Problem 114 - Thomas Calculus

Question

Evaluate the integrals. \begin{equation}\frac{1}{\ln a} \int_{1}^{x} \frac{1}{t} d t, \quad x>0\end{equation}

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\ln |x|}{\ln a} \).

Step 1: Recognize the Integral Form

The integral given is \( \int_{1}^{x} \frac{1}{t} \, dt \). This is a known integral that fits the form \( \int \frac{1}{t} \, dt = \ln |t| + C \).

Step 2: Evaluate the Indefinite Integral

Using the known result, evaluate the indefinite integral from 1 to \( x \): \( \int \frac{1}{t} \, dt = \ln |t| + C \).

Step 3: Apply the Definite Integral

Apply the limits of integration to the indefinite integral: \[ \left. \ln |t| \right|_1^x = \ln |x| - \ln 1 \].

Step 4: Simplify the Expression

Since \( \ln 1 = 0 \), the expression simplifies to \( \ln |x| \). This is because subtracting zero does not change the value.

Step 5: Multiply by the Constant Factor

The original expression has a factor \( \frac{1}{\ln a} \). Multiply this factor with the result from Step 4 to get \( \frac{1}{\ln a} \cdot \ln |x| \).

Key Concepts

Indefinite Integral

An indefinite integral is a fundamental concept in calculus, often symbolized by the integral sign without specified limits. It represents the collection of all antiderivatives of a function. To find an indefinite integral, you are essentially looking to reverse the process of differentiation.

In the case of the integral \( \int \frac{1}{t} \, dt \), the solution emerges as \( \ln |t| + C \). Here, \( C \) is the constant of integration, which accounts for the fact that indefinite integrals represent families of functions either shifted up or down a vertical axis.

When dealing with indefinite integrals, remember:

• The presence of a constant \( C \), allowing for any vertical shift.
• The concept of antiderivation which is the reverse of differentiation. For example, differentiating \( \ln |t| \) gives you \( \frac{1}{t} \).
• Indefinite integrals don't concern themselves with specific limits of integration.

Integration Limits

When transitioning from an indefinite to a definite integral, integration limits come into play. These are specified values, placed at the upper and lower boundaries of the integral symbol. They transform an indefinite problem into one with precise numerical outcomes.

For our integral \( \int_{1}^{x} \frac{1}{t} \, dt \), the limits are 1 and \( x \). You solve a definite integral by calculating the antiderivative at the upper limit (\( x \)) and subtracting the antiderivative evaluated at the lower limit (1):

• Upper limit: \( \ln |x| \)
• Lower limit: \( \ln 1 = 0 \)

The result is \( \ln |x| - \ln 1 = \ln |x| \).

Considering integration limits is crucial as they define the interval over which you are accumulating the area under the curve.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \approx 2.71828 \). It's highly beneficial in calculus because it's directly connected with the function \( y = e^x \).

A key property of the natural logarithm is that \( \ln 1 = 0 \), pivotal for simplifying expressions like \( \ln |x| - \ln 1 \). Another property used in integrals is that the derivative of \( \ln |x| \) is \( \frac{1}{x} \), which aids in identifying the antiderivative of \( \frac{1}{t} \).

Overall, knowing how to work with \( \ln \) is essential when solving integrals, especially those involving exponential growth and decay, or in this case, functions like \( \frac{1}{t} \). It's a foundational element in algebra and calculus.

Constant Factor Multiplication

Constant factor multiplication in integration allows you to simplify calculations by taking constants out of the integral. This maneuver enhances the ability to focus on the function's core behavior during the integration process.

In the given problem: \( \frac{1}{\ln a} \int_{1}^{x} \frac{1}{t} \, dt \), \( \frac{1}{\ln a} \) is the constant factor. Multiplying the final result of the definite integral by this factor, you end up with:

• \( \ln |x| \) from evaluation of the integral.
• Multiply by \( \frac{1}{\ln a} \), giving \( \frac{1}{\ln a} \cdot \ln |x| \).

This property aids in managing integration tasks by isolating any constant multiple, ensuring that you correctly scale the resultant value of the integral. Remember, constant factors remain outside unless you're integrating them by themselves (like \( \int c \, dx = cx + C \)).