Problem 9 - Calculus and its applications

Question

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

Step-by-Step Solution

Answer

The function \( f(x) = \frac{1}{3}x^2 \) satisfies Property 2 as \( \int_{-2}^{1} f(x) \, dx = 1 \).

Step 1: Understand Property 2 of a Probability Density Function

Property 2 states that the total probability over the given interval must equal 1. Mathematically, it involves integrating the function over the interval and checking whether the result is 1: \( \int_{-2}^{1} f(x) \, dx = 1 \).

Step 2: Set Up the Integral

To verify Property 2, we need to integrate the given function \( f(x) = \frac{1}{3}x^2 \) over the interval \([-2, 1]\). The integral to solve is: \[ \int_{-2}^{1} \frac{1}{3}x^2 \, dx \].

Step 3: Calculate the Indefinite Integral

First, find the indefinite integral of \( \frac{1}{3}x^2 \). This becomes \( \frac{1}{3} \int x^2 \, dx = \frac{1}{3} \times \frac{x^3}{3} = \frac{x^3}{9} \).

Step 4: Evaluate the Definite Integral

Evaluate the definite integral \( \int_{-2}^{1} \frac{x^3}{9} \, dx \) by finding \( \frac{1^3}{9} \) and \( \frac{(-2)^3}{9} \). Compute: \( \frac{1}{9} - \frac{-8}{9} \). This simplifies to \( \frac{1 + 8}{9} = \frac{9}{9} = 1 \).

Step 5: Verify Property

Since the integral evaluates to 1, Property 2 holds true for \( f(x) = \frac{1}{3}x^2 \) over \([-2, 1]\). Thus, \( f(x) \) is a valid probability density function over the specified interval.

Key Concepts

Integral Calculus

Integral calculus is a fundamental concept in mathematics that focuses on finding the total accumulation of quantities, such as areas and volumes. It's one of the two principal branches of calculus, with the other being differential calculus. When you hear the term "integral," it generally refers to the process of integration, which is an operation that combines infinitely small pieces to calculate a whole.

There are two main types of integrals: definite and indefinite. Both serve different purposes but are interconnected in the toolkit of calculus. Integration finds significant applications in various scientific and engineering fields by allowing us to solve complex problems related to geometry, physics, and probability. When working with probability density functions, integration is crucial as it helps determine the total probability over an interval.

• Indefinite integrals: Represent the general form of an antiderivative.
• Definite integrals: Provide the actual total value over an interval.

Definite Integral

In the realm of integral calculus, the definite integral is a powerful tool for calculating the total value of a function over a specified interval. It is represented by the integral sign with limits, which usually denotes the interval over which we are integrating.

For example, in our exercise, we calculated the definite integral of \( f(x) = \frac{1}{3}x^2 \) over the interval \([-2, 1]\). When we solved this, the result was \(1\), fulfilling one of the key properties needed for a function to qualify as a valid probability density function.

A definite integral accounts for the accumulation of the function's area under its curve in a specified range. To compute this, one must first determine the antiderivative, then apply the fundamental theorem of calculus:

• The key steps involve finding the antiderivative of the function.
• Substituting the upper and lower limits into the antiderivative.
• Subtracting these values to find the total area.

By mastering the definite integral, you gain the ability to evaluate quantities like probabilities, energies, and areas effectively.

Indefinite Integral

An indefinite integral, often termed as an antiderivative, is a function that represents a family of all possible antiderivatives given a function. It lacks the limits of integration, unlike the definite integral, thereby providing a general formula rather than a numeric value.

The process of finding an indefinite integral involves determining the original function from its derivative. This is often represented by adding an arbitrary constant "C" to the solution because differentiation of functions only determines their form up to an additive constant. In our exercise, we found the indefinite integral of \( \frac{1}{3}x^2 \), which was \( \frac{x^3}{9} \).

• Indefinite integrals provide the general solution to a derivative.
• The addition of the constant "C" accounts for the fact that multiple functions can have the same derivative.
• Used as the first step towards finding definite integrals.

Ultimately, mastering indefinite integrals equips you with the knowledge to reverse operations of differentiation, offering insights into the cumulative effects of functions.