Problem 73 - Thomas Calculus

Question

Evaluate the integrals. \(\int_{1}^{1 / x} \frac{1}{t} d t, \quad x>0\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Integral Limits

The integral given is \( \int_{1}^{1 / x} \frac{1}{t} \, dt \). This is a definite integral with variable limits of integration. The lower limit is 1 and the upper limit is \( \frac{1}{x} \).

Step 2: Identify the Integral Formula

The integrand \( \frac{1}{t} \) is a common function, and its antiderivative is \( \ln |t| \).

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, we evaluate the antiderivative \( \ln |t| \) from the limits 1 to \( \frac{1}{x} \):\[ \int_{1}^{1/x} \frac{1}{t} dt = \ln \left| \frac{1}{x} \right| - \ln |1| \]

Step 4: Simplify the Expression

Since \( \ln |1| = 0 \), simplify the expression:\[ \ln \left| \frac{1}{x} \right| = \ln \left| 1 \right| \] becomes\[ \ln \left| \frac{1}{x} \right| = \ln \left(\frac{1}{x}\right) \]Using logarithmic properties, \( \ln \left( \frac{A}{B} \right) = \ln A - \ln B \), it becomes:\[ -\ln x \]

Step 5: Final Answer

Since there is no division by zero and \( x > 0 \), the valid solution for the original integral is:\[ -\ln x \]

Key Concepts

Definite Integrals

A definite integral gives the area under a curve between two specific points on the x-axis. In the integral \( \int_{1}^{1/x} \frac{1}{t} \, dt \), the definite integral setup means that we calculate the total accumulation of the function \( \frac{1}{t} \) between the limits \( t = 1 \) and \( t = \frac{1}{x} \). These boundaries are the start and end points of the interval over which we are integrating.The result of a definite integral is a numerical value, providing a measure of the area under the curve. In our case, since the limits of integration are variable, the result reflects how this area changes with different values of \( x \). This changing area is what gives rise to a function, rather than a single number, as the result of the integration process.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the gap between differentiation and integration. Essentially, it shows that these two processes are inversely related. The theorem has two main parts. The first part states that if a function is continuous over an interval, then the integral of the function over that interval is related to its antiderivative.In practical terms, when calculating a definite integral like \( \int_{1}^{1/x} \frac{1}{t} \, dt \), the theorem allows us to evaluate the integral by finding the antiderivative of \( \frac{1}{t} \) and subtracting its values at the endpoints. This means we compute the antiderivative at \( \frac{1}{x} \) and \( 1 \), and then subtract:

• Evaluate at \( t = \frac{1}{x} \)
• Subtract the evaluation at \( t = 1 \)

By doing so, we transform the problem of finding an area into a simpler arithmetic task.

Antiderivatives

An antiderivative of a function is another function whose derivative is the initial function. For instance, if given a function \( \frac{1}{t} \), its antiderivative is \( \ln |t| \). This means that if we differentiate \( \ln |t| \), we return to \( \frac{1}{t} \).Finding the antiderivative is a key step in solving definite integrals. Once the antiderivative is determined, you can use it to apply the Fundamental Theorem of Calculus. In mathematical terms, you can express the antiderivative as:\[ F(t) = \ln |t| + C \]Where \( C \) is the constant of integration, though it disappears in definite integrals since it cancels out.

Logarithmic Functions

Logarithmic functions are functions of the form \( \ln(x) \), which are natural logarithms. These functions are the inverse operations of the exponential function \(e^x\). In the context of our integral \( \int_{1}^{1/x} \frac{1}{t} \, dt \), the presence of \( \frac{1}{t} \) directly leads to a logarithmic function via integration.The specific logarithmic function that arises from integrating \( \frac{1}{t} \) is \( \ln |t| \). Logarithms have unique properties, such as:

• \( \ln(a) + \ln(b) = \ln(ab) \)
• \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \)

These properties are instrumental in simplifying expressions during integral evaluations and are used in our solution when simplifying \( \ln \left( \frac{1}{x} \right) \) to \( -\ln x \). Understanding logarithmic behavior is essential in manipulating and understanding solutions involving logarithms.