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 \(\frac{1}{8} - \frac{1}{8}e^{-8}\).
Step by step solution
01
Review the Average Value Formula
The average value of a continuous function \( f(x) \) on an interval \([a, b]\) is given by the formula: \[ f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx \]. In this case, our function is \(f(x) = e^{-2x}\) and the interval is \([0, 4]\).
02
Set Up the Integral
Substitute the function \( f(x) = e^{-2x} \), \( a = 0 \), and \( b = 4 \) into the average value formula to set up the integral. This gives us: \[ f_{avg} = \frac{1}{4-0} \int_0^4 e^{-2x} \, dx \].
03
Evaluate the Integral
Find the antiderivative of \( e^{-2x} \). The antiderivative of \( e^{-2x} \) is \(-\frac{1}{2}e^{-2x} \). Evaluate this from 0 to 4: \[ \int_0^4 e^{-2x} \, dx = \left[-\frac{1}{2}e^{-2x}\right]_0^4 = -\frac{1}{2}e^{-8} + \frac{1}{2}e^0 \].
04
Simplify the Evaluated Integral
Substitute and simplify: \[ -\frac{1}{2}e^{-8} + \frac{1}{2}e^0 = -\frac{1}{2}e^{-8} + \frac{1}{2} = \frac{1}{2} - \frac{1}{2}e^{-8} \].
05
Calculate Average Value
Substitute the result from the integral back into the average value formula: \[ f_{avg} = \frac{1}{4} \left( \frac{1}{2} - \frac{1}{2}e^{-8} \right) = \frac{1}{8} - \frac{1}{8}e^{-8} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
The concept of a definite integral is fundamental in calculus. It allows us to find the accumulation of values, such as area under a curve, across a specified interval. The definite integral of a function \( f(x) \) from \( a \) to \( b \) is represented by \( \int_a^b f(x) \, dx \).
The symbol \( \int \) indicates integration, \( a \) and \( b \) are the limits of integration, with \( a \) as the lower limit and \( b \) as the upper limit. The function \( f(x) \) is the integrand. This integral calculates the net area between the function and the x-axis over the interval \([a, b]\).
The symbol \( \int \) indicates integration, \( a \) and \( b \) are the limits of integration, with \( a \) as the lower limit and \( b \) as the upper limit. The function \( f(x) \) is the integrand. This integral calculates the net area between the function and the x-axis over the interval \([a, b]\).
- Net Area: It accounts for areas above and below the axis, with areas below counted as negative.
- Bounds: The specified limits always dictate where the calculation begins and ends.
- Result: The result is a real number representing total accumulation over this interval.
Exponential Function
An exponential function is characterized by its constant base raised to a variable exponent. A common form of an exponential function is \( e^x \), where \( e \) is a mathematical constant approximately equal to 2.71828. In the exercise, \( f(x) = e^{-2x} \) represents an exponential decay due to the negative exponent.
In exponential decay functions like \( e^{-2x} \):
In exponential decay functions like \( e^{-2x} \):
- The function decreases as \( x \) increases.
- The rate of decrease is proportionate to its current value, which is a key property of exponential functions.
- These functions approach zero but never actually reach zero, indicating that they decay rapidly but never fully disappear.
Antiderivative
Finding the antiderivative, or the indefinite integral, of a function involves determining a function whose derivative matches the original function. In the exercise, we need to find the antiderivative of \( e^{-2x} \).
The antiderivative of \( e^{-kx} \) is \(-\frac{1}{k}e^{-kx} \), plus a constant \( C \). Specifically, for \( e^{-2x} \), the antiderivative becomes \( -\frac{1}{2}e^{-2x} \). This new function allows us to evaluate the definite integral by substituting the upper and lower bounds and calculating the difference.
Some key aspects regarding antiderivatives:
The antiderivative of \( e^{-kx} \) is \(-\frac{1}{k}e^{-kx} \), plus a constant \( C \). Specifically, for \( e^{-2x} \), the antiderivative becomes \( -\frac{1}{2}e^{-2x} \). This new function allows us to evaluate the definite integral by substituting the upper and lower bounds and calculating the difference.
Some key aspects regarding antiderivatives:
- They are the reverse process of differentiation.
- Unlike derivatives, a sawtooth pattern or wiggle can result from changing the constant \( C \).
- Knowing basic antiderivatives helps with complex problem solving in calculus.