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The exponential distribution is a continuous probability distribution often used to model the time between events in a Poisson process. This particular problem involves the age of a cell, which follows an exponential distribution due to its constant rate of division every 10 days.

For an exponential distribution, the probability density function (PDF) is given by:
\[ f(x) = \lambda e^{-\lambda x}, \ x \geq 0 \]
where \( \lambda \) is the rate parameter. In our case, the PDF is slightly modified to:
\[ f(x) = 2k e^{-kx}, 0 \leq x \leq 10 \]
with \( k = \frac{\ln 2}{10} \). The base of the PDF ensures that the total probability for the range [0, 10] equals 1.

Understanding how the exponential distribution works helps in solving problems involving the times between events. It is characterized by a lack of memory property, meaning the probability of an event occurring in the next interval is independent of how much time has already elapsed.

Integration is a fundamental concept in calculus that is used to calculate areas under curves, among other things. In the context of probability, integration helps us determine the probability of a continuous random variable falling within a certain range.

In this exercise, we are asked to find the probability that a cell is at most 5 days old. This involves integrating the given PDF from 0 to 5:
\[ P(X \leq 5) = \int_{0}^{5} 2k e^{-kx} \, dx \]

To solve this, we substitute \( k = \frac{\ln 2}{10} \) into the integral:
\[ P(X \leq 5) = 2 \left( \frac{\ln 2}{10} \right) \int_{0}^{5} e^{- \left(\frac{\ln 2}{10} \right) x} \, dx \]
Recognize that this is an integral of an exponential function, which has a standard solution:
\[ \int e^{-ax} \, dx = -\frac{1}{a} e^{-ax} \]
Applying this, we integrate, substitute the limits, and simplify to find the desired probability.

Probability calculations involve finding the likelihood that a random variable falls within a specific range. Here, we want to calculate the probability that the cell's age is at most 5 days.

Using the integral we set up earlier, we find:
\[ P(X \leq 5) = 2 \left( \frac{\ln 2}{10} \right) \left[ - \frac{10}{\ln 2} \left( e^{- \left( \frac{\ln 2}{10} \right) 5} - e^{- \left( \frac{\ln 2}{10} \right) 0} \right) \right] \]
Evaluating this at the bounds and simplifying, we get:
\[ P(X \leq 5) = 2 \left( \frac{\sqrt{2} - 1}{\ln 2} \right) \]
Numerically, this is approximately 0.58496.

This calculation shows that there is about a 58.5% chance that a randomly selected cell in this population is at most 5 days old. Understanding how to perform these calculations is crucial for solving many probability-related problems.