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Magazine Chapter 7: Problem 16
In Problems 1-36, use integration by parts to evaluate each integral. $$ \int_{2}^{3} \frac{\ln 2 x^{5}}{x^{2}} d x $$
Short Answer
Expert verified
\(- \frac{1}{3} \ln(486) + \frac{1}{2} \ln(32) + \frac{5}{6}\)
Step by step solution
01
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 \( dv = x^{-2} \, dx \).
02
Differentiate and Integrate Parts
Differentiate \( u \) to find \( du \):\( u = \ln(2x^5) \)\( du = 5x^{-1} \, dx \).Integrate \( dv \) to find \( v \):\( dv = x^{-2} \, dx \)\( v = -x^{-1} \).
03
Apply Integration by Parts Formula
Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:\[ \int \ln(2x^5)x^{-2} \, dx = - \ln(2x^5) x^{-1} \bigg|_2^3 - \int_2^3 (-x^{-1}) (5x^{-1} \, dx) \].
04
Simplify the Integral
The integral becomes:\(- \ln(2x^5) x^{-1} \bigg|_2^3 + \int_2^3 5x^{-2} \, dx \).Evaluate the simpler integral:\( \int 5x^{-2} \, dx = -5x^{-1} \).
05
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.
06
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:\[- \frac{1}{3} \ln(486) + \frac{1}{2} \ln(32) + \frac{5}{6} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( x = 3 \).
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 \( u = \ln(2x^5) \).
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.
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.
