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Chapter 5: Problem 58

Evaluate each definite integral. \(\int_{-3}^{-1}\left(1+x^{-1}\right) d x\)

Short Answer

Expert verified
The value of the definite integral is \(2 - \ln(3)\).

Step by step solution

01

Split the Integral

Split the integral into two separate integrals to simplify the evaluation:\[\int_{-3}^{-1} \left(1 + x^{-1}\right) \, dx = \int_{-3}^{-1} 1 \, dx + \int_{-3}^{-1} x^{-1} \, dx.\]
02

Evaluate the First Integral

Find the integral of the constant function:\[\int_{-3}^{-1} 1 \, dx = \left[ x \right]_{-3}^{-1} = (-1) - (-3) = 2.\]
03

Evaluate the Second Integral

Determine the integral of the variable function:\[\int_{-3}^{-1} x^{-1} \, dx = \left[\ln|x|\right]_{-3}^{-1} = \ln|-1| - \ln|-3| = 0 - \ln(3) = -\ln(3).\]
04

Combine Results

Combine the results from the two integrals:\[\int_{-3}^{-1} \left(1 + x^{-1}\right) \, dx = 2 - \ln(3).\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Splitting
When you face a definite integral with multiple terms, splitting the integral is a handy technique. This method involves breaking down a complex integral into simpler, more manageable parts. For example, given \[ \int_{-3}^{-1} (1 + x^{-1}) \, dx \]you can split it into:
  • \( \int_{-3}^{-1} 1 \, dx \)
  • \( \int_{-3}^{-1} x^{-1} \, dx \)
By separating the integral, it is easier to evaluate each part individually. This often simplifies calculations and reduces the potential for error. Each integral can be solved independently, after which you can combine the results for the final solution. Remember, maintaining the same limits of integration for each part is crucial to ensure precision in your calculations.
Constant Function Integration
Integral calculus with constant functions is straightforward and usually the easiest part of solving many problems. When integrating a constant like 1 over an interval, the process reflects finding the area of a rectangle with the length equal to the integration limits. Thus, \[ \int_{-3}^{-1} 1 \, dx \]translates to evaluating \[ \left[ x \right]_{-3}^{-1} = x \text{ from } -3 \text{ to } -1 \]This results in the computation:
  • \( -1 - (-3) = 2 \)
This simply means that the area is equal to the difference between the upper and lower limits. When dealing with constant functions, integration merely involves multiplying the constant by the width of the integration interval.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is crucial for integrating rational functions like \( x^{-1} \). In calculus, the integral of \( x^{-1} \) is \[ \ln|x| + C \]where \( C \) is the constant of integration. For definite integrals, the constant \( C \) is not included since we derive a specific numerical result. Therefore, \[ \int_{-3}^{-1} x^{-1} \, dx \]results in evaluating
  • \( \left[ \ln|x| \right]_{-3}^{-1} = \ln|-1| - \ln|-3| \)
  • \( \ln(1) - \ln(3) = 0 - \ln(3) = -\ln(3) \)
The natural logarithm of 1 is always 0, simplifying calculations. Understanding the behavior of logarithms helps prevent mistakes, especially when handling negative numbers or absolute values within integrals.

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