Problem 48 - Calculus Single Variable

Question

Evaluate each of the integrals. $$ \int 9 \sqrt{x} \ln (x) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(6 x^{3/2} \ln(x) - 4 x^{3/2} + C\).

Step 1: Identify the Form of Integration Required

This integral is a product of the form \( \int u \, dv \). For this specific problem, we will utilize integration by parts because it involves a product of two functions: \( \sqrt{x} \) and \( \ln(x) \).

Step 2: Choose Functions for Integration by Parts

According to integration by parts, \( \int u \, dv = uv - \int v \, du \). Let \( u = \ln(x) \) so that \( du = \frac{1}{x} \, dx \), and \( dv = 9 \sqrt{x} \, dx \) so that \( v \) will be obtained by integrating \( dv \).

Step 3: Compute v by Integrating dv

Integrate \( dv = 9 \sqrt{x} \, dx \) to find \( v \). Since \( \sqrt{x} = x^{1/2} \), integrating gives:\[\int 9 x^{1/2} \, dx = 9 \cdot \frac{x^{3/2}}{3/2} = 6 x^{3/2}.\] Thus, \( v = 6 x^{3/2} \).

Step 4: Apply Integration by Parts Formula

Substitute \( u, du, v, \) and \( dv \) into the integration by parts formula:\[\int 9 \sqrt{x} \ln(x) \, dx = \ln(x) \cdot 6 x^{3/2} - \int 6 x^{3/2} \cdot \frac{1}{x} \, dx.\] Simplify the second term by rewriting it as:\[6 \int x^{1/2} \, dx = 6 \cdot \frac{x^{3/2}}{3/2} = 4 x^{3/2}.\]

Step 5: Simplify and Write Final Answer

Combine the terms from the integration by parts result:\[6 x^{3/2} \ln(x) - 4 x^{3/2} + C,\] where \( C \) represents the constant of integration. The final answer is:\[6 x^{3/2} \ln(x) - 4 x^{3/2} + C.\]

Key Concepts

Integration by Parts

Integration by parts is a powerful method to solve integrals that involve the product of two functions. The basic idea stems from the product rule for differentiation, which can be turned around to serve as a method for integration. The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

In this formula, you choose one part of the integral to differentiate (\( u \)) and another part to integrate (\( dv \)). This method is particularly useful when dealing with a product like \( 9 \sqrt{x} \ln(x) \).
In the exercise, we identified \( u \) as \( \ln(x) \) because its derivative, \( \frac{1}{x} \), is simpler to handle. Meanwhile, \( dv \) was chosen as \( 9 \sqrt{x} \, dx \), making it easier to find \( v \) through integration.
This strategic choice simplifies the integral significantly, leveraging the symmetry of the functions involved. The integration by parts method transforms a challenging integral into simpler terms that can be evaluated directly or require less demanding integration techniques.

Definite and Indefinite Integrals

Integrals come in two primary forms: definite and indefinite. Indefinite integrals represent a family of functions and include an arbitrary constant \( C \):

• \( \int f(x) \, dx = F(x) + C \)

This indefinite nature arises because when you differentiate a constant, it vanishes; hence, integration can only recover the anti-derivatives up to a constant.
In contrast, a definite integral computes the exact area under a curve between two points, producing a numerical value. It is evaluated over a specified interval \([a, b]\):

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)

In the exercise, we computed an indefinite integral since no limits were given. This results in an expression plus the constant \( C \). If limits were provided, we would use them to find the numerical value of the area under the curve.As integrating functions like \( 9 \sqrt{x} \ln(x) \) often results in complex expressions, specifying the integral type is crucial for understanding the problem and its context.

Logarithmic Integration

Logarithmic integration is a technique used when a logarithm appears as part of the function to integrate. Often, this method ties closely with integration by parts, especially when dealing with products involving \( \ln(x) \).
The integral of \( \ln(x) \) itself, \( \int \ln(x) \, dx \), doesn't have a straightforward antiderivative in elementary functions. However, it can be elegantly solved using integration by parts by setting:

• \( u = \ln(x) \)
• \( dv = dx \)
• \( du = \frac{1}{x} \, dx \)
• \( v = x \)

Applying integration by parts, we find:

• \( \int \ln(x) \, dx = x \ln(x) - \int x \cdot \frac{1}{x} \, dx = x \ln(x) - x + C \)

In our specific exercise, \( \ln(x) \) served as \( u \), simplifying our work with the integration by parts formula. Recognizing these patterns can make tackling integrals with logarithmic functions much more manageable.