Problem 27 - Applied Calculus

Question

Use a calculator or computer to evaluate the integral. $$ \int_{1}^{3} \ln x d x $$

Step-by-Step Solution

Answer

The integral evaluates to approximately 1.2958.

Step 1: Understand the Problem

We are asked to evaluate the definite integral of the natural logarithm of \( x \) from \( x = 1 \) to \( x = 3 \). This can be found using integration techniques and technology for exact evaluation.

Step 2: Set Up the Integral Using the Fundamental Theorem of Calculus

First, identify an antiderivative for \( \ln(x) \). The antiderivative of \( \ln(x) \) is \( x \ln(x) - x + C \), where \( C \) is a constant of integration. Using the Fundamental Theorem of Calculus, we will compute the definite integral by evaluating the antiderivative at the bounds.

Step 3: Evaluate the Antiderivative at the Upper Limit

Substitute \( x = 3 \) into the antiderivative: \( F(3) = 3\ln(3) - 3 \). Calculate this value using a calculator if necessary.

Step 4: Evaluate the Antiderivative at the Lower Limit

Substitute \( x = 1 \) into the antiderivative: \( F(1) = 1\ln(1) - 1 = -1 \). Use the property that \( \ln(1) = 0 \).

Step 5: Calculate the Definite Integral

Apply the Fundamental Theorem of Calculus: the value of the integral is \( F(3) - F(1) = (3\ln(3) - 3) - (-1) \). Simplify to get \( 3\ln(3) - 2 \).

Step 6: Use Technology for the Final Evaluation

Use a calculator or computer software to evaluate \( 3\ln(3) - 2 \). This should give approximately \( 1.2958 \).

Key Concepts

Definite Integral

A definite integral is a powerful concept in integral calculus, representing the area under a curve within a specified interval. In this context, we deal with the curve defined by the function \( f(x) = \ln x \) and the interval from \( x = 1 \) to \( x = 3 \).

• Interval of Integration: The specific segment along the x-axis for which we want to calculate the area under the curve is crucial. In our case, it starts at \( x = 1 \) and ends at \( x = 3 \).
• Area under the Curve: The definite integral gives us this area. It's important to understand that the value can be negative if the curve lies below the x-axis. However, with \( \ln x \), this is not an issue within our interval.

Definite integrals help encapsulate complex information about functions within fixed bounds, revealing various properties such as accumulated change, total distance, or area.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the gap between differentiation and integration. It has two main parts: the derivative of an integral providing the original function, and the evaluation of a definite integral via antiderivatives.

• Antiderivative Connection: By finding an antiderivative of the function \( f(x) \), we can compute the definite integral using its values at the boundaries of the interval.
• Evaluation Process: By evaluating this antiderivative at the upper and lower bounds, we use: \[ F(b) - F(a) \] where \( a \) and \( b \) are the limits of integration.

In our example, the theorem allows us to turn the problem of finding the integral into simply evaluating \( x \ln x - x \) at \( x = 3 \) and \( x = 1 \), providing a straightforward way to calculate the area.

Antiderivatives

Antiderivatives, or indefinite integrals, are essential for solving definite integrals. An antiderivative of a function \( f(x) \) is another function \( F(x) \) such that \( F'(x) = f(x) \). It's like running differentiation in reverse.

• Finding Antiderivatives: These are not always straightforward, but with practice, identifying common patterns becomes easier. For instance, the antiderivative of \( \ln x \), though not immediately obvious, is \( x \ln x - x + C \).
• Role in Definite Integrals: Once we have \( F(x) \), the antiderivative, we can use it in the Fundamental Theorem of Calculus to aid in our evaluation of the definite integral.

Antiderivatives are often a key step in working with integral calculus, effectively reducing complex problems to simpler arithmetic evaluations.