Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the Comparison Test of Problem 46 to show that \(\int_{1}^{\infty} e^{-x^{2}} d x\) converges. Hint: \(e^{-x^{2}} \leq e^{-x}\) on \([1, \infty)\)

Step-by-Step Solution

Answer

The integral \( \int_{1}^{\infty} e^{-x^{2}} dx \) converges by the Comparison Test.

Step 1: Identify the Original Integral

The given integral to evaluate for convergence is \( \int_{1}^{\infty} e^{-x^{2}} dx \). We need to assess whether this integral converges.

Step 2: State the Comparison Test

The Comparison Test for integrals states that if \( 0 \leq f(x) \leq g(x) \) for all \( x \) in the interval, and \( \int_{1}^{\infty} g(x) dx \) converges, then \( \int_{1}^{\infty} f(x) dx \) also converges.

Step 3: Establish the Comparison Function

We are given that \( e^{-x^{2}} \leq e^{-x} \) for \( x \geq 1 \). Thus, we can use \( g(x) = e^{-x} \) as the comparison function.

Step 4: Evaluate the Comparison Integral

First, consider the integral \( \int_{1}^{\infty} e^{-x} dx \). This can be evaluated as follows: \[ \int_{1}^{\infty} e^{-x} dx = \lim_{b \to \infty} \left[ -e^{-x} \right]_{1}^{b} = \lim_{b \to \infty} \left( -e^{-b} + e^{-1} \right) = e^{-1}. \] Since \( e^{-b} \to 0 \) as \( b \to \infty \), the integral \( \int_{1}^{\infty} e^{-x} dx \) converges.

Step 5: Apply the Comparison Test

Since \( e^{-x^{2}} \leq e^{-x} \) on \([1, \infty)\) and we have shown that \( \int_{1}^{\infty} e^{-x} dx \) converges, by the Comparison Test, the original integral \( \int_{1}^{\infty} e^{-x^{2}} dx \) also converges.

Key Concepts

Integral Convergence

Integral convergence is a core topic in calculus, particularly when dealing with improper integrals. Essentially, an integral converges if it produces a finite result. For improper integrals, which have infinite limits of integration, assessing convergence tells you if the integral settles to a specific value even across an infinite range. For example, consider the integral \( \int_{1}^{\infty} e^{-x^{2}} dx \). Here, we're tasked with determining convergence due to the infinity in the upper limit. The focus is on whether this integral approaches a precise value as the upper limit extends indefinitely.Several testing methods can help determine convergence, with the Comparison Test being one of them. It’s crucial to understand the behavior of the function we are integrating over its domain, which in this specific problem is \([1, \infty)\). Recognizing convergence allows us to deal with functions that appear when modeling real-world behaviors such as diffusion processes, predicting that they remain stable over time.

Comparison Function

The comparison function is an essential tool when using the Comparison Test for convergence. This method involves finding a function that is easier to analyze and compare to your integral of interest. In our exercise, establish that \( e^{-x^{2}} \leq e^{-x} \) for \( x \geq 1 \). This comparison is rooted in the fact that \( x^{2} \geq x \) when \( x \geq 1 \), making \( e^{-x^{2}} \leq e^{-x} \).From this, we see that the function \( g(x) = e^{-x} \) acts as our comparison function. Why? Because it simplifies our analysis. Assessing \( \int_{1}^{\infty} e^{-x} dx \) confirms that this integral is simpler and converges to a finite value. This simplification provides a questioning framework: if \( e^{-x} \) can be integrated over the interval to produce a finite value, then the more complex function is also likely to converge.

Improper Integrals

Improper integrals involve functions or limits that are infinite. These integrals extend our understanding beyond basic integration by testing the boundaries of calculus. When evaluating an integral like \( \int_{1}^{\infty} e^{-x^{2}} dx \), we encounter an improper integral due to the infinite upper limit.To solve an improper integral, it’s common to employ a limit process. Essentially, you replace the infinity symbol with a variable that approaches infinity, such as the integral \( \int_{1}^{b} e^{-x^{2}} dx \) as \( b \to \infty \). This transformation enables us to hone in on whether the integral converges or diverges based on the results of the evaluated limit.Understanding improper integrals allows us to assess complex situations effectively, whether in mathematical contexts or applied fields. Recognizing when to transform and compare such integrals is key to mastering more advanced calculus topics.