Problem 11 - Calculus and its applications

Question

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

Step-by-Step Solution

Answer

The integral of the function over \([0, \infty)\) equals 1, verifying it's a valid probability density function.

Step 1: Identify the Property to Verify

Property 2 of a probability density function states that the integral of the function over its entire range must equal 1. For this problem, this means: \[ \int_{0}^{\infty} f(x) \, dx = 1. \]

Step 2: Write Down the Integral

To verify this property, we need to compute the definite integral of the function \( f(x) = 3e^{-3x} \) over the interval \([0, \infty)\). The integral is: \[ \int_{0}^{\infty} 3e^{-3x} \, dx. \]

Step 3: Perform a Substitution

Use the substitution \( u = -3x \), which implies \( du = -3 \, dx \) or \( dx = -\frac{1}{3} \, du \). Moreover, when \( x = 0 \), \( u = 0 \), and when \( x = \infty \), \( u = -\infty \). So, the new integral becomes: \[ \int_{0}^{-\infty} 3 e^{u} \left(-\frac{1}{3}\right) \, du = \int_{-\infty}^{0} -e^{u} \, du. \]

Step 4: Evaluate the Integral

Evaluate the integral \( \int_{-\infty}^{0} -e^{u} \, du \). This can be simplified to: \[ \int_{-\infty}^{0} e^{u} \, du = \left[ e^{u} \right]_{-\infty}^{0}. \]

Step 5: Calculate the Definite Integral

Substitute the limits of integration into the result: \( \left[ e^{u} \right]_{-\infty}^{0} = e^{0} - \lim_{u \to -\infty} e^{u} = 1 - 0 = 1 \).

Step 6: Verify the Result

Since the integral evaluates to 1, which matches the requirement of Property 2 of a probability density function, we have verified that \( f(x) = 3e^{-3x} \) is indeed a valid probability density function over the interval \([0, \infty)\).

Key Concepts

Definite Integral

A definite integral of a function quantifies the area under the curve of that function over a specific interval. In the context of probability density functions, the definite integral is essential to verify certain properties, such as ensuring the total probability sums up to 1. Think of the definite integral as a tool that not only computes total quantities but also evaluates consistent behaviors over given intervals. It's like measuring the total weight of an object by summing up the small bits, ensuring the total is accurate. In mathematical terms, the definite integral from a to b of a function f(x) is \[ \int_{a}^{b} f(x) \, dx. \]This mathematical expression captures the total area between the function and the x-axis over the interval \( [a, b] \).

• Used to find total quantities like area or probability.
• Measures how a function behaves over a range.
• Essential to verifying properties of functions like probability density functions.

To correctly use a definite integral, you need boundaries (like 0 and infinity in many probability contexts) and the function itself, ensuring calculations adhere to these limits.

Exponential Function

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. These functions are pivotal in modeling growth or decay processes, such as populations increasing or radioactive substances decaying over time.The general form of an exponential function is n\[ f(x) = a \cdot e^{bx}, \]where \( e \) is Euler's number, approximately 2.718. In probability density contexts, these functions often describe time until an event occurs, like failure times or intervals between events, because of their unique property of consistent growth and decay rates.

• Characterized by constant growth or decay rate.
• \( e \) is the natural base, crucial for natural processes.
• Frequently encountered in statistical models and probability theory.

When used in probability, exponential functions are critical because they can define continuous distributions that describe the time until an event happens, crucial in fields like reliability engineering or survival analysis.

Substitution Method

The substitution method is a technique for solving integrals and simplifying integral expressions by changing variables. It's a way to re-frame a complex problem into one that's easier to solve.In this method, parts of the equation under the integral sign are replaced with a new variable, making integration more feasible. Consider the integral of an exponential function, for instance, where the exponent itself could be a complex expression.Say you have \[ \int 3e^{-3x} \, dx. \]Using substitution, you set \( u = -3x \), thus simplifying the exponential part. With \( dx = -\frac{1}{3} du \), the integral simplifies considerably:\[ \int e^{u} (-\frac{1}{3}) du, \]This newfound integral structure is often simpler to resolve.

• Simplifies complex expressions within integrals.
• Changes the variable of integration to ease calculations.
• Crucial for functions that do not integrate cleanly at first glance.

This technique is like translating difficult problems into easier languages, creating a path toward correct solutions. It's particularly useful when direct integration is cumbersome or when the function inside the integral is tangled with difficult polynomials or exponentials.