/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Assuming that \(u\) and \(v\) ca... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 53

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 \(\bar{u} \leq \bar{v}\).

Short Answer

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

Step by step solution

01

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.
02

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, \(\bar{u} + \bar{v} = \overline{u+v}\).
03

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, \(k \bar{u} = \overline{ku}\).
04

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 \(\bar{u} \leq \bar{v}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Household electric current can be modeled by the voltage \(V=\hat{V} \sin (120 \pi t+\phi)\), where \(t\) is measured in seconds, \(\hat{V}\) is the maximum value that \(V\) can attain, and \(\phi\) is the phase angle. Such a voltage is usually said to be 60 -cycle, since in 1 second the voltage goes through 60 oscillations. The root-mean-square voltage, usually denoted by \(V_{\mathrm{rms}}\) is defined to be the square root of the average of \(V^{2} .\) Hence $$ V_{\mathrm{rms}}=\sqrt{\int_{\phi}^{1+\phi}(\hat{V} \sin (120 \pi t+\phi))^{2} d t} $$ A good measure of how much heat a given voltage can produce is given by \(V_{\mathrm{rms}}\) (a) Compute the average voltage over 1 second. (b) Compute the average voltage over \(1 / 60\) of a second. (c) Show that \(V_{\mathrm{rms}}=\frac{\hat{V} \sqrt{2}}{2}\) by computing the integral for \(V_{\mathrm{rms}}\) Hint: \(\int \sin ^{2} t d t=-\frac{1}{2} \cos t \sin t+\frac{1}{2} t+C .\) (d) If the \(V_{\mathrm{rms}}\) for household current is usually 120 volts, what is the value \(\hat{V}\) in this case?

Suppose that an object, moving along the \(x\) -axis, has velocity \(v=t^{2}\) meters per second at time \(t\) seconds. How far did it travel between \(t=3\) and \(t=5\) ? See Problem 61 .

Oil is leaking at the rate of \(V^{\prime}(t)=1-t / 110\) from a storage tank that is initially full of 55 gallons. How much leaks out during the first hour? During the tenth hour? How long until the entire tank is drained?

, use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{0}^{\pi / 2} \cos ^{2} x \sin x d x $$

Find a nice formula for the sum $$ \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)} $$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.