Chapter 7: Problem 4
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.
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.
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.
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.
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.