91Ó°ÊÓ

Integration by parts is a technique used to solve complex integral problems, and it is derived from the product rule of differentiation. This method is especially useful when dealing with integrals of a product of functions. The formula for integration by parts is given by:\[ \int u \, dv = uv - \int v \, du \]where:

• \( u \) is a function of \( x \)
• \( dv \) is another function of \( x \) times \( dx \)
• \( du \) is the derivative of \( u \)
• \( v \) is the antiderivative of \( dv \)

In the given problem, this method helps solve the integral \( \int x e^{-x/2} \, dx \). First, identify \( u = x \) and \( dv = e^{-x/2} \, dx \). Differentiating \( u \), we get \( du = dx \). Integrating \( dv \), we find \( v = -2e^{-x/2} \). This substitution simplifies the integral allowing us to proceed with the calculation. The process of integration by parts helps break down otherwise difficult integrals into simpler components.

The normalization condition refers to the principle that the total probability of all possible outcomes must equal 1. This ensures that a probability density function (PDF) represents a valid probability distribution. For a continuous random variable with density function \( f(x) \), the normalization condition requires:\[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \]In practice, this often simplifies to integrating over the range where \( f(x) \) is non-zero. In our exercise, \( f(x) = Cx e^{-x/2} \) is non-zero for \( x > 0 \). Therefore, the integral:\[ \int_0^\infty Cx e^{-x/2} \, dx = 1 \]must hold. Using techniques such as integration by parts, we evaluate this integral. Solving it allows us to find the constant \( C \) that ensures the PDF adds up to 1. In this case, solving gives \( C = \frac{1}{4} \), ensuring the PDF is normalized.

The exponential distribution is often used to model the time until an event occurs, making it ideal for analyzing the lifespan of components or systems. It is characterized by a PDF that decreases exponentially. The general form of the PDF for an exponential distribution with rate parameter \( \lambda \) is:\[ f(x) = \lambda e^{-\lambda x} \text{ for } x \geq 0 \]In this exercise, the distribution takes a modified exponential form \( f(x) = \frac{1}{4}x e^{-x/2} \) due to the presence of the \( x \) factor, making it related to what is known as a "gamma distribution" rather than a simple exponential. The rate of decrease and the shape are essential in determining how likely different time frames are.This distribution explains the variability of time until the failure of the system, represented by \( X \). Handling such models helps in practical situations like reliability testing, where understanding the probability of a system exceeding specific operational times (like the 5 months in the problem) is crucial.