Question

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

Short answer

k = \( \frac{12}{78125} \).

Step-by-step solution

Understand the PDF Condition

A probability density function (PDF) must satisfy two conditions: it must be non-negative for all values in its domain, and the integral over the entire space must equal 1. In this case, the function is non-negative due to the squares and range, so we need to focus on the second condition, which is integrating over the interval [0,5] and setting it equal to 1.

Set Up the Integral

Set up the integral of the function over the interval from 0 to 5. This is expressed as: \[ \int_{0}^{5} k x^{2}(5-x)^{2} \, dx = 1 \] Our goal is to solve for the constant \( k \).

Expand the Expression

Expand the expression \((5-x)^{2}\) to simplify the integral. This gives: \((5-x)^2 = 25 - 10x + x^2\). Therefore, the function becomes \(f(x) = kx^2 (25 - 10x + x^2)\).

Simplify and Integrate

Substitute the expanded expression into the integral: \[ \int_{0}^{5} k (25x^2 - 10x^3 + x^4) \, dx \] Compute each integral separately: \( \int_0^5 x^2 \, dx, \int_0^5 x^3 \, dx, \int_0^5 x^4 \, dx \), then combine.

Compute Indefinite Integrals

Compute the indefinite integrals: - For \(x^2\), \(\int x^2 \, dx = \frac{x^3}{3} \)- For \(x^3\), \(\int x^3 \, dx = \frac{x^4}{4}\)- For \(x^4\), \(\int x^4 \, dx = \frac{x^5}{5}\).

Evaluate Limits of Integration

Evaluate each integral from 0 to 5:- \(\int_{0}^{5} 25x^2 \, dx = 25 \times \left[ \frac{5^3}{3} - \frac{0^3}{3} \right] = \frac{25 \times 125}{3} = \frac{3125}{3}\)- \(\int_{0}^{5} -10x^3 \, dx = -10 \times \left[ \frac{5^4}{4} - \frac{0^4}{4} \right] = -10 \times \frac{625}{4} = -1562.5\)- \(\int_{0}^{5} x^4 \, dx = \left[ \frac{5^5}{5} - \frac{0^5}{5} \right] = 625\).

Sum Results and Solve for k

Substitute the evaluated results into the integral condition: \[ k \left( \frac{3125}{3} - 1562.5 + 625 \right) = 1 \]Simplify the expression: \[ k \left( \frac{3125}{3} - 1562.5 + 625 \right) = 1 \] Solve for \( k \).

Conclusion

Calculate \( \frac{3125}{3} + 625 - 1562.5 = \frac{78125}{12} \). Therefore, \[ k \cdot \frac{78125}{12} = 1 \] implies \[ k = \frac{12}{78125} \].

Key concepts

Calculus Integration

Calculus integration is a fundamental operation in mathematics that is used to find areas, volumes, and central points, among other things. When we talk about integration, we deal with the process of adding up infinitesimally small quantities to get cumulative values. In the context of probability, integration helps us determine probabilities over continuous spaces, such as the area under a probability density function (PDF).

There are two main types of integrals involved: definite and indefinite. But both share the same foundational steps, like setting up the integral, simplifying the expression, and evaluating it. For the function to be a valid PDF, the definite integral over its entire range must be 1, ensuring the total probability across the interval is complete.

• The integral should cover the whole domain being studied.
• The goal is often to solve for a constant that makes the total area equal to 1.
• This is crucial when working with PDFs as shown in the exercise.

Indefinite Integrals

Indefinite integrals are a type of calculus integration that involves finding the antiderivative of a function. Rather than giving a specific numerical result like definite integrals, indefinite integrals provide a general form, including a constant of integration, usually denoted as "+ C." This reflects the family of functions that differentiate to the original function.

When you find an indefinite integral, you essentially reverse the differentiation process. For example, if you integrate a power of x, such as \( x^n \), the result is \( \frac{x^{n+1}}{n+1} + C \). This process is vital for handling more complex integrals, where simplifying parts of a function is needed before combining them in a definite integral.

• Indefinite integrals lack bounds, unlike definite integrals.
• They cover all possible antiderivatives of a function.
• This is crucial when preparing for definite integration in the problem.

Definite Integrals

Definite integrals, unlike indefinite integrals, provide a specific numerical result representing the area under the curve of a function for a given interval. In this exercise, we compute these definite integrals to ensure the total area under the probability density function equals 1. This is a key condition for the function to be a valid PDF.

The process involves applying the fundamental theorem of calculus, which connects differentiation and integration. You first find the indefinite integral, and then evaluate it at the upper and lower bounds of the interval.

• The definite integral symbolizes accumulation of values over an interval.
• It allows us to solve for constants, such as \( k \), that standardize total values.
• This is used to confirm that the total probability is equal to 1, as required for PDFs.

Polynomial Expansion

Polynomial expansion is a technique used in algebra to make integrations or other operations simpler. It involves rewriting expressions like \((5-x)^2\) in a more expanded form by multiplying the factors.

This simplification helps in breaking down the function for easier integration. In this exercise, expanding \((5-x)^2\) into \(25 - 10x + x^2\) reveals a clearer view of the polynomial, making each term straightforward to integrate. This kind of simplification is vital in calculus to manage complex problems step by step.

• Polynomial expansion makes integration of each term feasible.
• It offers clarity when substituting into a larger expression.
• This is part of structuring a solution where each detail is necessary for solving integrals.
• Crucial for decomposing functions into simpler parts you can integrate separately.