Problem 43 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int_{1}^{2} \frac{2 \ln x}{x} d x$$

Step-by-Step Solution

Answer

The integral evaluates to \((\ln 2)^2\).

Step 1: Recognize the Integral Form

Identify that the integral \( \int \frac{\ln x}{x} \, dx \) suggests a substitution method, as it involves a function \( \ln x \) and its derivative divided by \( x \).

Step 2: Choose a Variable Substitution

To simplify the integral, use the substitution \( u = \ln x \). Then \( du = \frac{1}{x} \, dx \). Rewriting the differential, we have \( dx = x \, du \), but with the substitution, this simplifies easily to \( du = \frac{1}{x} \, dx \).

Step 3: Change the Limits of Integration

With the substitution \( u = \ln x \), calculate the new limits of integration: when \( x = 1 \), \( u = \ln 1 = 0 \); when \( x = 2 \), \( u = \ln 2 \).

Step 4: Convert the Integral into the New Variable

Substitute into the integral to get \( \int_{0}^{\ln 2} 2u \, du \). This transformation simplifies the original problem.

Step 5: Solve the New Integral

Evaluate the new integral \( \int 2u \, du \) as follows: The integral of \( 2u \) is \( u^2 + C \), thus evaluate \( \left[u^2\right]_{0}^{\ln 2} \).

Step 6: Apply the Fundamental Theorem of Calculus

Compute \( \left[u^2\right]_{0}^{\ln 2} = \left(\ln 2\right)^2 - (0)^2 \). Since \( (0)^2 = 0 \), the expression simplifies to \( (\ln 2)^2 \).

Step 7: Finalize the Solution

Thus, the evaluated integral is \( (\ln 2)^2 \).

Key Concepts

Definite Integral

A definite integral represents the area under a curve within specific limits on the x-axis. It is usually written in the form \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the limits of integration. This specific exercise involves evaluating a definite integral with limits from 1 to 2. The definite integral calculates the net area bounded by the curve defined by the integrand, which is the function inside the integral, and the x-axis, between these two points.

As the name suggests, a definite integral results in a specific value, unlike an indefinite integral which gives a function (and includes a constant of integration). When computing, it's crucial to change the limits appropriately when using substitution methods, to reflect the new variable in use. This ensures the area is calculated correctly across the transformed coordinates.

Substitution Method

The substitution method is a common technique used in calculus to simplify integrands and make them more manageable. This method is particularly useful when dealing with functions that are compositions, such as a function and its derivative.To apply substitution effectively:

• Identify parts of the integral that resemble a derivative and its base function.
• Select a substitution variable, often denoted as \( u \), to simplify the expression. For example, in this exercise, by choosing \( u = \ln x \), the integral transforms significantly.
• Determine the differential \( du \) using \( du = \frac{d}{dx}(\ln x) \cdot dx \), leading here to \( du = \frac{1}{x} \, dx \).
• Changing the limits of integration according to your substitution is crucial to evaluate the new definite integral correctly.

This results in a much simpler integral \( \int_{0}^{\ln 2} 2u \, du \), which is easier to evaluate.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a key principle in calculus that connects differentiation with integration. It has two main parts:

• The first part allows us to evaluate definite integrals using antiderivatives. It states that if \( F \) is an antiderivative of \( f \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).
• The second part states that if \( f \) is a continuous function on an interval \([a, b]\), and \( F \) is its antiderivative, the derivative of \( F \) with respect to \( x \) gives back \( f(x) \).

In the exercise, after transforming the integral with substitution, the Fundamental Theorem of Calculus helps to find \( [u^2]_{0}^{\ln 2} \). By evaluating the antiderivative at the new limits, we find \( (\ln 2)^2 \) as the solution, which shows how integration accumulates results from derivative functions.