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Chapter 10: Problem 4

Verify the two properties of a probability density function over the given interval. $$ f(x)=\frac{1}{x+1},[0, e-1] $$

Short Answer

Expert verified
The non-negative function integrates to 1, so it's a valid probability density function.

Step by step solution

01

Verify the function is non-negative

First, ensure that the given function is non-negative over the interval [0, e-1]. Since the function is \( f(x)=\frac{1}{x+1} \), for any \( x \) in [0, e-1], \( x+1 \) is positive, hence \( \frac{1}{x+1} \) is non-negative.
02

Integrate the function over the interval

To verify if the function \( f(x) \) is a probability density function, \( \int_{0}^{e-1} f(x) \, dx \) should equal 1. We need to calculate the integral of \( f(x)=\frac{1}{x+1} \) over the interval [0, e-1]:\[ \int_{0}^{e-1} \frac{1}{x+1} \, dx \] Using the substitution \( u = x + 1 \), therefore \( du = dx \), the integral becomes \[ \int_{1}^{e} \frac{1}{u} \, du \] which is equal to \( \ln |u| \) from 1 to e.
03

Evaluate the integral

Evaluating the integral, we get:\[ \ln e - \ln 1 = 1 - 0 = 1 \] Thus, the integral equals 1.
04

Conclusion

Since \( f(x) \) is non-negative on [0, e-1] and \( \int_{0}^{e-1} f(x) \, dx = 1 \), \( f(x)=\frac{1}{x+1} \) satisfies both properties of a probability density function over the given interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Non-Negative Function
A function is non-negative if it is always greater than or equal to zero over its domain.
In this problem, our function is given as \( f(x) = \frac{1}{x+1} \). To check if it is non-negative over the interval [0, e-1], we need to examine the term in the denominator.
For any value of \( x \) between 0 and \( e-1 \), \( x + 1 \) is always positive, since even the smallest value, which is 0, makes the denominator 1 (0+1).
Thus, \( \frac{1}{x+1} \) will always be positive and therefore non-negative within our interval.
This confirms that the function meets the first condition required for a probability density function (PDF).
Integration
Integration is a fundamental concept in calculus that allows us to find the area under a curve.
To verify that a function \( f(x) \) is a probability density function, its integral over the specified interval must equal 1.
For our function, we need to compute the integral from 0 to \( e-1 \):
\[ \int_{0}^{e-1} \frac{1}{x+1} \, dx \].
If the result of this integration is 1, it means that the total area under the curve over the interval [0, e-1] is 1, which is a requirement for a PDF.
Substitution Method
The substitution method is a powerful technique in integration, especially for functions that don't integrate easily in their given form.
In this problem, we use the substitution \( u = x + 1 \). This simplifies our integral significantly:
\[ \int_{0}^{e-1} \frac{1}{x+1} \, dx \] becomes \[ \int_{1}^{e} \frac{1}{u} \, du \].
By rewriting the limits of integration in terms of \( u \), we transform the integral into a much simpler form.
The integral now becomes the natural logarithm of \( u \) evaluated from 1 to \( e \).
Natural Logarithm
A natural logarithm (\( \ln \)) is the logarithm to the base \( e \), where \( e \) is an irrational number approximately equal to 2.718.
When we integrate \( \frac{1}{u} \) with respect to \( u \), the result is \( \ln |u| \).
After performing the integral, we evaluate this from 1 to \( e \):
\[ \ln e - \ln 1 \]
Since \( \ln e = 1 \) and \( \ln 1 = 0 \), the result is 1 - 0 which equals 1.
This confirms that our integral over the specified interval equals 1, satisfying the probability density function requirement.
Therefore, the function \( f(x) = \frac{1}{x+1} \) meets both necessary properties.

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