Problem 6 - Calculus and its applications

Question

Verify Property 2 of the definition of a probability density function over the given interval. $$ f(x)=\frac{3}{64} x^{2}, \quad[0,4] $$

Step-by-Step Solution

Answer

The integral of the function over [0, 4] is 1, verifying the PDF property.

Step 1: Understand Property 2 of a Probability Density Function (PDF)

Property 2 of a PDF states that the total area under the probability density function over its entire range must be equal to 1. This is expressed as: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] For our function, we need to integrate \( f(x) = \frac{3}{64} x^{2} \) over the interval from 0 to 4 and ensure the result is 1.

Step 2: Set Up the Integral for the Given Interval

We need to calculate the following integral to verify Property 2 over the interval \([0, 4]\):\[ \int_{0}^{4} \frac{3}{64} x^{2} \, dx \] This requires finding the antiderivative of the function \( f(x) = \frac{3}{64} x^{2} \).

Step 3: Find the Antiderivative of \( f(x) = \frac{3}{64} x^{2} \)

The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \). Therefore, integrating \( \frac{3}{64} x^2 \) gives:\[ \int \frac{3}{64} x^{2} \, dx = \frac{3}{64} \cdot \frac{x^3}{3} = \frac{x^3}{64} \] This function will be used to evaluate the integral over [0, 4].

Step 4: Evaluate the Definite Integral from 0 to 4

Now, apply the limits of integration to the antiderivative:\[ \left[ \frac{x^3}{64} \right]_{0}^{4} = \frac{4^3}{64} - \frac{0^3}{64} \]Calculate the expression:\[ \frac{64}{64} - 0 = 1 \]

Step 5: Confirm the Result

The result of the integral is 1, which confirms Property 2 of a probability density function. Thus, the total area under the curve \( f(x) = \frac{3}{64} x^{2} \) over the interval [0, 4] is indeed equal to 1.

Key Concepts

Antiderivative

An antiderivative is a fundamental concept in calculus. It is a function that "undoes" differentiation, commonly referred to as the indefinite integral. When you differentiate a function and then take an antiderivative, you return to your original function, up to a constant. In essence, the antiderivative of a function is a new function whose derivative is the original function. Let's look at the antiderivative for the function given in the exercise:

• The function is given as \( f(x) = \frac{3}{64} x^{2} \).
• We seek an antiderivative \( F(x) \) such that when differentiated, \( F'(x) = f(x) \).
• For \( x^2 \), the antiderivative is \( \frac{x^3}{3} \), so by multiplying through by \( \frac{3}{64} \), the antiderivative becomes \( F(x) = \frac{x^3}{64} \).

The purpose of finding an antiderivative in this context is to prepare us for evaluating a definite integral. We'll use \( F(x) \) in the definite integration process over the interval \([0, 4]\).

Definite Integral

A definite integral helps to find the area under a curve over a particular interval, specifically between two points on the x-axis. For the exercise, we use it to find the total area under \( f(x) \) from 0 to 4 and verify a property of probability density functions. The definite integral of \( f(x) = \frac{3}{64} x^2 \) from 0 to 4 is calculated as follows:

• Begin by finding the antiderivative, which we identified as \( \frac{x^3}{64} \).
• Evaluate this antiderivative at the bounds of our interval: \([0, 4]\).

With the formula for a definite integral, once you have \( F(x) = \frac{x^3}{64} \), you compute: \[ \left[ \frac{x^3}{64} \right]_0^4 = \frac{4^3}{64} - \frac{0^3}{64} \] This gives us \( \frac{64}{64} - 0 = 1 \), demonstrating that the total area under the curve within this interval is 1. This confirms that the function satisfies an essential property of probability density functions.

Property of PDF

A probability density function (PDF) describes the likelihood of a random variable within a particular range. It has specific properties that must be satisfied: one of which is that the total area under the curve of a PDF over its defined interval must equal 1. This property ensures the function is properly normalized and can represent probabilities. In our exercise, this property is verified by:

• Integrating the function \( f(x) = \frac{3}{64} x^{2} \) over the interval \([0, 4]\).
• Computing the area using the definite integral. This requires the antiderivative, which converts the function describing the slope back into a function describing area under the curve.

Once we've carried out these steps and found the integral equals to 1, we affirm that \( f(x) \) is a valid probability density function over the specified interval. This validation is crucial as it allows \( f(x) \) to be used in statistical applications involving continuous data.