Problem 46 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(37-54\). $$ \int_{2}^{16} \frac{d x}{2 x \sqrt{\ln x}} $$

Step-by-Step Solution

Answer

\( \sqrt{\ln 2} \)

Step 1: Identify the Integral

We are given the integral \[ \int_{2}^{16} \frac{d x}{2 x \sqrt{\ln x}} \] with limits from 2 to 16 defined by the substitution rule.

Step 2: Choose a Substitution

Let us set \( u = \ln x \). This implies that the derivative is \( dx = \frac{1}{x} du \). The substitution thus becomes \[ dx = \frac{du}{x} \] and we can express \( \ln x \) as \( u \).

Step 3: Convert the Limits of Integration

When \( x = 2 \), \( u = \ln 2 \). When \( x = 16 \), \( u = \ln 16 \). Hence, the limits of integration in terms of \( u \) are from \( \ln 2 \) to \( \ln 16 \).

Step 4: Substitute into the Integral

Substitute \( u \) into the integral:\[ \int_{\ln 2}^{\ln 16} \frac{1}{2\sqrt{u}} du \]. In this form, the integral is simplified to be solved in the next steps.

Step 5: Simplify the Integral

Factor out the constant before the integral:\[ \frac{1}{2} \int_{\ln 2}^{\ln 16} u^{-1/2} du \]. This makes it easier to integrate.

Step 6: Integrate

Apply the power rule for integrals. Recall that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \):\[ \frac{1}{2} \left[ 2u^{1/2} \right]_{\ln 2}^{\ln 16} \].

Step 7: Evaluate the Definite Integral

Evaluate the antiderivative, first by plugging in the upper limit, and then subtracting the result of the lower limit:\[ = \left[ \sqrt{\ln 16} - \sqrt{\ln 2} \right]. \]

Step 8: Simplify the Expression

Recognize that \( \ln 16 = 4\ln 2 \) since 16 is \( 2^4 \). Therefore, the integral evaluates to:\[ \sqrt{4}\ln 2 - \sqrt{\ln 2} = 2\sqrt{\ln 2} - \sqrt{\ln 2} = \sqrt{\ln 2}. \]

Key Concepts

Definite Integrals

Definite integrals represent the area under a curve between two points on the x-axis. Unlike indefinite integrals, they promise a single numerical answer. When you see a definite integral, it is expressed with limits of integration at the top and bottom of the integral sign. In our exercise, these limits are 2 and 16.
Definite integrals differ from indefinite ones because they don't include a constant of integration, as they compute an exact value. This value can represent quantities like distance, area, or volume. In essence, definite integrals help capture what lies between specified boundaries on a function.
When solving definite integrals, follow these steps:

• Integrate the function as you would for an indefinite integral.
• Plug in the upper limit into the antiderivative (result of integration).
• Subtract the result of plugging the lower limit into the antiderivative from the upper limit's result.

Upon finishing these steps, you’ll achieve an exact numerical value that represents the area or other quantity you are evaluating.

Substitution Rule

The substitution rule is a handy tool for integrating complex functions. It allows you to transform a tough integral into a simpler one by using a clever change of variables.
In the given exercise, choosing a substitution makes the integral much more approachable. By setting a difficult part of the function to a new variable (like setting \( u = \ln x \) in our example), you change the entire form of the integral.
Here’s how to apply the substitution rule:

• Select a substitution \( u \) for a part of the function so that \( du \) accounts for the integral’s complexity.
• Change the differential \( dx \) to \( du \) using the derivative of \( u \).
• Convert original integral limits into limits in terms of \( u \).
• Substitute and transform the entire integral into terms of \( u \). Then solve the simpler integral.
• Don’t forget to switch back to the original variable after integrating if dealing with indefinite integrals.

This method not only simplifies the computation but also broadens your tools for tackling diverse integral problems.

Power Rule for Integrals

The power rule for integrals is a fundamental technique for solving a wide array of indefinite integrals. It simplifies integrating functions of the form \( u^n \). In the exercise, after using substitution, the integral becomes a form ideal for the power rule.
The generic form of the power rule is:

• \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), where \( n eq -1 \).

It’s a quick and effective way to find the antiderivative of powers of \( u \). Here's how to apply it:

• Identify the exponent \( n \) and increment it by one to get \( n+1 \).
• Divide \( u^{n+1} \) by \( n+1 \).
• When dealing with a definite integral, evaluate at the boundaries using the fundamental theorem of calculus.

By mastering the power rule, you become well-equipped to solve numerous integration challenges, simplifying otherwise complex tasks.