Problem 66 - Applied Calculus

Question

True or False: If \(f(x)\) is continuous, nonnegative, and \(\lim _{x \rightarrow \infty} f(x)=2,\) then \(\int_{1}^{\infty} f(x) d x\) diverges.

Step-by-Step Solution

Answer

True: \( \int_{1}^{\infty} f(x) \ dx \) diverges.

Step 1: Understanding the Problem

We need to determine whether the integral \( \int_{1}^{\infty} f(x) \, dx \) diverges under the condition that \( f(x) \) is continuous, nonnegative, and it approaches 2 as \( x \to \infty \).

Step 2: Limiting Behavior of f(x)

Since \( \lim_{x \to \infty} f(x) = 2 \), this implies \( f(x) \) approaches 2 as \( x \) becomes very large. \( f(x) \) stays close to 2 after some point. This means on an interval from some large \( a \) to \( \infty \), \( f(x) \) is approximately 2.

Step 3: Integral Comparison

Since \( f(x) \geq 0 \) and approaches 2, we can compare \( \int_{1}^{\infty} f(x) \, dx \) with \( \int_{1}^{\infty} 2 \, dx \). The integral \( \int_{1}^{\infty} 2 \, dx = \lim_{b \to \infty} [2x]_1^b = \lim_{b \to \infty} (2b - 2) \), which clearly diverges.

Step 4: Conclusion

Since \( \int_{1}^{\infty} 2 \, dx \) diverges and \( f(x) \) behaves similarly to 2, the integral \( \int_{1}^{\infty} f(x) \, dx \) also diverges. This comparison confirms that despite \( f(x) \) approaching a finite number, the integral itself does not converge.

Key Concepts

Continuous Functions

Functions that are continuous play a fundamental role in calculus and analysis. A continuous function is a type of function where small changes in the input result in small changes in the output. More formally, a function \( f(x) \) is continuous at a point \( x = c \) if and only if the following three conditions are met:

• \( f(c) \) is defined.
• The limit \( \lim_{x \to c} f(x) \) exists.
• The limit \( \lim_{x \to c} f(x) = f(c) \).

This means there are no sudden jumps, breaks, or oscillations in the graph of the function over its domain. When applied to improper integrals, continuity ensures that the function can be integrated over a given interval. Moreover, if a function is continuous everywhere in its domain, it greatly simplifies the process of calculating integrals since there are no discontinuities to account for during integration. Thus, analyzing an integral of a continuous function becomes much more straightforward.

Limit Behavior

The behavior of a function as its input approaches a certain value, especially infinity, is crucial in understanding long-term trends and integrals. The limit behavior is detailed as follows: if \( \lim_{x \to \infty} f(x) = L \), it implies that as \( x \) becomes very large, \( f(x) \) settles or tends to remain close to \( L \).
In the context of the original problem, knowing \( \lim_{x \to \infty} f(x) = 2 \) informs us that the function does not taper off to zero as \( x \to \infty \).
Instead, it approaches the constant value 2, which suggests that the total area under the curve from 1 to infinity will continue to grow without bound. Hence, this implies that the integral \( \int_{1}^{\infty} f(x) \, dx \) is more likely to diverge, rather than converge, since the function approaches a non-zero constant at infinity.

Integral Comparison Test

The Integral Comparison Test is a powerful tool for determining the convergence or divergence of improper integrals by comparing them with integrals whose behavior is known. The test works by finding a function \( g(x) \) such that:

• For all \( x \ge c \), \( f(x) \ge g(x) \ge 0 \) where \( f(x) \le g(x) \ge 0 \) for convergence studies.
• If \( \int_c^{\infty} g(x) \, dx \) converges, then \( \int_c^{\infty} f(x) \, dx \) also converges, and vice versa if it diverges.

By comparing the given integral \( \int_{1}^{\infty} f(x) \, dx \) with \( \int_{1}^{\infty} 2 \, dx \), we know that the latter diverges, as it evaluates to \( \lim_{b \to \infty} (2b - 2) \), which tends towards infinity.
This means the original integral must also diverge, confirming that the integral of a function mirroring the behavior of a non-zero constant will not result in convergence, due to the ever-increasing area under the curve as \( x \) extends to infinity.