Problem 37 - Calculus for Biology and Medicine

Question

(a) Show that $$0 \leq e^{-x^{2}} \leq e^{-x}$$ for \(x \geq 1\). (b) Use your result in (a) to show that $$\int_{1}^{\infty} e^{-x^{2}} d x$$ is convergent.

Step-by-Step Solution

Answer

(a) For \(x \geq 1\), \(0 \leq e^{-x^2} \leq e^{-x}\). (b) \(\int_{1}^{\infty} e^{-x^{2}} \, dx\) is convergent.

Step 1: Prove the lower bound 0 ≤ e^{-x^2}

Since exponential functions are always positive, we know that for any real number \(x\), \(e^{-x^2} > 0\). Thus, \(0 \leq e^{-x^2}\) is true for all \(x\), including when \(x \geq 1\).

Step 2: Compare e^{-x^2} and e^{-x}

To show \(e^{-x^{2}} \leq e^{-x}\) for \(x \geq 1\), consider the function \(f(x) = e^{-x} - e^{-x^2}\).For \(x \geq 1\), we need to show that \(f(x) \geq 0\) which means \(e^{-x} \geq e^{-x^2}\).For this inequality, note that \(-x^2 \leq -x\), thus \(e^{-x^2} \leq e^{-x}\) when \(x \geq 1\).

Step 3: Identify region of absolute convergence for the integral

We have determined that \(0 \leq e^{-x^2} \leq e^{-x} \) for \(x \geq 1\). We know that \( \int_{1}^{\infty} e^{-x} \ dx \) converges since the exponential function \(e^{-x}\) has an exponential decay rate at infinity that assures convergence.

Step 4: Use comparison test to evaluate convergence of integral

Using the comparison test, since \(0 \leq e^{-x^2} \leq e^{-x} \) and \( \int_{1}^{\infty} e^{-x} \ dx \) converges, \( \int_{1}^{\infty} e^{-x^2} \ dx \) also converges. Thus, \(\int_{1}^{\infty} e^{-x^{2}} \, dx\) is convergent by the comparison with a known convergent integral.

Key Concepts

Exponential Decay

When we talk about exponential decay, we're describing a process where quantities decrease at a rate proportional to their current value. This often comes up in natural phenomena like cooling, radioactive decay, or economic depreciation. In mathematical terms, an exponential decay function can be expressed as \( e^{-x} \), where as \( x \) increases, \( e^{-x} \) becomes very small very quickly.
This rapid decrease is what we term as 'decay'. It's like saying something melts away faster and faster over time.
- Key features of exponential decay include: - The rate of decrease is proportional to the value itself. - As \( x \) tends to infinity, \( e^{-x} \) approaches 0.In the context of our exercise, understanding the nature of exponential decay helps in grasping why \( e^{-x} \) and \( e^{-x^2} \) lead to convergent integrals. As \( x \to \infty \), both expressions shrink rapidly toward zero, preventing the integral from blowing up to infinity.

Comparison Test

The Comparison Test is a fundamental concept for determining the convergence of integrals or series. It can tell us whether an integral converges (settles at a specific value) or diverges (heads towards infinity) by comparing it to another known integral or series. Here's how it works:- If two functions, say \( g(x) \) and \( f(x) \), satisfy \( 0 \leq f(x) \leq g(x) \) for all \( x \) in some interval, then: - If \( \int_a^b g(x) \; dx \) converges, so does \( \int_a^b f(x) \; dx \). - Conversely, if \( \int_a^b f(x) \; dx \) diverges, then \( \int_a^b g(x) \; dx \) must also diverge. In the exercise, this test is key: we prove \( 0 \leq e^{-x^2} \leq e^{-x} \) for \( x \geq 1 \). We know that the integral of \( e^{-x} \) from 1 to infinity converges. Thus, by comparison, the integral of \( e^{-x^2} \) from 1 to infinity must also converge.

Convergent Integral

A convergent integral is one where the total area under a curve, often extending to infinity, adds up to a finite number. Understanding convergence is crucial for many areas of mathematics, especially calculus and analysis.Consider the integral \( \int_{a}^{b} f(x) \; dx \):- If this integral sums to a finite number, we describe \( f(x) \) over \( [a, b] \) as having a convergent integral.- Conversely, if the integral cannot reach a finite sum, it is divergent.For the integral \( \int_{1}^{\infty} e^{-x^2} \; dx \), convergence is determined using the comparison test:- Since \( 0 \leq e^{-x^2} \leq e^{-x} \) for \( x \geq 1 \), and because we know \( \int_{1}^{\infty} e^{-x} \; dx \) converges, so must \( \int_{1}^{\infty} e^{-x^2} \; dx \).
This proves that no matter how far \( x \) extends, the integral accumulates to a bounded number, showing that \( e^{-x^2} \)'s contribution to the area remains finite.