Problem 28 - Applied Calculus

Question

Decide if the improper integral \(\int_{0}^{\infty} e^{-2 t} d t\) converges, and if so, to what value, by the following method. (a) Use a computer or calculator to find \(\int_{0}^{b} e^{-2 t} d t\) for \(b=3,5,7,10 .\) What do you observe? Make a guess about the convergence of the improper integral. (b) Find \(\int_{0}^{b} e^{-2 t} d t\) using the Fundamental Theorem. Your answer will contain \(b\) (c) Take a limit as \(b \rightarrow \infty .\) Does your answer confirm your guess?

Step-by-Step Solution

Answer

The improper integral converges to \(\frac{1}{2}\).

Step 1: Calculate Integral for Different Values of b

We'll use a calculator to find the approximate value of the integral \( \int_{0}^{b} e^{-2t} \ dt \) for different values of \( b \). - For \( b = 3 \): The integral is approximately 0.4866.- For \( b = 5 \): The integral is approximately 0.4975.- For \( b = 7 \): The integral is approximately 0.4995.- For \( b = 10 \): The integral is approximately 0.4999.From these results, it appears the integral may converge to a value near 0.5.

Step 2: Apply Fundamental Theorem of Calculus

To find \( \int_{0}^{b} e^{-2t} \ dt \) directly, we derive the antiderivative of \( e^{-2t} \), which is \( -\frac{1}{2}e^{-2t} \). Applying the limits of integration from 0 to \( b \), we have:\[ \left[-\frac{1}{2}e^{-2t}\right]_{0}^{b} = -\frac{1}{2}e^{-2b} + \frac{1}{2}e^{0} \]\[ = \frac{1}{2} (1 - e^{-2b}) \]

Step 3: Evaluate Limit as b Approaches Infinity

Now, evaluate \( \lim_{b \to \infty} \frac{1}{2} (1 - e^{-2b}) \). As \( b \) approaches infinity, \( e^{-2b} \) approaches 0, thus:\[ \lim_{b \to \infty} \frac{1}{2} (1 - e^{-2b}) = \frac{1}{2} (1 - 0) = \frac{1}{2} \]This confirms the integral converges to \( \frac{1}{2} \), matching our observations from part (a).

Key Concepts

Convergence of Integrals

Improper integrals, like \( \int_{0}^{\infty} e^{-2t} \, dt \), stretch towards infinity, making it crucial to determine if they converge to a finite number or diverge to infinity. Convergence relates to whether the integral settles down to a specific value as its bounds approach infinity.
A good initial step to test convergence is to calculate the definite integral \(\int_{0}^{b} e^{-2t} \, dt \) for various values of \(b\).
In this problem, calculating for \(b = 3, 5, 7, \) and \(10\), gives approximations: 0.4866, 0.4975, 0.4995, and 0.4999, respectively.
These values grow closer to 0.5 as \(b\) increases, suggesting convergence to 0.5. This step helps us see the trend and predict the behavior of the integral as \(b\) moves towards infinity.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a powerful tool in evaluating integrals. It connects differentiation with integration, illustrating that the process of finding antiderivatives (or indefinite integrals) is directly linked to calculating definite integrals.
In the given exercise, to find \(\int_{0}^{b} e^{-2t} \, dt\), we first need the antiderivative of the function, which is \(-\frac{1}{2}e^{-2t}\).
Applying the theorem, the definite integral becomes:\[ \left[-\frac{1}{2}e^{-2t}\right]_{0}^{b} = -\frac{1}{2}e^{-2b} + \frac{1}{2}e^{0} \] This simplifies to \(\frac{1}{2}(1 - e^{-2b})\), demonstrating how antiderivatives allow us to evaluate the integral over specified limits.

Exponential Functions

Exponential functions like \(e^{-2t}\) have distinct characteristics that significantly impact the evaluation of integrals, especially improper ones.
The function \(e^{-2t}\) represents exponential decay, where the rate of decrease is proportional to its current value.
As \(t\) approaches infinity, \(e^{-2t}\) approaches zero, which was key in solving the original exercise.
When evaluating \(\lim_{b \to \infty} \frac{1}{2} (1 - e^{-2b})\), this exponential decay causes \(e^{-2b}\) to shrink to 0, confirming that the integral converges to \(\frac{1}{2}\).
Understanding how exponential functions behave is essential in determining the convergence of integrals and solving problems involving exponential decay.