Problem 2 - Calculus and its applications

Question

Verify Property 2 of the definition of a probability density function over the given interval. $$ f(x)=\frac{1}{4} x, \quad[1,3] $$

Step-by-Step Solution

Answer

The function is a valid PDF over [1,3], as the integral of f(x) equals 1.

Step 1: Understand Property 2 of PDF

Property 2 of a probability density function (PDF) states that the total probability over an interval must equal 1. This is checked by integrating the function over the given interval and ensuring it equals 1.

Step 2: Set Up the Integral

To verify this property, we need to compute the integral of the function \( f(x) = \frac{1}{4} x \) over the interval \([1, 3]\). The integral will be: \[ \int_1^3 \frac{1}{4} x \, dx \]

Step 3: Integrate the Function

Calculate the integral of \( \frac{1}{4} x \). The antiderivative of \( \frac{1}{4} x \) is \( \frac{1}{8} x^2 \). Applying the limits from 1 to 3, we have: \[ \left. \frac{1}{8} x^2 \right|_1^3 = \frac{1}{8}(3^2) - \frac{1}{8}(1^2) \]

Step 4: Simplify the Result

Solve the expression from the previous step: \[ \frac{1}{8}(9) - \frac{1}{8}(1) = \frac{9}{8} - \frac{1}{8} = \frac{8}{8} \]

Step 5: Verify the Integral Result

The result from the integral is 1, as it should be for a valid probability density function. This means that the function \( f(x) = \frac{1}{4} x \), over the interval \([1, 3]\), meets the requirement for the total probability to be 1.

Key Concepts

Understanding Integration

Integration is a core concept in calculus that involves finding the area under a curve. In the context of this exercise, we have a function \( f(x) = \frac{1}{4}x \) and the task is to integrate this function over a specific interval, which is [1, 3].
By calculating the area under the function from \( x = 1 \) to \( x = 3 \), integration helps us determine the total accumulation of values across this interval.

• The integration process involves setting up an integral: \( \int_1^3 \frac{1}{4} x \, dx \).
• This represents taking the sum of infinitely small pieces of area under the curve from start to end of the interval.

Essentially, integration allows us to find out how much the function increases from one point to another, acting as a tool to verify properties like total probability in a probability density function (PDF). Understanding how and why to integrate is fundamental in solving calculus-based problems.

The Concept of an Antiderivative

Finding an antiderivative, or an indefinite integral, is about reversing differentiation. An antiderivative of a function is a new function whose derivative is the original function. This means if you take the derivative of the antiderivative, you get back to the original function.
In the exercise, the antiderivative of \( \frac{1}{4}x \) is determined to be \( \frac{1}{8}x^2 \). This is because differentiating \( \frac{1}{8}x^2 \) gives you \( \frac{1}{4}x \).

• We use basic rules of integration to find antiderivatives, such as adding one to the exponent and dividing by the new exponent.
• For example, for a function \( ax^n \), the antiderivative is \( \frac{a}{n+1}x^{n+1} \).

So, understanding antiderivatives is crucial, as they are pivotal in calculating definite integrals, like ensuring our probability density function meets Property 2.

Application of Calculus in Probability

Calculus, and specifically integration, plays a vital role in probability and statistics. When dealing with continuous random variables, we often use calculus to define probability density functions (PDFs). These functions describe the likelihood of different outcomes within a certain range.

• A PDF is used to find the probability that a random variable falls within a specific interval.
• Property 2 of a PDF states that the integral of the PDF over its entire range equals 1, which represents total certainty that the variable must exist somewhere in that interval.

In our exercise, by using calculus, we confirmed that the area under the curve from \( x = 1 \) to \( x = 3 \) is equal to 1, signifying that all possible outcomes within this range have been accounted for.
Understanding how calculus underpins these concepts is key for using mathematics effectively in real-world problems, such as verifying the conditions of a probability density function.