Question

Finding an Area Determine whether the area between the graph of \(f(x)=\frac{1}{x}\) and the \(x\) -axis over the interval \([1,+\infty)\) is finite or infinite.

Short answer

The area is infinite.

Step-by-step solution

Understand the Function and Interval

We need to analyze the function \( f(x) = \frac{1}{x} \) over the interval \([1, \infty)\). The area under a curve represented by a function \(f(x)\) from \(x = a\) to \(x = b\) is given by the definite integral \( \int_a^b f(x) \, dx \).

Set Up the Improper Integral

Since the interval is infinite, we use an improper integral to find the area. We write this as the limit of a proper integral: \[ \int_{1}^{\infty} \frac{1}{x} \, dx = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x} \, dx. \]

Evaluate the Definite Integral

Evaluate the definite integral \( \int_{1}^{b} \frac{1}{x} \, dx \). The antiderivative of \( \frac{1}{x} \) is \( \ln|x| \). So, \[ \int_{1}^{b} \frac{1}{x} \, dx = [\ln|x|]_1^b = \ln|b| - \ln|1|. \]

Calculate the Limit

Substitute the result from Step 3 into the limit:\[ \lim_{b \to \infty} (\ln|b| - \ln|1|). \]This simplifies to:\[ \lim_{b \to \infty} \ln b. \]

Determine the Convergence

The expression \( \ln b \) tends to \( +\infty \) as \( b \to \infty \). Therefore, the limit is infinite, indicating the integral diverges.

Key concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus. They represent the area under the curve of a function over a closed interval. This means that if you have a function, like in this case, \( f(x) = \frac{1}{x} \), you can visualize the area between its curve and the x-axis on a specific range. In simpler terms, it’s like trying to figure out how much surface lies beneath the curve from one point to another on the horizontal axis.
For a definite integral, you have the starting point \( a \) and the endpoint \( b \). You write it out as \( \int_a^b f(x) \, dx \), which asks, "What's the area under \( f(x) \) between \( a \) and \( b \)?"
Key points to remember about definite integrals:

• You need both a lower limit \( a \) and an upper limit \( b \).
• The integral calculates a fixed number representing the area.
• If the interval is finite and the function is well-behaved (continuous), the integral gives a finite result.

In this problem, we start by setting up a definite integral, but due to the nature of the interval (which isn’t finite), we move towards working with an improper integral.

Limits

Limits help us understand behavior as we approach a certain point—in this case, as we move towards infinity. When dealing with integrals over infinite intervals, you're dealing with what's called an "improper integral."
The problem here uses a limit to evaluate an integral that goes to infinity. Specifically, it computes \( \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x} \, dx \). This limit indicates that instead of calculating up to infinity directly, we consider what happens when we extend the interval further and further out.
This involves two main considerations:

• You're looking at what happens towards the very end of the interval—what does the function's behavior add up to?
• This often simplifies to identifying if the area becomes stable, a phenomenon known as convergence, or becomes unrestrained, hence diverging.

In using limits for improper integrals, you substitute increasingly large values for \( b \), examining if the integral settles on a particular value as \( b \) tends to infinity.

Convergence and Divergence

Convergence and divergence are terms used to describe the behavior of series or integrals as they extend towards infinity. Understanding these terms helps in deciding whether an integral results in a finite area or an "unbounded" one.
For the integral \( \int_{1}^{\infty} \frac{1}{x} \, dx \), we want to see if the area from 1 to infinity is finite (convergent) or infinite (divergent).
The integral was converted into the expression \( \lim_{b \to \infty} \ln b \), as seen in the solution steps. At this point, knowing properties of the natural logarithm, \( \ln b \) grows without bound as \( b \to \infty \).
What this means for convergence and divergence:

• If the limit approaches a specific number, it's convergent, meaning the integral represents finite area.
• If the limit continues to grow without bound, it's divergent, indicating an infinite area under the curve.

For this exercise, since \( \ln b \) heads to infinity, the integral is divergent, confirming that the area under \( f(x) = \frac{1}{x} \) from 1 to infinity is indeed infinite.