Problem 100 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(93-106.\) $$\int_{1}^{e} \frac{2 \ln 10 \log _{10} x}{x} d x$$

Step-by-Step Solution

Answer

The integral evaluates to 2.

Step 1: Understand the Integrand

The integrand given is \( \frac{2 \ln 10 \log_{10} x}{x} \). Notice that \( \log_{10} x \) can be rewritten using the change of base formula as \( \frac{\ln x}{\ln 10} \). Hence, the integrand becomes \( \frac{2 \ln 10}{x} \cdot \frac{\ln x}{\ln 10} = \frac{2 \ln x}{x} \). This simplifies the integration process.

Step 2: Set Up the Integral

The problem now translates to evaluating the integral \( \int_{1}^{e} \frac{2 \ln x}{x} \, dx \). This is a standard integral involving the natural logarithm divided by \( x \).

Step 3: Identify the Integration Technique

Recognize that the integrand \( \frac{2 \ln x}{x} \) is suitable for using integration by parts. Let \( u = \ln x \) and \( dv = \frac{2}{x} \, dx \).

Step 4: Apply Integration by Parts

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we first calculate \( du = \frac{1}{x} \, dx \) and \( v = 2 \ln x \). Substitute these into the formula to get \( 2 \int \ln x \, d(\ln x) = 2(\ln x)^2 - 2 \int (\ln x) \, d(\ln x) \).

Step 5: Solve the Integral

The integral of \( \ln x \, d(\ln x) \) is \( \frac{1}{2} (\ln x)^2 \), hence \( 2 \left[ \ln x \right]_{1}^{e} = 2 \left[ (\ln e)^2 - (\ln 1)^2 \right] \). Evaluate the bounds: when \( x = e \), \( \ln e = 1 \); when \( x = 1 \), \( \ln 1 = 0 \).

Step 6: Calculate the Definite Integral

Plug in the evaluated value from Step 5: \( 2 \left[ 1^2 - 0^2 \right] = 2 \cdot 1 = 2 \). The final value of the integral is 2.

Key Concepts

Integration by Parts

Integration by parts is a crucial technique used to solve integrals that involve the product of two functions. The essence of this method is derived from the product rule for differentiation. When solving an integral using integration by parts, we typically choose two parts from the integrand: one to differentiate (denoted as \( u \)) and the other to integrate (denoted as \( dv \)). The formula for integration by parts is given by:\[\int u \, dv = uv - \int v \, du\]Choosing the correct \( u \) and \( dv \) is important and can sometimes be trial and error:

• Pick \( u \): Often, \( u \) is a function whose derivative simplifies the integrand, like \( \, \ln x \, \).
• Pick \( dv \): The remaining part of the integrand is chosen as \( dv \), so it should be easily integrable.

In our example, taking \( u = \ln x \) simplifies the differentiation, \( du = \frac{1}{x} \, dx \), while \( dv = \frac{2}{x} \, dx \) integrates easily to get \( v = 2 \ln x \). This approach simplifies the original integral, allowing you to replace it with easier sub-problems.
The solution then proceeds by subsituting these into our integration by parts formula to simplify and calculate step-by-step.

Natural Logarithms

Natural logarithms, denoted as \( \ln \, \), are logarithms to the base \( e \), where \( e \approx 2.718281828 \, \). The natural logarithm is a foundational concept in calculus and appears frequently in integration and differentiation processes.A few key properties of natural logarithms include:

• \( \ln 1 = 0 \): because \( e^0 = 1 \).
• \( \ln e = 1 \): because \( e^1 = e \).
• \( \ln(ab) = \ln a + \ln b \): log of a product is the sum of the logs.
• \( \ln\left( \frac{a}{b} \right) = \ln a - \ln b \): log of a quotient is the difference of the logs.
• \( \ln(a^b) = b \ln a \): log of a power is the power times the log.

When dealing with integrals involving \( \ln x \), such as in our example, these properties and rules for differentiation and integration of natural logs can greatly simplify the process. The derivative of \( \ln x \) is \( \frac{1}{x} \), which aids in solving integrals through integration by parts, making complex problems more manageable.

Change of Base for Logarithms

Changing the base of logarithms is often necessary for simplifying expressions and solving integrals, as seen in the original exercise problem. The change of base formula provides a method to convert logarithms from one base to another, typically moving to the natural logarithm base:\[\log_b a = \frac{\ln a}{\ln b}\]This formula allows us to rewrite any base logarithm in terms of the natural logarithm, which is often easier to handle due to its frequent occurrence in calculus and natural growth processes. In the problem, the term \( \log_{10} x \) becomes:
\[\frac{\ln x}{\ln 10}\]This transformation simplifies integration, allowing us to use known techniques around \( \ln x \). By converting base-10 logarithms into natural logs, we unify our expressions under one commonly used base, facilitating easier manipulation and computation within the realm of calculus. This clever change transforms complex integrands into a more tractable form, enhancing our ability to evaluate and solve integrals efficiently.