Problem 67
Question
Use a graphing utility to graph the function over the interval. Find the average value of the function over the interval. Then find all \(x\) -values in the interval for which the function is equal to its average value. $$ f(x)=2 e^{x} \quad[-1,1] $$
Step-by-Step Solution
Verified Answer
The average value of \(f(x) = 2e^x\) over the interval \([-1,1]\) is \(e - e^{-1}\). The specific x-values for which the function equals its average value can be found by solving the equation \(2e^x = e - e^{-1}\).
1Step 1: Graphing the function
Use a graphing tool or program to graph the function \(f(x)=2e^x\) over the interval \([-1,1]\). This gives us a clear visualization of how the function behaves in this interval.
2Step 2: Calculating the average value of the function
The average value of a function \(f(x)\) over the interval \([a, b]\) can be found using the formula: \[\frac{1}{b-a}\int_{a}^{b}f(x)dx.\] In this case, \(f(x) = 2e^x\) and the interval is [-1, 1]. So plugging in these values, we get the average value as \[\frac{1}{1-(-1)}\int_{-1}^{1}2e^xdx = e - e^{-1}.\]
3Step 3: Finding x-values at which the function equal its average value
Set the function \(f(x)\) equal to its average value \(e - e^{-1}\) and solve for \(x\). That is, solve the equation \(2e^x = e - e^{-1}\). The solutions for \(x\) will be the x-values for which the function is equal to its average value.
Key Concepts
Average Value of a FunctionExponential FunctionsIntegration in Calculus
Average Value of a Function
When we talk about the average value of a function over a specific interval, it essentially means what the "mean height" of the function graph is when viewed over that interval. To find the average value, we use the formula:
In our problem, the function is \(f(x)=2e^x\) over \([-1,1]\). By plugging it into the formula, we derive \( e - e^{-1} \) as the average value. This formula skillfully decomposes the total area through integration, thus providing the average height or value over a given interval.
- \( \frac{1}{b-a}\int_{a}^{b}f(x)dx \).
In our problem, the function is \(f(x)=2e^x\) over \([-1,1]\). By plugging it into the formula, we derive \( e - e^{-1} \) as the average value. This formula skillfully decomposes the total area through integration, thus providing the average height or value over a given interval.
Exponential Functions
Exponential functions are a type of mathematical function of the form \( f(x) = a \cdot e^{bx} \) where \( e \) is the base of natural logarithms, approximately equal to 2.71828. These functions exhibit rapid growth or decay, making them significant in modeling natural phenomena such as population growth or radioactive decay.
In our exercise, the function \( f(x) = 2e^x \) increases exponentially as \( x \) goes from \(-1\) to \(1\). A distinctive trait of exponential growth is that the rate of increase depends proportionally on the function's current value. Thus, even small increments in \( x \) lead to substantial changes in \( f(x) \).
Graphing these functions provides a powerful visual aid as they curve upwards, showing the nature of their continuous and unbounded growth. This characteristic makes exponential functions both fascinating and crucial for various practical applications.
In our exercise, the function \( f(x) = 2e^x \) increases exponentially as \( x \) goes from \(-1\) to \(1\). A distinctive trait of exponential growth is that the rate of increase depends proportionally on the function's current value. Thus, even small increments in \( x \) lead to substantial changes in \( f(x) \).
Graphing these functions provides a powerful visual aid as they curve upwards, showing the nature of their continuous and unbounded growth. This characteristic makes exponential functions both fascinating and crucial for various practical applications.
Integration in Calculus
Integration is a fundamental concept in calculus that involves finding the total accumulation of quantities, such as areas under curves, volumes, or other measures. It is essentially the reverse process of differentiation and an exquisite method to solve real-world problems where accumulating changes are involved.
In our example, the integration process helps determine the average value of the function \(2e^x\) over the interval \([-1, 1]\). By performing the integral \( \int_{-1}^{1}2e^xdx \), we achieve an expression that represents the total aggregate of areas beneath the curve, which is then used in our average value formula.
Understanding how to integrate effectively is key to making sense of calculus's powerful predictive and analytical capabilities. It provides us tools to decode complex systems and behaviors analytically, translating calculus into a practical and enriching toolkit for solving diverse problems.
In our example, the integration process helps determine the average value of the function \(2e^x\) over the interval \([-1, 1]\). By performing the integral \( \int_{-1}^{1}2e^xdx \), we achieve an expression that represents the total aggregate of areas beneath the curve, which is then used in our average value formula.
Understanding how to integrate effectively is key to making sense of calculus's powerful predictive and analytical capabilities. It provides us tools to decode complex systems and behaviors analytically, translating calculus into a practical and enriching toolkit for solving diverse problems.
Other exercises in this chapter
Problem 64
Find the area of the region bounded by the graphs of the equations. Use a graphing utility to verify your results. $$ y=e^{x}, \quad y=0, \quad x=0, \quad \text
View solution Problem 65
Use a graphing utility to graph the function over the interval. Find the average value of the function over the interval. Then find all \(x\) -values in the int
View solution Problem 68
Use a graphing utility to graph the function over the interval. Find the average value of the function over the interval. Then find all \(x\) -values in the int
View solution Problem 69
Find the profit function for the given marginal profit and initial condition. $$ \frac{d P}{d x}=-18 x+1650 \quad P(15)=\$ 22,725 $$
View solution