Problem 48 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(39-56\) $$ \int_{2}^{16} \frac{d x}{2 x \sqrt{\ln x}} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \sqrt{\ln 2} \).

Step 1: Substitution

Let's perform a substitution to simplify the integral. Set \( u = \ln x \), which implies \( du = \frac{1}{x} dx \), and thus \( dx = x du \). Also, \( x = e^u \). Consequently, when \( x = 2 \), \( u = \ln 2 \), and when \( x = 16 \), \( u = \ln 16 = 4 \ln 2 \).

Step 2: Change of Variable

Substitute the expressions in the integral: \[ \int_{\ln 2}^{4 \ln 2} \frac{udu}{ 2 e^u \sqrt{u}}. \]This simplifies to \[ \int_{\ln 2}^{4 \ln 2} \frac{du}{2 \sqrt{u}} \] because \( e^u = x \) and \( x \) cancels out.

Step 3: Simplifying the Integral

Rewriting the integral, we have:\[ \int_{\ln 2}^{4 \ln 2} \frac{du}{2 \sqrt{u}} = \frac{1}{2} \int_{\ln 2}^{4 \ln 2} u^{-1/2} du. \]This is a standard integral form and can be solved using the power rule.

Step 4: Applying the Power Rule

The integral now is:\[ \frac{1}{2} \int u^{-1/2} du. \]Using the power rule for integration, \( \int u^n du = \frac{u^{n+1}}{n+1} + C \), this integral becomes:\[ \frac{1}{2} \left( 2u^{1/2} \right) = u^{1/2}. \]

Step 5: Evaluate the Definite Integral

Evaluate the definite integral from \( \ln 2 \) to \( 4 \ln 2 \):\[ \left[ u^{1/2} \right]_{\ln 2}^{4 \ln 2} = (4 \ln 2)^{1/2} - (\ln 2)^{1/2}. \]

Step 6: Simplifying the Result

Simplify the expression: \[ \sqrt{4 \ln 2} - \sqrt{\ln 2} = 2\sqrt{\ln 2} - \sqrt{\ln 2} = \sqrt{\ln 2}. \]

Key Concepts

Definite Integrals

In the world of calculus, definite integrals help in measuring the area under a curve within certain bounds. These bounds are the limits of integration, meaning they tell us from where to where we calculate the area. When we solve a definite integral, we always have two boundary values to consider—the lower limit and the upper limit.

• Definite integral notation: \( \int_{a}^{b} f(x) \, dx \) defines the area under the function \( f(x) \) from \( x = a \) to \( x = b \).
• The result is a numerical value that represents this area.

To efficiently solve definite integrals, it's crucial to select the correct limits. In our solution, as we perform substitution, our limits also change according to our new variable. Notice how we converted limits from \( x \) values to \( u \) values appropriately.
For example, in the exercise, the integration from \( x = 2 \) to \( x = 16 \) translates to \( u = \ln 2 \) to \( u = 4 \ln 2 \) after substitution.

Substitution Method

The substitution method is an essential tool for solving integrals, especially when dealing with complex expressions. Also known as the change of variable technique, it simplifies the integration process by transforming challenging integrals into a more manageable form.

• Identify a part of the integral to substitute. Choose \( u \) such that it simplifies the expression, often setting \( u = g(x) \).
• Compute \( du \) by differentiating your choice of \( u \). Replace \( dx \) in the integral with \( du \).
• Don't forget to convert the limits of integration if computing a definite integral. Substitution affects these limits.

In our step-by-step solution, we chose \( u = \ln x \), which eases the integration due to its presence in the original function. Substituting transformed the original limits from \( x \) values to \( u \) values, leading to a simpler form of the integral that we later evaluated.

Power Rule for Integration

Once we have simplified an integral, the power rule for integration is a straightforward method to solve it, provided the function is in a power form. This rule is highly applicable when dealing with polynomials or variables raised to a power.

• The power rule states: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \).
• This formula allows us to integrate terms like \( u^{-1/2} \), transforming it into \( u^{1/2}/(1/2+1) \).

In the exercise solution, after substitution, we rewrite \( \int \frac{1}{2 \sqrt{u}} \, du \) as \( \frac{1}{2} \int u^{-1/2} \, du \). By applying the power rule, we easily find the integral's result. Finally, we calculate the definite integral using this resolved form by evaluating it at the limits. This application of the power rule greatly simplifies finding the answer to a complex-looking integral.