/*! 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 4 Find the average value of the fu... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the average value of the function on the given interval. Use equation (4.8) if it applies. If an average value is zero, you may be able to decide this from a quick sketch which shows you that the areas above and below the \(x\) axis are the same. $$1-e^{-x} \text { on }(0,1)$$

Short Answer

Expert verified
The average value of the function on \( (0,1) \) is \( \frac{1}{e} \).

Step by step solution

01

Identify the Interval and Function

The given function is \(1 - e^{-x}\). The interval provided is \( (0,1) \).
02

Write Down the Average Value Formula

The average value of a function \(f(x)\) on the interval \( [a, b] \) is given by equation (4.8): \[ f_{avg} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx \]. In this case, \((a,b) = (0,1)\).
03

Set Up the Integral

Substitute the given function and interval into the formula: \[ f_{avg} = \frac{1}{1-0} \int_{0}^{1} (1 - e^{-x}) \, dx \]. It simplifies to \[ f_{avg} = \int_{0}^{1} (1 - e^{-x}) \, dx \].
04

Evaluate the Integral

The integral can be split and evaluated separately: \[ \int_{0}^{1} (1 - e^{-x}) \, dx = \int_{0}^{1} 1 \, dx - \int_{0}^{1} e^{-x} \, dx \].
05

Compute Each Part

\[ \int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1 - 0 = 1 \] and \[ \int_{0}^{1} e^{-x} \, dx = [-e^{-x}]_{0}^{1} = -e^{-1} - (-e^{0}) = 1 - \frac{1}{e} \].
06

Combine the Results

Subtract the second result from the first: \[ f_{avg} = 1 - (1 - \frac{1}{e}) = \frac{1}{e} \].

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Key Concepts

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

integral calculus
Integral calculus is a branch of mathematics that deals with the concept of integration. Integration is essentially the process of finding the area under a curve. This area is represented by an integral, usually written in the form \( \int f(x) \, dx \) where \( f(x) \) is the function being integrated and \( x \) represents the variable.

In the given exercise, we use integral calculus to compute the average value of a function over a specific interval. The integral provides a way to sum up infinitely many small parts to find the whole. It's especially useful when dealing with continuous functions.

Learning integral calculus is vital as it applies to various fields like physics, engineering, and economics. It helps solve problems related to area, volume, displacement, and many other concepts.
definite integrals
A definite integral is a type of integral that evaluates the area under the curve of a function between two specific points, known as the bounds. For the given interval \( [a, b] \), the definite integral is written as \( \int_{a}^{b} f(x) \, dx \). This notation tells us to calculate the integral from \((a) \) to \((b) \).

In our exercise, the interval is \( (0, 1) \). So, we look at the function \(1 - e^{-x}\) from 0 to 1. The value of this integral consists of summing up all small changes from 0 to 1.

Unlike indefinite integrals, which include a constant of integration, definite integrals yield a numerical value, representing the total accumulation within those bounds.
exponential functions
An exponential function is a mathematical function of the form \( f(x) = a^x \), where the base \( a \) is a constant and the exponent \( x \) is a variable. One of the most common exponential functions encountered in calculus is \( e^x\), where \ e\ is Euler's number, approximately equal to 2.71828.

In the given exercise, we have the function \(1 - e^{-x} \). Here, \( e^{-x} \) means the inverse or decreasing exponential function. Exponential functions are important as they model various real-world phenomena such as population growth, radioactive decay, and interest calculations.

Understanding how to integrate exponential functions is key in solving many calculus problems, as they frequently appear in various types of integrals.
interval notation
Interval notation is a way of representing subsets of real numbers and it is often used in calculus problems to describe the domain or range of functions, or the bounds of integrals. It uses brackets \( [ \) and \( ] \) to indicate closed intervals (including the endpoints) and parentheses \( ( \) and \( ) \) to indicate open intervals (excluding the endpoints).

In the given exercise, the interval \( (0, 1) \) tells us that we are looking at all the x-values between 0 and 1, but not including the points 0 and 1 themselves. This is helpful in integration because it tells us exactly where we should start and stop calculating the area under the curve.

Understanding interval notation is crucial when working with integrals as it precisely specifies the limits or boundaries within which you need to compute the integral.

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Most popular questions from this chapter

(a) Let \(f(x)\) on \((0,2 l)\) satisfy \(f(2 l-x)=f(x),\) that is, \(f(x)\) is symmetric about \(x=l .\) If you expand \(f(x)\) on \((0,2 l)\) in a sine series \(\sum b_{n} \sin \frac{n \pi x}{2 l},\) show that for even \(n, b_{n}=0 .\) Hint: Note that the period of the sines is \(4 l .\) Sketch an \(f(x)\) which is symmetric about \(x=l,\) and on the same axes sketch a few sines to see that the even ones are antisymmetric about \(x=l\). Alternatively, write the integral for \(b_{n}\) as an integral from 0 to \(l\) plus an integral from \(l\) to \(2 l,\) and replace \(x\) by \(2 l-x\) in the second integral. (b) Similarly, show that if we define \(f(2 l-x)=-f(x),\) the cosine series has \(a_{n}=0\) for even \(n\).

In each of the following problems you are given a function on the interval \(-\pi < x < \pi\) Sketch several periods of the corresponding periodic function of period \(2 \pi\). Expand the periodic function in a sine-cosine Fourier series. $$f(x)=\left\\{\begin{array}{lr} \pi+x, & -\pi < x < 0, \\ \pi-x, & 0 < x < \pi. \end{array}\right.$$

Find the amplitude, period, frequency, wave velocity, and wavelength of the given wave. By computer, plot on the same axes, \(y\) as a function of \(x\) for the given values of \(t,\) and label each graph with its value of \(t .\) Similarly, plot on the same axes, \(y\) as a function of \(t\) for the given values of \(x,\) and label each curve with its value of \(x.\) $$y=\cos 2 \pi\left(x-\frac{1}{4} t\right) ; \quad t=0,1,2,3 ; \quad x=0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}$$

The charge \(q\) on a capacitor in a simple a-c circuit varies with time according to the equation \(q=3 \sin (120 \pi t+\pi / 4) .\) Find the amplitude, period, and frequency of this oscillation. By definition, the current flowing in the circuit at time \(t\) is \(I=d q / d t\) Show that \(I\) is also a sinusoidal function of \(t,\) and find its amplitude, period, and frequency.

A simple pendulum consists of a point mass \(m\) suspended by a (weightless) cord or rod of length \(l,\) as shown, and swinging in a vertical plane under the action of gravity. Show that for small oscillations (small \(\theta\) ), both \(\theta\) and \(x\) are sinusoidal functions of time, that is, the motion is simple harmonic. Hint: Write the differential equation \(\mathbf{F}=m \mathbf{a}\) for the particle \(m .\) Use the approximation \(\sin \theta=\theta\) for small \(\theta,\) and show that \(\theta=A \sin \omega t\) is a solution of your equation. What are \(A\) and \(\omega ?\)

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