Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-36, use integration by parts to evaluate each integral. $$ \int_{2}^{3} \frac{\ln 2 x^{5}}{x^{2}} d x $$

Step-by-Step Solution

Answer

\(- \frac{1}{3} \ln(486) + \frac{1}{2} \ln(32) + \frac{5}{6}\)

Step 1: Identify Parts for Integration By Parts

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). We need to choose \( u \) and \( dv \) such that the integration becomes simpler. Here, we choose \( u = \ln(2x^5) \) and \( dv = x^{-2} \, dx \).

Step 2: Differentiate and Integrate Parts

Differentiate \( u \) to find \( du \):\( u = \ln(2x^5) \)\( du = 5x^{-1} \, dx \).Integrate \( dv \) to find \( v \):\( dv = x^{-2} \, dx \)\( v = -x^{-1} \).

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:\[ \int \ln(2x^5)x^{-2} \, dx = - \ln(2x^5) x^{-1} \bigg|_2^3 - \int_2^3 (-x^{-1}) (5x^{-1} \, dx) \].

Step 4: Simplify the Integral

The integral becomes:\(- \ln(2x^5) x^{-1} \bigg|_2^3 + \int_2^3 5x^{-2} \, dx \).Evaluate the simpler integral:\( \int 5x^{-2} \, dx = -5x^{-1} \).

Step 5: Evaluate Definite Integrals

Evaluate the expression:\( - \ln(2(3)^5) \times \frac{1}{3} + \ln(2(2)^5) \times \frac{1}{2}\) and\( + [-5x^{-1}]_2^3 = [-5 \cdot \frac{1}{3} + 5 \cdot \frac{1}{2}] \).Evaluate both to get the final numerical result.

Step 6: Combine Results for Final Calculation

Substitute the values to get\(- \frac{1}{3} \ln(486) + \frac{1}{2} \ln(32) + \left( -\frac{5}{3} + \frac{5}{2} \right) \).Calculate these expressions to get the result:\[- \frac{1}{3} \ln(486) + \frac{1}{2} \ln(32) + \frac{5}{6} \].

Key Concepts

Definite Integration

Definite integration involves calculating the area under a curve from one point to another. In the given exercise, we want to find the definite integral of the function \( \frac{\ln 2 x^5}{x^2} \) from \( x = 2 \) to \( x = 3 \). This means we are looking at the total change or accumulation of the logarithmic function over this interval.

• The definite integration distinguishes from indefinite integration because it provides a numerical value rather than a function plus a constant.
• In this exercise, definite integration is handled with additional steps to evaluate the bounds at \( x = 2 \) and \( x = 3 \).

Using integration by parts, the definite integral becomes easier to solve, because it's about breaking down the problem into simpler parts that can be calculated independently and then combined.

Logarithmic Functions

Logarithmic functions, such as \( \ln(2x^5) \) in this problem, represent the inverse operations of exponential functions. They can be tricky due to their properties and the presence of the natural logarithm \( \ln \).

• Logarithms have a base \( e \), where \( e \approx 2.71828 \), and are commonly used in continuous growth or decay problems.
• One important property of logarithms used here is the chain rule for differentiation, which allows us to find \( du \) when we differentiate \( u = \ln(2x^5) \).

Knowing how to work with logarithmic functions is essential, especially for integration by parts, as they often appear as \( u \) due to their simplified derivative properties.

Calculus Problem Solving

Problem-solving in calculus requires a strategic approach. Here, breaking down complex integrals by using integration techniques exemplifies such strategies.

• Start with assessing the integral's components to identify \( u \) and \( dv \), which is crucial in integration by parts.
• Thoroughly check each differential and integral step ensures a correct application of the integration by parts formula.

By incrementally solving each subproblem, calculus problem-solving builds a path to the final result, as seen when solving definite integrals with challenging components. This approach minimizes errors and improves understanding.

Integration Techniques

Integration by parts is a powerful technique to integrate products of functions. It follows the formula: \( \int u \, dv = uv - \int v \, du \).

• Choosing appropriate \( u \) and \( dv \) simplifies the process by manipulating the complexity of the integral.
• In this problem, \( u = \ln(2x^5) \) and \( dv = x^{-2} \, dx \) were chosen.
• This choice helps in finding \( v \) and \( du \), making the integral more straightforward to solve.

This technique is especially useful when dealing with integrals of logarithmic and polynomial functions, allowing the integral to evolve into a simpler format for solving.