91Ó°ÊÓ

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.

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 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.