Problem 47 - Calculus and its applications

Question

Find \(c\) such that \(f(x)=c x e^{2 x},\) for \(1 \leq x \leq 2,\) is a probability density function.

Step-by-Step Solution

Answer

The value of \(c\) is \( c = \frac{4}{e^{4} - e^{2}} \).

Step 1: Understand the PDF Condition

For a function to be a probability density function (PDF) over the interval \([a, b]\), the integral of the function over that interval must equal 1. We need to find \(c\) such that \( \int_{1}^{2} cxe^{2x} \, dx = 1 \).

Step 2: Set Up the Integral Equation

Set up the equation: \( \int_{1}^{2} cxe^{2x} \, dx = 1 \). To find \(c\), we first need to evaluate the indefinite integral \( \int x e^{2x} \, dx \).

Step 3: Use Integration by Parts

To integrate \( \int x e^{2x} \, dx \), use integration by parts. Let \(u = x\) and \(dv = e^{2x} dx\). Then \(du = dx\) and \(v = \frac{1}{2}e^{2x}\). Apply integration by parts formula: \( \int u \, dv = uv - \int v \, du \).

Step 4: Integrate Using Parts

Following Step 3, we have: \( \int x e^{2x} \, dx = x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \, dx \). Simplify this to \( \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} + C \).

Step 5: Evaluate the Definite Integral

Now, find the definite integral: \( \int_{1}^{2} x e^{2x} \, dx = \left[ \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} \right]_{1}^{2} \). Calculate it to find the numerical value.

Step 6: Calculate the Definite Integral

First, evaluate at \(x=2\): \( \frac{2}{2}e^{4} - \frac{1}{4}e^{4} = \frac{1}{2}e^{4} - \frac{1}{4}e^{4} = \frac{1}{4}e^{4} \). Next, evaluate at \(x=1\): \( \frac{1}{2}e^{2} - \frac{1}{4}e^{2} = \frac{1}{4}e^{2} \). Now subtract: \( \frac{1}{4}e^{4} - \frac{1}{4}e^{2} \).

Step 7: Solve for c

From the previous step, the integral \( \int_{1}^{2} x e^{2x} \, dx \) simplifies to \( \frac{1}{4}(e^{4} - e^{2}) \). Set this equal to \( \frac{1}{c} \), to satisfy the PDF condition: \( c \cdot \frac{1}{4}(e^{4} - e^{2}) = 1 \). Solve for \(c\): \( c = \frac{4}{e^{4} - e^{2}} \).

Key Concepts

Integration by Parts

Integration by parts is a critical technique in calculus, particularly when dealing with products of functions within an integral. This method transforms the integral of a product of functions into an expression involving the integral of their derivative and anti-derivative.
Here is how it works:

• Choose two parts of the integrand, assigning them to functions \( u \) and \( dv \).
• Differentiate \( u \) to find \( du \), and integrate \( dv \) to get \( v \).
• Utilize the integration by parts formula: \( \int u \, dv = uv - \int v \, du \).

In our example, we needed to find \( \int x e^{2x} \, dx \). By setting \( u = x \) and \( dv = e^{2x} \, dx \), we compute \( du = dx \) and \( v = \frac{1}{2}e^{2x} \).
This simplifies the problem into manageable integrals, enabling us to compute the definite integral necessary for determining the probability density function constant, \( c \). This strategy of breaking down complex products greatly aids in solving many integral problems.

Definite Integral

A definite integral is a fundamental concept used in calculus to calculate the area under a curve between two bounds. Unlike indefinite integrals, which involve adding an integration constant \( C \), definite integrals yield a numerical result.
To find the value of a definite integral:

• First determine the antiderivative of the function.
• Evaluate this antiderivative at the upper bound and lower bound of the interval.
• Subtract the value at the lower bound from the value at the upper bound.

In the problem we addressed, we needed to calculate \( \int_{1}^{2} x e^{2x} \, dx \). After applying integration by parts, we evaluated the resulting expression at the limits of 2 and 1. This led to calculating the expression \( \frac{1}{4}(e^{4} - e^{2}) \).
This evaluation provides the essential value while solving for \( c \) to confirm that the function meets the criteria of a probability density function.

Continuous Probability Distributions

Continuous probability distributions describe random variables that can take an infinite number of potential values. These variables are often represented by a probability density function (PDF), which encapsulates how probabilities are distributed across possible outcomes.
For a function to qualify as a PDF:

• The area under its curve, over the entire range of possible values, must equal 1. This ensures all possible outcomes are accounted for.
• The value of the function should be non-negative for all \( x \) within the range.

In our example, \( f(x) = cx e^{2x} \) was provided as a candidate PDF over the interval \([1, 2]\). To affirm it meets the PDF condition, we ran the integral \( \int_{1}^{2} cxe^{2x} \, dx = 1 \). Solving this allowed us to find \( c \), ensuring that the outlined conditions are satisfactorily met.
This process illustrates how integration ensures that a function defines a valid probability distribution over a given interval, providing an essential tool in statistical analysis and probability theory.