval, then their integrals (and thereby their average values) will share this order.

If \( u(x) \leq v(x) \) for all \( x \) in the interval \([a, b]\), integration behaves logically, maintaining this inequality:

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

When you translate this to their average values, dividing through by \( b-a \) does not change the direction of the inequality:

• \( \bar{u} = \frac{1}{b-a}\int_a^b u(x)\,dx \leq \frac{1}{b-a}\int_a^b v(x)\,dx = \bar{v} \)

This concept asserts the notion of a "fair" comparison between the averages of two functions based on their hierarchy in value at each point over the interval. If one function is always at least as large as another, this relationship is preserved in their average values as well.