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When dealing with probability density functions (PDFs), integration is key. A PDF, like the one we're discussing, describes a continuous probability distribution. To determine if a function is a PDF over a specific interval, we must ensure that the total area under the curve equals 1 within that interval. This is where integration comes into play, as it allows us to calculate the area under the curve. For the function given,

• We need to integrate \( \frac{k}{x} \) over the interval \( [1, 2] \)
• This integral calculates the weighted area between the curve and the x-axis over the specified interval.

Evaluating the integral ensures that the function correctly models a probability distribution over the given domain. The integral of \( \frac{k}{x} \) results in \( k \cdot \left[ \ln(x) \right] \), showcasing the beauty of linking integration with logarithmic functions.

The natural logarithm, represented as \( \ln \), is a mathematical function that provides insights into exponential growth and functions. In the context of our problem, the natural logarithm arises from the integral of \( \frac{1}{x} \). Why is this important?

• The integral of \( \frac{1}{x} \) is \( \ln|x| \), indicating that the natural logarithm plays a critical role in solving such integrals.
• Natural logs help convert multiplicative relationships into additive ones, simplifying calculations.

In our example, by evaluating \( \ln(x) \) from 1 to 2, we find \( \ln(2) - \ln(1) \). Knowing that \( \ln(1) = 0 \), this simplifies to \( \ln(2) \), which is crucial for defining the PDF correctly. This illustrates the usefulness of natural logarithms in mathematical problem-solving.

Continuous probability deals with outcomes that can take any value within a specified range. Unlike discrete probability, where outcomes are distinct and countable, continuous probability involves distributions over intervals. A probability density function (PDF) models this continuous nature when finding probabilities over a range. In our scenario,

• The PDF, \( f(x) = \frac{k}{x} \), represents a continuous distribution over the interval \( [1, 2] \).
• To qualify as a PDF, the integral over this interval must equate to 1.

This condition ensures that the total probability covered by the interval accounts for all possible outcomes in that range. Understanding this is essential for correctly interpreting continuous variables and their resulting probabilities. When we solve for \( k = \frac{1}{\ln(2)} \), we ensure that the function satisfies all requirements of a PDF, effectively capturing continuous probabilities over our selected interval.