Problem 16
Question
Find each indefinite integral. \(\int \frac{d x}{2 x}\)
Step-by-Step Solution
Verified Answer
The indefinite integral is \( \frac{1}{2} \ln |x| + C \).
1Step 1: Identify the Integral Form
The integral given is \( \int \frac{d x}{2 x} \). This is a simple rational function where we need to integrate an expression of the form \( \int \frac{1}{x} \, dx \), but in this case, it includes a constant denominator.
2Step 2: Factor Out the Constant
We can factor out the constant \( \frac{1}{2} \) from the integral. This simplifies the integral to \( \frac{1}{2} \int \frac{1}{x} \, dx \).
3Step 3: Integrate the Simplified Function
The integral \( \int \frac{1}{x} \, dx \) is a standard integral that equals \( \ln |x| + C \), where \( C \) is the constant of integration.
4Step 4: Apply the Integration Result
Applying this result to our expression, we get \( \frac{1}{2} (\ln |x| + C) = \frac{1}{2} \ln |x| + \frac{1}{2} C \). Since \( C \) is an arbitrary constant, we typically write it simply as \( C \).
5Step 5: Write the Final Solution
The final solution of the integral is \( \frac{1}{2} \ln |x| + C \), where \( C \) is the constant of integration.
Key Concepts
Rational FunctionsIntegration TechniquesConstant Factor Rule
Rational Functions
In mathematics, rational functions are expressions of the form \( \frac{f(x)}{g(x)} \) where \( f(x) \) and \( g(x) \) are polynomials. These functions are interesting because they can represent complex relationships using simpler elements like polynomials, and they are often smooth and continuous except at points where the denominator is zero. For example, the function \( \frac{1}{2x} \) is a simple rational function with a constant numerator and the linear polynomial \( 2x \) in the denominator.
To integrate rational functions, we aim to simplify the expression whenever possible, such as factoring out constants or breaking it into simpler parts that are easier to manage. Understanding the characteristics of rational functions assists in identifying possible methods for integration, which is crucial when tackling integration problems.
Rational functions often appear in many calculus problems, and the techniques learned here are applicable to broader functions, especially when pursuing higher-level calculus topics.
To integrate rational functions, we aim to simplify the expression whenever possible, such as factoring out constants or breaking it into simpler parts that are easier to manage. Understanding the characteristics of rational functions assists in identifying possible methods for integration, which is crucial when tackling integration problems.
Rational functions often appear in many calculus problems, and the techniques learned here are applicable to broader functions, especially when pursuing higher-level calculus topics.
Integration Techniques
Integration techniques are strategies that help us find the antiderivative of a function. Some of these include substitution, integration by parts, partial fractions, and recognizing standard integrals.
In the given exercise, the integral \( \int \frac{1}{2x} \, dx \) simplifies due to the nature of the function \( \frac{1}{x} \), which is a well-known derivative. We recognize this as a standard integral form, allowing us to directly apply our knowledge that \( \int \frac{1}{x} \, dx = \ln |x| + C \).
Choosing the appropriate technique is often key to solving integrals efficiently. For functions like \( \frac{1}{2x} \), understanding the characteristics of logarithmic derivatives plays a vital role. Mastery of these techniques not only streamlines problem-solving but also enlarges one's mathematical toolkit for dealing with increasingly complex integrals.
In the given exercise, the integral \( \int \frac{1}{2x} \, dx \) simplifies due to the nature of the function \( \frac{1}{x} \), which is a well-known derivative. We recognize this as a standard integral form, allowing us to directly apply our knowledge that \( \int \frac{1}{x} \, dx = \ln |x| + C \).
Choosing the appropriate technique is often key to solving integrals efficiently. For functions like \( \frac{1}{2x} \), understanding the characteristics of logarithmic derivatives plays a vital role. Mastery of these techniques not only streamlines problem-solving but also enlarges one's mathematical toolkit for dealing with increasingly complex integrals.
Constant Factor Rule
The Constant Factor Rule is a straightforward yet powerful concept in integral calculus. It states that constants can be factored out from under the integral sign when integrating with respect to a variable. This rule is represented as \( \int k \cdot f(x) \, dx = k \cdot \int f(x) \, dx \) for any constant \( k \).
In our problem, \( \int \frac{1}{2x} \, dx \), this rule allows us to simplify the integration process by taking \( \frac{1}{2} \) out of the integral: \( \frac{1}{2} \int \frac{1}{x} \, dx \). This makes the problem significantly easier and quicker to solve, as we are left with the straightforward integral \( \int \frac{1}{x} \, dx \).
By recognizing and applying the Constant Factor Rule, students can focus on more intricate parts of the integrand without being bogged down by constant multipliers. Remember, when applying this rule, it is crucial to keep track of the constants throughout the calculation, ensuring they are included in the final result.
In our problem, \( \int \frac{1}{2x} \, dx \), this rule allows us to simplify the integration process by taking \( \frac{1}{2} \) out of the integral: \( \frac{1}{2} \int \frac{1}{x} \, dx \). This makes the problem significantly easier and quicker to solve, as we are left with the straightforward integral \( \int \frac{1}{x} \, dx \).
By recognizing and applying the Constant Factor Rule, students can focus on more intricate parts of the integrand without being bogged down by constant multipliers. Remember, when applying this rule, it is crucial to keep track of the constants throughout the calculation, ensuring they are included in the final result.
Other exercises in this chapter
Problem 16
Find each indefinite integral by the substitution method or state that it cannot be found by our substitution formulas. $$ \int e^{-x^{4}} x^{3} d x $$
View solution Problem 16
For each demand function \(d(x)\) and supply function \(s(x)\) : a. Find the market demand (the positive value of \(x\) at which the demand function intersects
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Find the average value of each function over the given interval. \(f(x)=e^{-2 x}\) on [0,1]
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Use a graphing calculator Riemann Sum program from the Internet (see page 332 ) to find the following Riemann sums. i. Calculate the Riemann sum for each functi
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