Problem 65 - Applied Calculus

Question

True or False: If \(f(x)\) is continuous, nonnegative, and \(\lim _{x \rightarrow \infty} f(x)=0,\) then \(\int_{1}^{\infty} f(x) d x\) converges.

Step-by-Step Solution

Answer

False, a counterexample (e.g., \(f(x) = \frac{1}{x}\)) shows divergence.

Step 1: Understanding the Problem

We need to determine if the statement is true or false: "If \(f(x)\) is continuous, nonnegative, and \(\lim _{x \rightarrow \infty} f(x)=0,\) then \(\int_{1}^{\infty} f(x) d x\) converges." This involves analyzing the behavior of an improper integral based on the behavior of a function.

Step 2: Review Key Concepts

Recall that for an improper integral \(\int_{1}^{\infty} f(x) \, dx\) to converge, the function \(f(x)\) must approach zero fast enough as \(x\) approaches infinity. A limit at infinity of zero alone does not guarantee convergence.

Step 3: Counterexample Exploration

Consider the function \(f(x) = \frac{1}{x}\). This function is continuous, nonnegative, and \(\lim_{x \rightarrow \infty} \frac{1}{x} = 0\). However, the integral \(\int_{1}^{\infty} \frac{1}{x} \, dx\) is known to diverge.

Step 4: Evaluate the Counterexample

Evaluate the integral \(\int_{1}^{\infty} \frac{1}{x} \, dx\). We find \(\int_{1}^{t} \frac{1}{x} \, dx = [\ln(x)]_1^t = \ln(t) - \ln(1) = \ln(t)\). As \(t\) approaches infinity, \(\ln(t)\) goes to infinity, demonstrating the integral diverges.

Step 5: Conclusion

Since \(f(x) = \frac{1}{x}\) meets all conditions but results in a divergent integral, the statement is false. The fact that \(f(x)\) approaches 0 is not sufficient for ensuring the convergence of the integral.

Key Concepts

continuity

Continuity is a fundamental concept in calculus and is essential when analyzing functions and their integrals. A function is continuous if its graph has no breaks, jumps, or holes. In formal terms, for a function \( f(x) \) to be continuous at a point \( x = a \), it must satisfy three conditions:

• The function \( f(x) \) is defined at \( x = a \).
• The limit of \( f(x) \) as \( x \) approaches \( a \) exists.
• The limit of \( f(x) \) as \( x \) approaches \( a \) is equal to \( f(a) \).

In the context of improper integrals, especially when dealing with limits at infinity, continuity ensures that the function behaves predictably over the interval of integration. For the given problem, the continuity of \( f(x) \) assures us that \( f(x) \) doesn't suddenly change behavior in a way that might complicate the analysis of the integral.

divergence

Divergence occurs when an integral does not settle at a finite value as its bounds extend to infinity. Even if a function \( f(x) \) is continuous, nonnegative, and approaches zero as \( x \) approaches infinity, this does not necessarily mean the integral converges. In the exercise, we explore

• a counterexample with \( f(x) = \frac{1}{x} \).
• This function meets all the initial conditions.

However, evaluating the integral \( \int_{1}^{\infty} \frac{1}{x} \, dx \) reveals it diverges. This example illustrates the critical distinction between the behavior of a function and its integral. While \( \lim_{x \to \infty} f(x) = 0 \) provides information about the function's end behavior, it doesn't ensure that the accumulated area under the curve approaches a finite amount. Understanding divergence helps students recognize the limitations of limits alone in predicting the behavior of improper integrals.

limits at infinity

Limits at infinity focus on understanding the behavior of a function as the input \( x \) grows larger and larger beyond all bounds. When we say \( \lim_{x \to \infty} f(x) = L \), it means as \( x \) becomes very large, \( f(x) \) gets arbitrarily close to \( L \). In improper integrals, these limits help us explore if the area under a curve extends infinitely or settles at a finite value.

• In the exercise, \( \lim_{x \to \infty} f(x) = 0 \) suggests that the function value decreases continuously toward zero.
• However, the critical mistake is assuming this always results in a convergent integral.

The example with \( \int_{1}^{\infty} \frac{1}{x} \, dx \) illustrates how limits alone cannot always determine the convergence of an integral. Understanding limits at infinity and their role in integrals helps distinguish between the diminishing values of a function and the potential infinite sum of those values when considered over an infinite interval.