Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the value of \(k\) that makes \(f(x)=k x^{2}(5-x)^{2}\), \(0 \leq x \leq 5\), a valid PDF.

Step-by-Step Solution

Answer

The value of \(k\) is approximately 0.00192.

Step 1: Understand the PDF Condition

A function is a valid probability density function (PDF) if its integral over the entire range is equal to 1. For this problem, we need to ensure that the integral of \(f(x) = k x^{2}(5-x)^{2}\) from 0 to 5 is equal to 1.

Step 2: Set Up the Integral Equation

The integral we need to evaluate is \(\int_{0}^{5} k x^{2}(5-x)^{2} \, dx = 1\). We will solve this integral and then solve for \(k\).

Step 3: Simplify the Function and Integrate

Expand \((5-x)^2\) to get \(25 - 10x + x^2\). Thus, \(f(x) = k x^2 (25 - 10x + x^2)\). Distribute \(x^2\) to get \(kx^2 \cdot 25 - kx^2 \cdot 10x + kx^4\). Integrate each term separately from 0 to 5.

Step 4: Compute Each Integral

Evaluate these integrals: 1. \(\int_{0}^{5} 25x^2 \, dx = 25 \cdot \frac{x^3}{3} \bigg|_0^5 = \frac{25 \cdot 125}{3}\)2. \(\int_{0}^{5} 10x^3 \, dx = 10 \cdot \frac{x^4}{4} \bigg|_0^5 = \frac{10 \cdot 625}{4}\)3. \(\int_{0}^{5} x^4 \, dx = \frac{x^5}{5} \bigg|_0^5 = \frac{3125}{5}\).Substitute these results back in: \(k \left( \frac{25 \cdot 125}{3} - \frac{10 \cdot 625}{4} + \frac{3125}{5} \right) = 1\).

Step 5: Solve for the Constant k

Simplify the expression inside the parentheses to find the value of \(k\). This simplifies to:\(- \frac{1250}{4} + \frac{3125}{5} = 520.83\).Therefore, the equation becomes:\[k \cdot 520.83 = 1\]Solve for \(k\): \(k = \frac{1}{520.83}\).

Step 6: Calculate the Value of k

Perform the division: \(k\approx 0.00192\). This is the constant that makes \(f(x)\) a valid PDF.

Key Concepts

Integral Calculation

In order to understand integral calculation, it's essential to grasp why integrals are used in probability density functions (PDFs). Integrals help determine the area under a curve, which in probability terms, represents the total probability. Your goal is to ensure this area equals 1 for a legitimate PDF.

To solve the exercise, you set up the integral equation with the function provided, ensuring the limits of integration cover the function's domain. Here, the function is integrated from 0 to 5. The integral we set up is \( \int_{0}^{5} k x^{2}(5-x)^{2} \, dx = 1 \).

Once the equation is set, the next step is to solve the integral. This involves both simplification and calculation steps, broken down into manageable parts. You first need to expand the polynomial, making the integration process more straightforward.

• Simplify the function if necessary.
• Perform the integral calculation for each term.
• Sum the evaluated integrals to solve for constants like \(k\).

Expansion of Polynomial Functions

Expanding polynomial functions is a crucial step in many mathematical problems, particularly when dealing with integrals. Let's break down why it's necessary in our context and how it's done.

For the given function \(f(x) = k x^{2}(5-x)^{2}\), it is key to simplify \((5-x)^{2}\) into a form that's easier to integrate. You start with expansion: \((5-x)^2 = 25 - 10x + x^2\). Now, with the expanded polynomial, it's easier to perform distribution across the function:

\(f(x) = k x^2 (25 - 10x + x^2)\) turns into:

• \(kx^2 \cdot 25\)
• \(- kx^2 \cdot 10x\)
• \(+ kx^4\)

Each component is necessary to calculate individual integrals in the next step. This expansion transforms the problem into simpler pieces solved independently and summed together.

Solving for Constants

Finding the constant in equations, like our task of determining the value of \(k\), requires attention and precision. The ultimate aim is to scale the function correctly to ensure it satisfies the condition of being a PDF—necessitating that the integral equals 1.

After evaluating the integrals individually, substitute results into the equation: \(k \left( \frac{25 \cdot 125}{3} - \frac{10 \cdot 625}{4} + \frac{3125}{5} \right) = 1\).

Simplify the expression inside the parentheses thoroughly. In this problem, the simplification yields a constant, which sums the contributions of all integral terms:

- Calculate each term separately
- Sum these to find the total size of the area
- Solve for \(k\) using the equation \(k \cdot 520.83 = 1\) by isolating \(k\) and finding its value: \(k \approx 0.00192\).

This process confirms the initial assumption that your calculated \(k\) ensures the function is a valid PDF from range 0 to 5.