Problem 71 - Thomas Calculus

Question

Evaluate the integrals. \(\int_{1}^{\ln x} \frac{1}{t} d t, \quad x>1\)

Step-by-Step Solution

Answer

The integral evaluates to \( \ln(\ln(x)) \).

Step 1: Understand the Problem

The problem asks for the evaluation of a definite integral with variable limits. The integrand is \( \frac{1}{t} \) and the limits are from 1 to \( \ln(x) \).

Step 2: Identify the Antiderivative

The antiderivative of \( \frac{1}{t} \) is \( \ln|t| \). Since \( t > 0 \) in the range from 1 to \( \ln(x) \), we can use \( \ln(t) \) directly.

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, \( \int_{a}^{b} f(t) \, dt = F(b) - F(a) \), where \( F(t) \) is the antiderivative of \( f(t) \). Here, \( F(t) = \ln(t) \).

Step 4: Evaluate the Antiderivative at the Upper Limit

Substitute \( \ln(x) \) into the antiderivative: \( F(\ln(x)) = \ln(\ln(x)) \).

Step 5: Evaluate the Antiderivative at the Lower Limit

Substitute 1 into the antiderivative: \( F(1) = \ln(1) = 0 \), since the logarithm of 1 is 0.

Step 6: Calculate the Definite Integral

Subtract the result from Step 5 from the result of Step 4: \( \ln(\ln(x)) - 0 = \ln(\ln(x)) \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a key idea connecting differentiation and integration. It consists of two main parts: first, it shows how an antiderivative can reverse differentiation, and second, it relates the total accumulation of a quantity to the values of its antiderivative at specific points.
In practice, this theorem states that if you have a continuous function, like our integrand \( \frac{1}{t} \), and you know its antiderivative \( F(t) \), then the definite integral from \( a \) to \( b \) of this function is simply the difference \( F(b) - F(a) \).

• \( \int_{a}^{b} f(t) \, dt = F(b) - F(a) \)

This is incredibly useful because it allows us to turn an infinite process into a simple subtraction, making calculations manageable. In the problem, using the antiderivative \( F(t) = \ln(t) \) of \( \frac{1}{t} \), we evaluate at \( \ln(x) \) and 1 to find the solution.

Antiderivative

An antiderivative of a function is another function that reverses its differentiation. For instance, if differentiating \( F(t) \) gives you \( f(t) \), then \( F(t) \) is an antiderivative of \( f(t) \). In our exercise, \( \frac{1}{t} \) is the function, and its antiderivative is \( \ln|t| \).
The concept is crucial because it provides a way to simplify the computation of integrals. Once you know that \( \ln(t) \) is an antiderivative of \( \frac{1}{t} \), integrating over any interval can be done through evaluation of this antiderivative at the boundary points.

• Antiderivatives provide a direct path to calculating definite integrals.
• Knowing \( F(t) \) lets us use it in the Fundamental Theorem of Calculus.

It's helpful to think of an antiderivative as the opposite of finding a derivative, effectively reversing the process to find a "primitive" form of the function.

Logarithmic Functions

Logarithmic functions are functions that use the logarithm as their basis. For instance, the natural logarithm, \( \ln(x) \), is the inverse of the exponential function with the base \( e \).
In our context, the problem evaluates integrals using the natural logarithm, because \( \ln(t) \) is an antiderivative of \( \frac{1}{t} \). This use highlights a key property of logarithms: they are useful in expressing the area under curves for functions of the form \( \frac{1}{t} \).

• Natural logarithms are identified by \( \ln \) and deal with base \( e \), where \( e \approx 2.718 \).
• The function \( \ln(x) \) is essential for antiderivatives of \( \frac{1}{t} \), making it a recurring element in calculus problems.

In simpler terms, if you need to recapture the accumulated change represented by \( \frac{1}{t} \), knowing logarithmic functions will provide the answer immediately—just calculate the natural logarithm of the boundary points.