f(x)=kx². [-1,1] = 1 (Type an exact answer.) he probability density function is f(x) = (Jyn
f(x)=kx². [-1,1] = 1 (Type an exact answer.) he probability density function is f(x) = (Jyn
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![### Finding the Probability Density Function
In this problem, we aim to determine the constant \( k \) such that the function \( f(x) \) is a probability density function over the interval \([-1, 1]\). The given function is:
\[ f(x) = kx^2 \]
#### Steps to Determine \( k \)
1. **Integrate \( f(x) \) over the interval \( [-1, 1] \)**:
\[
\int_{-1}^{1} kx^2 \, dx
\]
2. **Ensure the integral equals 1** (as this is a requirement for the function to be a probability density function):
\[
\int_{-1}^{1} kx^2 \, dx = 1
\]
Perform the integral calculation:
\[
\int_{-1}^{1} kx^2 \, dx = k \left[ \frac{x^3}{3} \right]_{-1}^{1} = k \left( \frac{1^3}{3} - \frac{(-1)^3}{3} \right) = k \left( \frac{1}{3} - \left( \frac{-1}{3} \right) \right) = k \left( \frac{1}{3} + \frac{1}{3} \right) = k \left( \frac{2}{3} \right)
\]
Thus, setting the integral equal to 1 for it to be a probability density function:
\[
k \left( \frac{2}{3} \right) = 1 \implies k = \frac{3}{2}
\]
#### Final Probability Density Function
Given \( k = \frac{3}{2} \), the probability density function \( f(x) \) is:
\[ f(x) = \frac{3}{2} x^2 \]
### Summary
After solving, we find:
- \( k = \frac{3}{2} \)
- The probability density function is \( f(x) = \frac{3}{2} x^2 \)
This ensures that \( f(x) \) satisfies the properties of a probability density function over the interval \([-1, 1]\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc56b3bff-26b7-4f28-9b8d-5c65e8bb1b58%2F7cf1269c-585e-4845-943e-df69d91396eb%2F5b002f_processed.png&w=3840&q=75)
Transcribed Image Text:### Finding the Probability Density Function
In this problem, we aim to determine the constant \( k \) such that the function \( f(x) \) is a probability density function over the interval \([-1, 1]\). The given function is:
\[ f(x) = kx^2 \]
#### Steps to Determine \( k \)
1. **Integrate \( f(x) \) over the interval \( [-1, 1] \)**:
\[
\int_{-1}^{1} kx^2 \, dx
\]
2. **Ensure the integral equals 1** (as this is a requirement for the function to be a probability density function):
\[
\int_{-1}^{1} kx^2 \, dx = 1
\]
Perform the integral calculation:
\[
\int_{-1}^{1} kx^2 \, dx = k \left[ \frac{x^3}{3} \right]_{-1}^{1} = k \left( \frac{1^3}{3} - \frac{(-1)^3}{3} \right) = k \left( \frac{1}{3} - \left( \frac{-1}{3} \right) \right) = k \left( \frac{1}{3} + \frac{1}{3} \right) = k \left( \frac{2}{3} \right)
\]
Thus, setting the integral equal to 1 for it to be a probability density function:
\[
k \left( \frac{2}{3} \right) = 1 \implies k = \frac{3}{2}
\]
#### Final Probability Density Function
Given \( k = \frac{3}{2} \), the probability density function \( f(x) \) is:
\[ f(x) = \frac{3}{2} x^2 \]
### Summary
After solving, we find:
- \( k = \frac{3}{2} \)
- The probability density function is \( f(x) = \frac{3}{2} x^2 \)
This ensures that \( f(x) \) satisfies the properties of a probability density function over the interval \([-1, 1]\).
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