/*! 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 87 Find the average value of the fu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 87

Find the average value of the function over the given interval. $$ f(x)=\frac{8}{x^{2}}, \quad[2,4] $$

Short Answer

Expert verified
The average value of the function \(f(x) = \frac{8}{x^{2}}\) over the interval [2, 4] is 1.

Step by step solution

01

Identifying a, b and f(x)

First, clearly identify the values of a, b and f(x). Here, a is the left endpoint of the interval which is 2; b is the right endpoint of the interval which is 4 and the function f(x) is \(\frac{8}{x^{2}}\).
02

Applying the Formula for the Average Value of a Function

Next, set up the integral using the formula for the average value of a function, which is \(f_{ave} = \frac{1}{b-a}\int_{a}^{b} f(x) dx\). Substituting our values, we get: \(f_{ave} = \frac{1}{4-2}\int_{2}^{4} \frac{8}{x^{2}} dx\).
03

Solving the Integral

Change \(\frac{8}{x^{2}}\) to \(8x^{-2}\), and then apply the power rule for integrals, which states \(\int x^n dx = \frac{1}{n + 1}x^{n + 1}\). The result is \(f_{ave} = \frac{1}{2}[-8x^{-1}]\Big|_{2}^{4}\).
04

Calculating the value

Next, evaluate the expression at the upper limit and at the lower limit and subtract the two results to get the average value of the function: \(f_{ave} = \frac{1}{2}[-8(4^{-1}) - -8(2^{-1})] = \frac{1}{2}[-2+4] = 1\).

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.

Integral Calculus
Integral calculus is a branch of calculus that focuses on the concept of integration. Integration is essentially the process of finding the area under a curve. This is extremely useful when dealing with various physical problems, such as calculating dimensions or any total change.
To understand integration:
  • Imagine the area under a curve is divided into numerous small sub-areas.
  • The total area, which is the integral, is determined by adding up these tiny areas.
In simpler terms, if differentiation tells us the rate of change, integration does the reverse, giving us the total accumulation over an interval.
Power Rule for Integrals
The power rule for integrals is an easy and direct technique for integrating functions of the form \(x^n\). It helps us to calculate integrals without much hassle.
The rule states:
  • For a function \(x^n\), the integral is given by \(\int x^n dx = \frac{1}{n + 1} x^{n + 1} + C\), where \(C\) is the constant of integration.
  • This rule applies as long as \(n eq -1\).
In our exercise, we applied the power rule to \(8x^{-2}\) by transforming it into the integral \(-8x^{-1}\), giving us an intermediate result necessary for further calculations. This demonstrates the application of the power rule in simplifying complex expressions.
Definite Integral
The definite integral provides the accumulation or total change of a function over a specified interval \([a, b]\). It is represented as \(\int_{a}^{b} f(x) dx\). Unlike indefinite integrals, the definite integral results in a numerical value rather than a function.
To compute a definite integral:
  • First, integrate the function as you would with an indefinite integral.
  • Second, evaluate the resulting expression at the upper bound \(b\) and at the lower bound \(a\).
  • Finally, subtract the value at \(a\) from the value at \(b\).
In the current example, the definite integral is used to find the total area under \(\frac{8}{x^2}\) between \(x=2\) and \(x=4\). This step is crucial for determining the average value of the function over the interval.
Function Evaluation
Function evaluation is the process of finding function values at specific inputs. This is essential, especially when dealing with integrals and other calculus concepts. After integrating using definite integrals, we often need to calculate the function's value at the boundaries of the interval.
For our exercise:
  • We evaluate the expression \(-8x^{-1}\) at \(x=4\) and \(x=2\).
  • By substituting these boundaries into the evaluated integral, we subtract to find the total change over the interval.
This step-by-step evaluation helps determine the average value of the function, confirming our understanding through direct calculation. It's a practical application of dealing with real values instead of just symbolic results.

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

Prove that a function has an inverse function if and only if it is one-to-one.

Solve the differential equation. $$ \frac{d y}{d x}=\frac{x^{3}-21 x}{5+4 x-x^{2}} $$

Describe how to find the inverse function of a one-to-one function given by an equation in \(x\) and \(y\). Give an example.

Find the integral. $$ \int \cosh ^{2}(x-1) \sinh (x-1) d x $$

An object is projected upward from ground level with an initial velocity of 500 feet per second. In this exercise, the goal is to analyze the motion of the object during its upward flight. (a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function. (b) Use the result in part (a) to find the position function and determine the maximum height attained by the object. (c) If the air resistance is proportional to the square of the velocity, you obtain the equation $$ \frac{d v}{d t}=-\left(32+k v^{2}\right) $$ where \(-32\) feet per second per second is the acceleration due to gravity and \(k\) is a constant. Find the velocity as a function of time by solving the equation $$ \int \frac{d v}{32+k v^{2}}=-\int d t $$ (d) Use a graphing utility to graph the velocity function \(v(t)\) in part (c) if \(k=0.001\). Use the graph to approximate the time \(t_{0}\) at which the object reaches its maximum height. (e) Use the integration capabilities of a graphing utility to approximate the integral \(\int_{0}^{t_{0}} v(t) d t\) where \(v(t)\) and \(t_{0}\) are those found in part (d). This is the approximation of the maximum height of the object. (f) Explain the difference between the results in parts (b) and (e).
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied 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.