Problem 23 - Calculus Single Variable

Question

In each of Exercises \(23-28,\) a function \(g\) and an interval I are specified. The function \(g\) is nonnegative on \(I .\) Find a number \(c\) such that \(f(x)=\operatorname{cg}(x)\) is a probability density function on \(I .\) $$ g(x)=\sec ^{2}(x) \quad I=[0, \pi / 3] $$

Step-by-Step Solution

Answer

The value of \( c \) is \( \frac{1}{\sqrt{3}} \).

Step 1: Understand the Given Information

We are given a function \( g(x) = \sec^2(x) \) and an interval \( I = [0, \pi/3] \). Our goal is to find a constant \( c \) such that \( f(x) = c \cdot g(x) \) becomes a probability density function (PDF) on the interval \( I \).

Step 2: Determine the Integral Over the Interval

For \( f(x) = c \cdot \sec^2(x) \) to be a PDF, it must satisfy the condition: \( \int_{0}^{\pi/3} f(x) \, dx = 1 \). Plugging in the function, we have \( \int_{0}^{\pi/3} c \cdot \sec^2(x) \, dx = 1 \).

Step 3: Solve the Integral

We solve the integral \( \int_{0}^{\pi/3} \sec^2(x) \, dx \). The antiderivative of \( \sec^2(x) \) is \( \tan(x) \). Thus, \[ \int_{0}^{\pi/3} \sec^2(x) \, dx = \left[ \tan(x) \right]_{0}^{\pi/3} = \tan(\pi/3) - \tan(0) = \sqrt{3} - 0 = \sqrt{3}. \]

Step 4: Determine the Value of c

Substitute the result of the integral back into the equation from Step 2: \( c \cdot \sqrt{3} = 1 \). Solve for \( c \): \[ c = \frac{1}{\sqrt{3}}. \]

Key Concepts

Integral calculus and its role in probability

Integral calculus is a fundamental concept in mathematics that helps us understand the accumulation of quantities. In this exercise, integral calculus is applied to ensure that a given function can represent a probability density function (PDF). A PDF must satisfy the condition that its integral over its entire domain equals 1, ensuring that the probability of all events in a given space sums up to a total certainty of 1.

In our specific exercise, we are dealing with the function \( g(x) = \sec^2(x) \) on the interval \( [0, \pi/3] \). The task is to find a constant \( c \) such that \( c\cdot g(x) \) becomes a PDF. This means we must compute the integral of the function over the interval and use this result to determine \( c \). Once integrated, the condition \[ \int_{0}^{\pi/3} f(x) \, dx = 1 \] must hold true for \( f(x) \) to be a PDF.

What is the Secant function and its application here

The secant function, denoted as \( \sec(x) \), is one of the trigonometric functions that is the reciprocal of the cosine function: \( \sec(x) = \frac{1}{\cos(x)} \). Its squared version, \( \sec^2(x) \), often appears in calculus problems, particularly in integrals related to trigonometric identities and functions.

In this exercise, the function \( g(x) = \sec^2(x) \) is given to you, and you are to find its integral over the interval \( [0, \pi/3] \). The secant and cosine functions are essential in calculus because they allow for the transformation and simplification of equations involving angles, making them easier to solve. Here, the secant function helps define the shape and scaling needed for \( f(x) \) to meet the criteria for being a probability density function on the specified interval.

Understanding the Antiderivative in solving integrals

The antiderivative, also known as the indefinite integral, is a crucial concept in integral calculus. It refers to a function whose derivative yields the original function. In simpler terms, it's the process of reversing differentiation, finding the function \( F(x) \) whose derivative is \( f(x) \).

For this exercise, we find the antiderivative of \( \sec^2(x) \) to solve the integral. The antiderivative of \( \sec^2(x) \) is \( \tan(x) \), a well-known result in trigonometry and calculus. By evaluating this antiderivative from \( 0 \) to \( \pi/3 \), we determine the integral over the interval. The calculation yields \( \left[ \tan(x) \right]_{0}^{\pi/3} = \tan(\pi/3) - \tan(0) = \sqrt{3} \). This simplifies determining the constant \( c \) necessary for the function \( f(x) = c\cdot \sec^2(x) \) to qualify as a PDF by ensuring its integral over \( I \) equals 1.