Chapter 28: Problem 32
Evaluate \(\int_{1}^{2} x^{-1} d x\) and \(\int_{2}^{4} x^{-1} d x .\) Give a geometric interpretation of these two results.
Short Answer
Expert verified
Both integrals equal \( \ln 2 \), representing the area under the curve \( y = \frac{1}{x} \) for their respective intervals.
Step by step solution
01
Recognize the Integral Function
The given integral is of the form \( \int x^{-1} \, dx \), which simplifies to \( \int \frac{1}{x} \, dx \). The integral of \( \frac{1}{x} \) is a standard integral and results in the natural logarithmic function.
02
Determine the Integral Formula
The antiderivative of \( \frac{1}{x} \) is \( \ln |x| + C \), where \( C \) is the constant of integration.
03
Evaluate the Definite Integral from 1 to 2
Substitute the bounds 1 and 2 into the integral: \[ \int_{1}^{2} \frac{1}{x} \, dx = \left[ \ln |x| \right]_{1}^{2} = \ln |2| - \ln |1|. \] Since \( \ln |1| = 0 \), the result is \( \ln 2 \).
04
Evaluate the Definite Integral from 2 to 4
Substitute the bounds 2 and 4 into the integral: \[ \int_{2}^{4} \frac{1}{x} \, dx = \left[ \ln |x| \right]_{2}^{4} = \ln |4| - \ln |2|. \] Since \( \ln |4| = 2\ln |2| \), the result is \( 2\ln 2 - \ln 2 = \ln 2 \).
05
Geometric Interpretation
The result \( \ln 2 \) in both integrals is the area under the curve \( y = \frac{1}{x} \) between x = 1 to x = 2 and x = 2 to x = 4. In terms of geometry, \( \ln 2 \) represents the same area for both intervals since the natural logarithm function grows slower for larger x values.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is a logarithmic function with the base \( e \), where \( e \approx 2.71828 \). This special number, \( e \), is the base of the natural logarithms and is an irrational constant that frequently appears in mathematics, especially when dealing with growth processes or in calculus.
- The function \( \ln(x) \) is defined for all positive values of \( x \).
- The natural logarithm of 1, \( \ln(1) \), is zero since \( e^0 = 1 \).
- As \( x \) increases, the function \( \ln(x) \) increases, albeit at a decreasing rate.
Antiderivative
In calculus, finding the antiderivative, also known as the indefinite integral, is an essential technique. It involves finding a function whose derivative is the given function. In the context of our problem, the function in question is \( \frac{1}{x} \). The antiderivative of \( \frac{1}{x} \) is known to be \( \ln|x| \), although, for indefinite integrals, we usually add a constant \( C \). This constant is important when dealing with indefinite integrals as it accounts for any vertical shifts in the graph of the original function.
- The process entails reversing differentiation, so it can be viewed as finding the 'original' function given its rate of change.
- Antiderivatives are crucial for solving differential equations and for finding the area under a curve using integral calculus.
- While computing definite integrals, where limits are specified, our focus is on the specific function values at these limits reducing the need to consider the constant \( C \).
Area Under Curve
The concept of the area under a curve is central to integral calculus. It refers to the space between the graph of a function and the x-axis within a specific interval. For functions like \( y = \frac{1}{x} \), this area can be defined using definite integrals. In the problem at hand, the definite integrals \( \int_{1}^{2} \frac{1}{x} \, dx \) and \( \int_{2}^{4} \frac{1}{x} \, dx \) represent the area under the curve \( y = \frac{1}{x} \) from 1 to 2, and 2 to 4, respectively.
- This area is calculated precisely using the definite integral, illustrative of the sum of infinitely thin rectangles spanning the interval.
- For \( y = \frac{1}{x} \), the rapid decay of the curve means that even though the intervals differ in width, the area amounts to \( \ln 2 \) in both cases.
- Understanding the area under a curve visually helps grasp the concept of accumulation, such as calculating total net distance or the sum of small changes.