Problem 46 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int_{2}^{16} \frac{d x}{2 x \sqrt{\ln x}}$$

Step-by-Step Solution

Answer

The value of the integral is \( \sqrt{\ln 2} \).

Step 1: Understand the Integral

We need to evaluate the integral \( \int_{2}^{16} \frac{d x}{2 x \sqrt{\ln x}} \). The integrand is \( \frac{1}{2 x \sqrt{\ln x}} \). Observing the integrand, we note that \( \ln x \) might be a candidate for substitution during integration.

Step 2: Perform a Substitution

Let \( u = \ln x \). Then, \( du = \frac{1}{x} \, dx \). This substitution simplifies the integration process because it relates \( dx \) with \( du \). Also, the limits of integration transform as follows: when \( x = 2 \), \( u = \ln 2 \), and when \( x = 16 \), \( u = \ln 16 = 4 \ln 2 \).

Step 3: Rewrite the Integral in Terms of u

Substitute \( u = \ln x \) into the integral. The integral becomes \( \int_{\ln 2}^{4 \ln 2} \frac{1}{2 \sqrt{u}} \, du \). This step simplifies the expression inside the integral.

Step 4: Evaluate the New Integral

Evaluate the integral \( \int \frac{1}{2 \sqrt{u}} \, du \). This is an integral of the form \( \int u^{-\frac{1}{2}} \, du \), which can be integrated to \( 2\sqrt{u} + C \). Therefore, \( \int \frac{1}{2 \sqrt{u}} \, du = \sqrt{u} + C \).

Step 5: Apply the Limits of Integration

Substitute the limits back into the integrated function. We calculate \( \left[ \sqrt{u} \right]_{\ln 2}^{4 \ln 2} = \sqrt{4 \ln 2} - \sqrt{\ln 2} \).

Step 6: Simplify the Result

Simplify the expression \( \sqrt{4 \ln 2} - \sqrt{\ln 2} \) to express it as \( \sqrt{4} \cdot \sqrt{\ln 2} - \sqrt{\ln 2} = 2\sqrt{\ln 2} - \sqrt{\ln 2} = \sqrt{\ln 2} \).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in integral calculus that allows us to calculate the accumulated value of a function within specific limits. Unlike indefinite integrals, which represent a family of functions, definite integrals evaluate the specific area under a curve between two points on the x-axis.
When you compute a definite integral, you find the net change in a function across the given interval, typically denoted \( \int_{a}^{b} f(x) \, dx \). Here, \( a \) and \( b \) are known as the limits of integration.

• The lower limit, \( a \), defines the starting point of the calculation.
• The upper limit, \( b \), specifies where the calculation ends.

The main property of a definite integral is that it is independent of the variable of integration; it solely depends on the limits and the function itself. In essence, the process of evaluating a definite integral involves performing an antiderivative and then applying the limits to find the accumulated area.
Understanding definite integrals is crucial, as they appear in various practical applications ranging from physics to economics, helping quantify things like displacement, total cost, or even probability.

Integration by Substitution

Integration by substitution is a powerful technique used to simplify the process of finding integrals. This method is particularly useful when dealing with complicated integrands that seem difficult to tackle directly. It is analogous to the chain rule in differentiation.
The main idea is to transform a complex integral into a simpler one by substituting part of the integrand with a new variable. This often turns a challenging problem into a basic one that can be easily solved. When applying substitution, follow these steps:

• Choose a substitution that simplifies the integral. Typically, choose expressions inside a square root, polynomial, or a function that appears repeatedly.
• Let \( u = g(x) \), where \( g(x) \) is the chosen substitution. Find \( du = g'(x) \, dx \). This relates the new variable \( u \) to the old variable \( x \).
• Adjust the limits of integration to reflect the new variable: transform \( x = a \) to \( u = g(a) \) and \( x = b \) to \( u = g(b) \).
• Rewrite the integral in terms of \( u \) and solve the simpler integral.
• Finally, replace \( u \) with the original expression to find the final answer if necessary.

In our problem, substituting \( u = \ln x \) transformed the initial complex integral into an easier one, allowing us to integrate it efficiently.

Logarithmic Functions

Logarithmic functions are a critical component of calculus and mathematics as a whole, frequently appearing in integration problems. The natural logarithm, denoted as \( \ln(x) \), is a logarithm with base \( e \), where \( e \) is approximately 2.718. It is essential for tackling growth problems and decay models.
Logarithmic functions are the inverse operations to exponential functions. This relationship means that if \( y = \ln x \), then \( x = e^y \). Here are some important properties:

• The domain of \( \ln x \) is \( (0, \infty) \), meaning it only accepts positive values.
• The natural logarithm of 1 is always 0, that is, \( \ln 1 = 0 \).
• The derivative of \( \ln x \) is \( \frac{1}{x} \), which signifies the rate of change of the logarithmic function.

When utilizing logarithmic functions in integrals, particularly in the context of the integration by substitution, they often help simplify the complexity of the problem at hand. For instance, knowing that \( \ln x \) was involved in the original complex integral offers a clear path for substitution, thereby easing the integration process.