Problem 47
Question
Evaluate the integrals in Exercises \(39-56\) $$ \int_{2}^{4} \frac{d x}{x(\ln x)^{2}} $$
Step-by-Step Solution
Verified Answer
\( \frac{1}{2 \ln 2} \)
1Step 1: Identify the Integral Form
The given integral is \( \int_{2}^{4} \frac{dx}{x(\ln x)^{2}} \). This can be recognized as an integral requiring substitution due to the complexity of \( x \ln x \).
2Step 2: Choose a Suitable Substitution
Let \( u = \ln x \). Consequently, \( \frac{du}{dx} = \frac{1}{x} \) or equivalently, \( dx = x \, du \). This means our integral becomes \( \int \frac{1}{x u^2} \, x \, du = \int \frac{1}{u^2} \, du \).
3Step 3: Change the Limits of Integration
Since \( u = \ln x \), the limits change accordingly. When \( x = 2 \), \( u = \ln 2 \), and when \( x = 4 \), \( u = \ln 4 \). The new integral is \( \int_{\ln 2}^{\ln 4} \frac{du}{u^2} \).
4Step 4: Integrate with Respect to \( u \)
The integral \( \int \frac{1}{u^2} \, du \) is \( -\frac{1}{u} + C \). Thus, the definite integral \( \int_{\ln 2}^{\ln 4} \frac{du}{u^2} \) becomes \(-\frac{1}{u} \bigg|_{\ln 2}^{\ln 4} \).
5Step 5: Evaluate the Definite Integral
Substitute the bounds into \(-\frac{1}{u} \) resulting in \(-\frac{1}{\ln 4} + \frac{1}{\ln 2} = \frac{1}{\ln 2} - \frac{1}{\ln 4} \).
6Step 6: Simplify the Expression
Note that \( \ln 4 = 2 \ln 2 \). Thus, \( \frac{1}{\ln 4} = \frac{1}{2 \ln 2} \). Therefore, the expression simplifies to \( \frac{1}{\ln 2} - \frac{1}{2 \ln 2} = \frac{1}{2 \ln 2} \).
Key Concepts
Substitution methodDefinite integralsLogarithmic functions
Substitution method
The substitution method is a powerful technique used in integration to simplify complex expressions into forms that are easier to solve. This method involves substituting a part of the integral with a single variable, usually denoted as \( u \), to reduce the complexity of the expression.
For the given exercise, we observe a challenging function \( \frac{1}{x(\ln x)^2} \). By letting \( u = \ln x \), the derivative \( \frac{du}{dx} = \frac{1}{x} \) allows us to substitute \( dx = x \, du \). This substitutes the integral into a simpler form, allowing us to focus on the substitution without the \( x \) variable cluttering the expression.
The original limits of integration also need to be adjusted in accordance with the substitution. When substituting variables in integrals, the original input bounds must change to correspond with the \( u \) variable:
For the given exercise, we observe a challenging function \( \frac{1}{x(\ln x)^2} \). By letting \( u = \ln x \), the derivative \( \frac{du}{dx} = \frac{1}{x} \) allows us to substitute \( dx = x \, du \). This substitutes the integral into a simpler form, allowing us to focus on the substitution without the \( x \) variable cluttering the expression.
The original limits of integration also need to be adjusted in accordance with the substitution. When substituting variables in integrals, the original input bounds must change to correspond with the \( u \) variable:
- When \( x = 2 \), \( u = \ln 2 \)
- When \( x = 4 \), \( u = \ln 4 \)
Definite integrals
Definite integrals, unlike indefinite integrals, provide a numerical value representing the area under a curve within specified bounds. Such integrals are written with limits of integration, denoted by numbers placed beside the integral sign. In our example, the definite integral is initially \( \int_{2}^{4} \), which after substitution changes to \( \int_{\ln 2}^{\ln 4} \).
Once the integration is performed, the results are evaluated at the bounds. Here, after integrating \( \int \frac{1}{u^2} \, du \), the function \(-\frac{1}{u} \) is evaluated from \( u=\ln 2 \) to \( u=\ln 4 \). This means:
\[ -\frac{1}{\ln 4} + \frac{1}{\ln 2} = \frac{1}{\ln 2} - \frac{1}{\ln 4} \]
The final step is the simplification: noting that \( \ln 4 = 2\ln 2 \), which refines the result to \( \frac{1}{2 \ln 2} \), illustrating the computation of the definite integral to find an exact value.
Once the integration is performed, the results are evaluated at the bounds. Here, after integrating \( \int \frac{1}{u^2} \, du \), the function \(-\frac{1}{u} \) is evaluated from \( u=\ln 2 \) to \( u=\ln 4 \). This means:
\[ -\frac{1}{\ln 4} + \frac{1}{\ln 2} = \frac{1}{\ln 2} - \frac{1}{\ln 4} \]
The final step is the simplification: noting that \( \ln 4 = 2\ln 2 \), which refines the result to \( \frac{1}{2 \ln 2} \), illustrating the computation of the definite integral to find an exact value.
Logarithmic functions
Logarithmic functions, fundamental in calculus, are the inverse of exponential functions and are crucial for integration, especially within contexts like our exercise. Here, the natural logarithm function \( \ln x \) played an essential role as it was chosen for substitution due to the derivative \( \frac{d}{dx}\ln x = \frac{1}{x} \).
This characteristic of logarithmic functions—having a derivative that simplifies multiplication inside the integral—is what permitted the transition from a complicated integrand involving \( x \) and \( \ln x \) to a much cleaner form in terms of \( u \).
Moreover, understanding the properties of logarithms, such as \( \ln 4 = 2\ln 2 \), helps in simplifying expressions after evaluation. These properties allow us to make precise calculations and reduce terms easily, ensuring that the final results are both accurate and simplified. Mastery of logarithmic identities and properties is therefore essential not only for integration but for calculus as a whole, enabling effective solutions to problems involving natural logs.
This characteristic of logarithmic functions—having a derivative that simplifies multiplication inside the integral—is what permitted the transition from a complicated integrand involving \( x \) and \( \ln x \) to a much cleaner form in terms of \( u \).
Moreover, understanding the properties of logarithms, such as \( \ln 4 = 2\ln 2 \), helps in simplifying expressions after evaluation. These properties allow us to make precise calculations and reduce terms easily, ensuring that the final results are both accurate and simplified. Mastery of logarithmic identities and properties is therefore essential not only for integration but for calculus as a whole, enabling effective solutions to problems involving natural logs.
Other exercises in this chapter
Problem 47
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