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Magazine Chapter 5: Problem 11
Find the average value of the function over the given interval. $$ f(x)=e^{-2 x} ;[0,4] $$
Short Answer
Expert verified
The average value is \( \frac{1}{8} (1 - e^{-8}) \).
Step by step solution
01
Understand the Problem
We need to find the average value of the function over a given interval. The formula for the average value of a function \( f(x) \) over the interval \([a,b]\) is given by \( \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \). In this case, \( f(x) = e^{-2x} \) and the interval is \([0, 4]\).
02
Set Up the Integral
Substitute the function \( f(x) = e^{-2x} \) and the interval \([0, 4]\) into the integral formula for average value: \[ \text{Average value} = \frac{1}{4 - 0} \int_{0}^{4} e^{-2x} \, dx. \]
03
Simplify the Expression
The expression \( \frac{1}{4} \int_{0}^{4} e^{-2x} \, dx \) simplifies the fraction \( \frac{1}{4} \) to multiply the integral, which we will now evaluate.
04
Evaluate the Integral
To evaluate \( \int_{0}^{4} e^{-2x} \, dx \), use the rule \( \int e^{ax} \, dx = \frac{1}{a} e^{ax} + C \). Here, \( a = -2 \), so the indefinite integral is \( \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x} + C \).
05
Apply the Limits of Integration
Evaluate the definite integral \( -\frac{1}{2} e^{-2x} \) from \( 0 \) to \( 4 \): \[ \left[ -\frac{1}{2}e^{-2(4)} \right] - \left[ -\frac{1}{2}e^{-2(0)} \right] = -\frac{1}{2}e^{-8} + \frac{1}{2}. \]
06
Substitute Results to Find Average
Multiply the result of the integral by \( \frac{1}{4} \): \[ \text{Average value} = \frac{1}{4} \left( \frac{1}{2} - \frac{1}{2}e^{-8} \right) = \frac{1}{8} \left( 1 - e^{-8} \right). \]
07
Simplify the Final Expression
The final average value of the function \( f(x) = e^{-2x} \) over the interval \([0, 4]\) is \( \frac{1}{8} \left( 1 - e^{-8} \right). \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
When we talk about the **definite integral**, we're dealing with a concept that allows us to calculate the "net area" beneath a curve over a given interval. Specifically, it considers the signed area, meaning it accounts for both areas above and below the x-axis. This is different from the indefinite integral, which includes a constant of integration.
- The process is represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits, respectively.
- For instance, the integral \( \int_{0}^{4} e^{-2x} \, dx \) is a definite integral where we find the area of the curve \( e^{-2x} \) from 0 to 4.
Exponential Function
An **exponential function** involves a constant raised to the power of a variable. It's essential to understand how these functions behave to tackle problems like finding their average value over an interval.
- In our example, the function is \( e^{-2x} \). The base \( e \) (approximately 2.718) is raised to the power of \(-2x \).
- The negative exponent indicates the function decreases as \( x \) increases. As \( x \) gets larger, \( e^{-2x} \) approaches zero.
Integration by Substitution
**Integration by substitution** is a powerful technique for solving integrals, especially when dealing with complex functions like exponential ones.
- This method involves substituting a part of the integrand with a new variable to simplify the integral before solving it.
- In our exercise, though direct integration suffices, the structure \( e^{-2x} \) can prompt substitution. Letting \( u = -2x \), the corresponding \( du = -2 dx \) transforms the integral into a simpler form.
