Textbook Question
Use any method to evaluate the integrals in Exercises 15–38. Most will require trigonometric substitutions, but some can be evaluated by other methods.
∫ dx / (x² √(x² + 1))
Ch. 8 - Techniques of Integration
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 8, Problem 8.9.18
Find the value of the constant c so that the given function is a probability density function for a random variable X over the specified interval.
f(x) = (1/x) over [c, c + 1]
Verified step by step guidance1
Recall that for a function \(f(x)\) to be a probability density function (pdf) over an interval \([a, b]\), it must satisfy the condition: \(\int_a^b f(x) \, dx = 1\).
In this problem, the function is \(f(x) = \frac{1}{x}\) defined over the interval \([c, c+1]\). We need to find the value of \(c\) such that \(\int_c^{c+1} \frac{1}{x} \, dx = 1\).
Set up the integral: \(\int_c^{c+1} \frac{1}{x} \, dx = \left[ \ln|x| \right]_c^{c+1} = \ln(c+1) - \ln(c)\).
Use the logarithm property to combine the difference: \(\ln(c+1) - \ln(c) = \ln\left( \frac{c+1}{c} \right)\).
Set the integral equal to 1 and solve for \(c\): \(\ln\left( \frac{c+1}{c} \right) = 1\). From here, exponentiate both sides to isolate \(c\) and solve the resulting equation.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Probability Density Function (PDF)
A probability density function describes the likelihood of a continuous random variable taking on a specific value. For a function to be a valid PDF, it must be non-negative over its domain and its total integral over the specified interval must equal 1.
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Properties of Functions
Definite Integral and Area Under the Curve
The definite integral of a function over an interval represents the area under its curve between two points. In the context of PDFs, integrating the function over its domain must yield 1, ensuring the total probability sums to one.
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Solving for Constants Using Integral Equations
When a function includes an unknown constant, we use integral equations to solve for it by setting the integral equal to 1 (for PDFs). This involves integrating the function with the constant over the interval and solving the resulting equation.
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Related Practice
Textbook Question
In Exercises 39–48, use an appropriate substitution and then a trigonometric substitution to evaluate the integrals.
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Textbook Question
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