Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the value of \(k\) that makes \(f(x)=k x(5-x)\), \(0 \leq x \leq 5\), a valid PDF. Hint: The PDF must integrate to \(1 .\)

Step-by-Step Solution

Answer

The value of \(k\) is \(\frac{6}{125}\).

Step 1: Understand the Problem

The problem is asking us to find the value of \(k\) that makes the function \(f(x) = kx(5-x)\) a valid probability density function (PDF) over the interval \(0 \leq x \leq 5\). A key property of a valid PDF is that the integral of the function over its defined range must equal 1.

Step 2: Set Up the Integral Equation

To find \(k\), we need to integrate the function \(f(x) = kx(5-x)\) over the interval from 0 to 5 and set the integral equal to 1. So, our integral equation is:\[\int_{0}^{5} kx(5-x) \, dx = 1\]

Step 3: Expand and Integrate the Function

First, expand the function: \(kx(5-x) = k(5x - x^2)\). Now, integrate term by term:\[\int_{0}^{5} (5kx - kx^2) \, dx = \int_{0}^{5} 5kx \, dx - \int_{0}^{5} kx^2 \, dx\]

Step 4: Perform the Integration

Find the integral of each term:- The integral of \(5kx\) from 0 to 5 is \(\frac{5k}{2}x^2\) evaluated from 0 to 5, which gives \(\frac{5k}{2}(25)\).- The integral of \(kx^2\) from 0 to 5 is \(\frac{k}{3}x^3\) evaluated from 0 to 5, which gives \(\frac{k}{3}(125)\).Combine these results:\[\frac{5k}{2}(25) - \frac{k}{3}(125)\]

Step 5: Solve for k

Set the combined expression equal to 1:\[\frac{5k}{2}(25) - \frac{k}{3}(125) = 1\]This simplifies to:\[\frac{125k}{2} - \frac{125k}{3} = 1\]Find a common denominator and solve for \(k\):\[\frac{375k - 250k}{6} = 1\]\[\frac{125k}{6} = 1\]Multiply both sides by 6 and divide by 125:\[k = \frac{6}{125}\]

Step 6: Verify the Solution

Ensure that the solution satisfies the original condition that the integral of the PDF over \([0, 5]\) equals 1:Calculate: \[\frac{125 \times \frac{6}{125}}{6} = 1\] which confirms our solution is correct.

Key Concepts

Integral Calculus

Integral calculus is a fundamental concept in mathematics that deals with finding the integral of a function. An integral is essentially the "accumulated" area under the curve of a function on a graph.
In simple terms, when you integrate a function, you are summing up an infinite number of infinitesimally small pieces to determine the total quantity, such as area or volume.
Integral calculus is widely used in various fields such as physics, engineering, and statistics. To grasp integral calculus, it's crucial to understand two key parts:

• Indefinite Integrals: These represent a family of functions and include a constant of integration (usually denoted as +C). They are anti-derivatives of a function.
• Definite Integrals: These calculate the net area between the curve and the x-axis over a specified interval, providing a numerical value.

For the given exercise, we focus on a definite integral because we need the specific value of the integral to be equal to one. This requirement will ensure the function satisfies the normalization condition for a probability density function.

Normalization Condition

The normalization condition is a crucial property for any valid probability density function (PDF).
In statistical terms, normalization ensures the total probability across the entire range of the random variable is 1. This is because the probability distribution of a function must account for all possible outcomes within its defined interval.
For the function to serve as a PDF, integrating it over the interval \[0, 5\] must yield exactly 1, thereby representing the total probability.To apply the normalization condition, students typically set up an integral equation: \[ \int_{a}^{b} f(x) \, dx = 1 \]In our exercise, this means we are solving: \[ \int_{0}^{5} kx(5-x) \, dx = 1 \]Through solving this equation, we determined the proper constant \( k \) that allows the function \( f(x) = k x (5-x) \) to be a valid PDF. This condition, therefore, ensures that the entire distribution of the given function sums up to the whole probability spectrum.

Definite Integrals

Definite integrals provide the exact area under a curve for a specified interval. They are essential in determining exact quantities, such as area under a curve or, in statistics, probabilities and overall expectations.
In the equation \[ \int_{a}^{b} f(x) \, dx \], - \(a\) and \(b\) are the boundaries of integration, indicating the interval over which we are calculating.- \(f(x)\) is the function under consideration.The process involves calculating the integral by evaluating the antiderivative of the function at both bounds and subtracting these values:

• Evaluate the antiderivative of \(f(x)\) at the upper bound \(b\).
• Evaluate the antiderivative at the lower bound \(a\).
• Subtract the lower bound evaluation from the upper bound evaluation.

In our exercise, finding the definite integral across \([0, 5]\) was necessary to determine \( k \).
This process captures the whole area under \(kx(5-x)\) over the given range, ensuring the probability function accurately sums to 1, thereby validating it as a proper PDF.