Problem 3 - Calculus for Biology and Medicine

Question

Determine \(c\) such that $$ f(x)=\frac{c}{1+x^{2}}, \quad x \in \mathbf{R} $$ is a density function.

Step-by-Step Solution

Answer

The constant \( c \) is \( \frac{1}{\pi} \).

Step 1: Understand the Properties of a Density Function

For a function to be a probability density function, it must satisfy two conditions: it must be non-negative for all values of its domain, and the integral over the entire real line must equal 1.

Step 2: Check Non-Negative Condition

The function given, \( f(x) = \frac{c}{1+x^2} \), is non-negative for all values of \( x \) since the denominator \( 1 + x^2 \) is always positive. Hence, \( f(x) \geq 0 \) for all \( x \in \mathbf{R} \).

Step 3: Set Up the Integral Condition

For \( f(x) \) to be a density function, the integral over its domain must equal 1: \[ \int_{-\infty}^{\infty} \frac{c}{1+x^2} \, dx = 1. \]

Step 4: Calculate the Integral

The integral \( \int \frac{1}{1+x^2} \, dx \) is a standard one, equal to \( \tan^{-1}(x) \). Therefore, \[ \int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx = \left[ \tan^{-1}(x) \right]_{-\infty}^{\infty} = \frac{\pi}{2} - \left(-\frac{\pi}{2} \right) = \pi. \]

Step 5: Solve for \(c\)

Using the integral condition, \[ c \times \pi = 1. \]Solve for \( c \) to find \[ c = \frac{1}{\pi}. \]

Key Concepts

Probability Density Function

A probability density function (PDF) is an essential concept in probability and statistics. It describes the likelihood of a continuous random variable occurring within a particular range of values. For a function to qualify as a PDF, it must fulfill two primary requirements:

• Non-negativity: The function must not take negative values. For any input from the domain, the output must be zero or positive.
• Normalization: The integral of the probability density function over the entire domain must equal one. This ensures that the total probability distributed over all possible values of the variable sums up to a complete certainty.

In our exercise, the function given is \( f(x) = \frac{c}{1+x^2} \). The denominator \(1 + x^2\) remains positive for all real numbers \(x\), thereby ensuring that the function is non-negative. Additionally, by ensuring that the integral over the whole real line equals one, we can determine a specific value for \(c\) that makes \(f(x)\) a legitimate probability density function.

Integral Calculus

Integral calculus plays a crucial role in determining the area under curves, which is a fundamental aspect when dealing with probability density functions. The integral of a function indicates the accumulated value or total area under the curve of that function.
For PDFs, integral calculus is used to ensure that the total area under the curve equals one. This is achieved through the process known as normalization. In our context, we need to integrate the function \( f(x) = \frac{c}{1+x^2} \) over the entire set of real numbers from \(-\infty\) to \(\infty\).
The integral\[ \int_{-\infty}^{\infty} \frac{c}{1+x^2} \, dx = 1 \]is solved by separating the constant \(c\) and addressing the remaining integral \( \int \frac{1}{1+x^2} \, dx \), which is known to equal \( \pi \). By solving the equation \( c \times \pi = 1 \), we find that \( c = \frac{1}{\pi} \), satisfying the normalization condition.

Arctangent Function

The arctangent function, often denoted as \( \tan^{-1}(x) \), arises in calculus as the inverse of the tangent function. It maps a given value back to an angle whose tangent is the number.
In the context of integral calculus, the derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1 + x^2} \). Thus, when integrating \( \frac{1}{1 + x^2} \), the result is \( \tan^{-1}(x) + C \), where \(C\) is the constant of integration. This characteristic makes the arctangent function pivotal in evaluating certain integrals that appear with this form.
For our problem, the integral \( \int \frac{1}{1+x^2} \, dx \) simplifies to finding \( \tan^{-1}(x) \) over the limits \(-\infty\) to \(\infty\). The evaluation of limits yields \( \frac{\pi}{2} - \left(-\frac{\pi}{2} \right) = \pi \). Recognizing this integral's result is crucial for solving the original exercise and helps derive the needed value of \(c\) to normalize the probability density function.