Problem 19 - Calculus and its applications

Question

Find \(k\) such that each function is a probability density function over the given interval. Then write the probability density function. $$ f(x)=k(2-x), \quad[0,2] $$

Step-by-Step Solution

Answer

The value of \( k \) is \( \frac{1}{2} \); the PDF is \( f(x) = \frac{1}{2}(2-x) \) for \( x \) in \([0, 2]\).

Step 1: Understand the PDF Condition

A probability density function (PDF) must satisfy the condition that the integral of the function over its entire domain is equal to 1. For this problem, we have the function \( f(x) = k(2-x) \) over the interval \([0, 2]\).

Step 2: Set Up the Integral Equation

To ensure \( f(x) \) is a probability density function, we need to set up an equation for the integral of \( f(x) \) over \([0, 2]\) and set it equal to 1. Compute:\[ \int_{0}^{2} k(2-x) \, dx = 1 \]

Step 3: Solve the Integral

Calculate the integral:\[ \int_{0}^{2} k(2-x) \, dx = k \left[ 2x - \frac{x^2}{2} \right]_{0}^{2} \]Substitute the limits:\[ k \left( [2(2) - \frac{2^2}{2}] - [2(0) - \frac{0^2}{2}] \right) = k(4 - 2) = 2k \].Thus, the equation is:\[ 2k = 1 \]

Step 4: Solve for k

Solve the equation for \( k \):\[ k = \frac{1}{2} \].

Step 5: Write the Probability Density Function

With \( k \) found, substitute back to express the probability density function:\[ f(x) = \frac{1}{2}(2-x) \] for \( x \) in \([0, 2]\). This satisfies the condition of a probability density function.

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that focuses on the concept of integration. It is used to find the total accumulation of quantities. In the context of this exercise, it helps us determine the constant \( k \) such that the function \( f(x) = k(2-x) \) satisfies the probability density function (PDF) requirements over the interval \([0, 2]\).
To begin, we set up an integral for the function over the given interval. This integral calculates the total area under the curve of the function, which must equal 1 to be considered a probability density function. The integral equation is set up as follows:

• \( \int_{0}^{2} k(2-x) \, dx = 1 \)

By solving this, we determine how \( k \) adjusts the height of the function to make the total area equal to 1. Calculating the integral and evaluating the limits result in the relationship \( 2k = 1 \), from which we find \( k = \frac{1}{2} \). This demonstrates that integral calculus is crucial in understanding and modeling the behavior of PDFs.

Probability Theory

Probability theory is the branch of mathematics that deals with the likelihood of events occurring. When discussing probability density functions, we're looking at continuous random variables, and how their probabilities are spread across an interval.
In a PDF, the probability of the random variable falling within a specific interval is equivalent to the area under the function's curve over that interval.A valid probability density function must follow some rules:

• The function must take non-negative values.
• The total area under the curve across the entire domain must be exactly 1.

In this exercise, we use these principles to confirm that \( f(x) = \frac{1}{2}(2-x) \) is indeed a valid PDF on the interval \([0, 2]\). By ensuring that the integral, or the total area under \( f(x) \), equals 1, we adhere to the fundamental laws of probability. This ensures that the function accurately represents a real-world continuous random variable.

Mathematical Functions

Mathematical functions describe relationships between sets of numbers. They are crucial in modeling real-world phenomena mathematically.
In this problem, the function \( f(x) = k(2-x) \) represents a linear relationship. Its graph is a straight line over the interval \([0, 2]\). The slope of this line is determined by the coefficient of \( x \), which in our probability context is influenced by \( k \) to meet PDF criteria.Key properties of functions in this context include:

• Continuity: For PDFs, the function should be continuous to avoid "gaps" in probability.
• Boundedness: Specifically for PDFs, the integration (total area under the curve) must be bounded to 1.

In our exercise, solving for \( k \) involved understanding these mathematical properties of the function to ensure it meets criteria for probability density functions. The attention to how mathematics structures such a simple function demonstrates the power of mathematical functions in probability theory and applied mathematics.