Problem 3 - Calculus and its applications

Question

Determine whether each improper integral is convergent or divergent, and find its value if it is convergent. $$ \int_{3}^{\infty} \frac{d x}{x} $$

Step-by-Step Solution

Answer

The integral diverges as it evaluates to infinity.

Step 1: Set up the improper integral

We start by expressing the given improper integral as a limit. The integral is from 3 to infinity, so we write it as:\[ \int_{3}^{ ext{b}} \frac{1}{x} \, dx \] where we take the limit as \( b \to \infty \).

Step 2: Integrate the integrand

Integrate the function \( \frac{1}{x} \). The antiderivative of \( \frac{1}{x} \) is \( \ln |x| \). So the integral becomes:\[ \int_{3}^{b} \frac{1}{x} \, dx = \left[ \ln |x| \right]_{3}^{b} \]

Step 3: Apply the limits

Now, apply the limits of integration from 3 to \( b \):\[ \left[ \ln |x| \right]_{3}^{b} = \ln(b) - \ln(3) \]

Step 4: Evaluate the limit as \( b \to \infty \)

Evaluate the limit of the expression as \( b \rightarrow \infty \):\[ \lim_{b \to \infty} (\ln(b) - \ln(3)) = \ln(\infty) - \ln(3) = \infty \]Since \( \ln(b) \rightarrow \infty \), the integral diverges.

Key Concepts

Integration Techniques

Improper integrals often require special techniques to evaluate due to their infinite limits or discontinuous integrands. These integrals are not standard because they extend to infinity or over discontinuities. To solve them, we typically express them as a limit problem. Consider the improper integral \( \int_{3}^{\infty} \frac{1}{x} \, dx \). Because the upper limit is infinite, we set it up using a limit:

• Replace infinity with a variable, like \( b \), so the integral becomes \( \int_{3}^{b} \frac{1}{x} \, dx \).
• Evaluate this integral in terms of the variable \( b \).
• Finally, take the limit as \( b \to \infty \).

This approach transforms the improper integral into a more manageable form, letting us apply standard integration techniques like finding antiderivatives. Here, the antiderivative of \( \frac{1}{x} \) is \( \ln|x| \), which simplifies the integration process.

Limits and Infinity

The concept of limits is fundamental to calculus and is especially vital when dealing with improper integrals that reach toward infinity. When examining the limit of a function as it approaches infinity, we're essentially exploring how the function behaves as it goes toward an extreme point. For the improper integral \( \int_{3}^{\infty} \frac{1}{x} \, dx \), after finding the antiderivative \( \ln|x| \), we evaluate it between \( 3 \) and \( b \):

• The result is \( \ln(b) - \ln(3) \).
• As \( b \) goes to infinity, \( \ln(b) \) increases without bound.
• This implies that the expression \( \ln(b) - \ln(3) \) tends towards infinity.

Since \( \ln(b) \rightarrow \infty \), the limit doesn't settle at a finite value. This indicates that the integral diverges and does not converge to a specific number, which is a key idea in the analysis of convergence and divergence.

Convergence and Divergence

Convergence and divergence are central concepts determining the behavior of improper integrals. An integral is convergent if, as you evaluate it over an infinite limit, it approaches a finite value. Conversely, it diverges if it becomes infinite or doesn't settle down to a single value. For the integral \( \int_{3}^{\infty} \frac{1}{x} \, dx \):

• We evaluated the integral and found that \( \ln(b) - \ln(3) \) goes to infinity as \( b \to \infty \).
• This means the integral doesn't approach a finite limit, indicating divergence.

Understanding whether an integral converges or diverges is important, as convergent integrals can yield useful real-world sums, while divergent ones signify processes that grow without bound. When dealing with such problems, carefully analyze the limits and behavior of your functions to make these determinations.