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Magazine Chapter 6: Problem 53
AREA Find the area between the curve \(y=e^{-a x}\) (for \(a>0\) ) and the \(x\) -axis from \(x=0\) to \(\infty\).
Short Answer
Expert verified
The area is \(\frac{1}{a}\).
Step by step solution
01
Interpret the Problem
We need to find the area between the curve given by the function \(y = e^{-ax}\) and the x-axis over the interval from \(x = 0\) to \(x = \infty\).
02
Set Up the Integral
The area between a curve \(y = f(x)\) and the x-axis over an interval \([a,b]\) is given by the integral \(\int_{a}^{b} f(x) \, dx\). Here, we need to evaluate the improper integral \(\int_{0}^{\infty} e^{-ax} \, dx\).
03
Evaluate the Improper Integral
An improper integral over \([0, \infty)\) is evaluated by taking the limit of a proper integral as the upper limit approaches infinity. So, compute \(\lim_{b \to \infty} \int_{0}^{b} e^{-ax} \, dx\).
04
Determine the Antiderivative
The antiderivative of \(e^{-ax}\) with respect to \(x\) is \(-\frac{1}{a}e^{-ax}\). This is based on the derivative rule for exponential functions.
05
Evaluate the Definite Integral
First, evaluate the definite integral \(\int_{0}^{b} e^{-ax} \, dx\):\[\left[-\frac{1}{a}e^{-ax}\right]_{0}^{b} = -\frac{1}{a}e^{-ab} + \frac{1}{a} \]
06
Take the Limit
Now, take the limit of the expression as \(b\) approaches infinity:\[\lim_{b \to \infty} \left(-\frac{1}{a}e^{-ab} + \frac{1}{a}\right)\]The term \(-\frac{1}{a}e^{-ab}\) approaches 0 as \(b\) goes to infinity because \(e^{-ab}\) goes to 0. Thus, the limit is \(\frac{1}{a}\).
07
Confirm the Solution
The area under the curve \(y = e^{-ax}\) from \(x = 0\) to \(x = \infty\) is \(\frac{1}{a}\). This is consistent with the characteristic behavior of exponential decay functions over an infinite interval.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Decay
Exponential decay describes how a quantity decreases rapidly at first and then slowly over time. In mathematical terms, if a function has the form \( y = e^{-ax} \), where \( a > 0 \), it models exponential decay. As \( x \) increases, the value of \( y \) becomes smaller because \( e^{-ax} \) quickly approaches zero.
Exponential functions are commonly used to represent phenomena such as radioactive decay, cooling of substances, and depreciation of value. These functions decrease at a rate proportional to their current value, making them a fundamental concept in scientific studies and real-world applications.
Exponential functions are commonly used to represent phenomena such as radioactive decay, cooling of substances, and depreciation of value. These functions decrease at a rate proportional to their current value, making them a fundamental concept in scientific studies and real-world applications.
- Rapid Start, Slow Finish: Initially, the process happens fast, but it slows down over time.
- Characteristic Shape: The curve is always decreasing and forms a downward slope.
- Half-Life Concept: Often used to analyze how long it takes for half the quantity to decay.
Definite Integrals
Definite integrals are used to calculate the total accumulation of a quantity over a specified interval. In this context, it is used to find the area between the curve \( y = e^{-ax} \) and the \( x \)-axis over a certain interval. To compute this, we use the integral notation \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
For our specific problem, the integral becomes an improper integral, as it needs evaluation from \( x = 0 \) to \( x = \infty \). This requires us to assess the behavior of the function as \( x \) heads towards infinity, ensuring the total area under the curve is found.
For our specific problem, the integral becomes an improper integral, as it needs evaluation from \( x = 0 \) to \( x = \infty \). This requires us to assess the behavior of the function as \( x \) heads towards infinity, ensuring the total area under the curve is found.
- Area Calculation: The integral gives the precise area under the curve between two points on the \( x \)-axis.
- Improper Integrals: When calculating over infinite intervals, we often convert to a limit process for analysis.
- Finite Result: Even with infinite limits, the area can be finite, as our example shows with a result of \( \frac{1}{a} \).
Antiderivatives
Antiderivatives are the inverse operation of derivation, turning a derivative back into its original function. In the context of our problem, we look at finding the antiderivative of \( e^{-ax} \). The formula for its antiderivative is \(-\frac{1}{a} e^{-ax} \), derived from the rule connecting exponential functions to their derivatives.
An antiderivative helps in solving integrals because if you can find the antiderivative of a function, then computing the integral becomes a matter of evaluating this at the interval boundaries. This is especially useful for definite integrals, simplifying them to a simple subtraction once the antiderivative is known.
An antiderivative helps in solving integrals because if you can find the antiderivative of a function, then computing the integral becomes a matter of evaluating this at the interval boundaries. This is especially useful for definite integrals, simplifying them to a simple subtraction once the antiderivative is known.
- Fundamental Calculus Tool: Understanding connections between derivatives and antiderivatives is crucial.
- Integration Simplified: Once found, an antiderivative makes solving integrals straightforward.
- Behavior Understanding: Recognizing how specific functions integrate provides deeper insight into their behavior.
