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Magazine Chapter 7: Problem 21
Prove that \(\int_{1}^{5} \frac{\lfloor x\rfloor}{x} \mathrm{~d} x=4 \log (5)-3 \log (2)-\log (3)\)
Short Answer
Expert verified
The integral evaluates to \(4 \log(5) - 3 \log(2) - \log(3)\), as required.
Step by step solution
01
Identify the function and its behavior
The function is \(\frac{\lfloor x \rfloor}{x}\), where \(\lfloor x \rfloor\) is the floor function which gives the greatest integer less than or equal to \(x\). We need to break the integration interval at each integer where \(\lfloor x \rfloor\) changes.
02
Define sub-intervals based on the floor function
The integration interval \([1, 5]\) must be split at integer points since the floor function \(\lfloor x \rfloor\) changes values at each integer. This gives sub-intervals: \([1, 2)\), \([2, 3)\), \([3, 4)\), and \([4, 5]\).
03
Evaluate the integral over each sub-interval
For each sub-interval, evaluate the integral:- For \([1, 2)\), \(\lfloor x \rfloor = 1\): \[ \int_{1}^{2} \frac{1}{x} \mathrm{~d}x = \log(2) - \log(1) = \log(2) \]- For \([2, 3)\), \(\lfloor x \rfloor = 2\): \[ \int_{2}^{3} \frac{2}{x} \mathrm{~d}x = 2 \log(3) - 2 \log(2) = 2 (\log(3) - \log(2)) \]- For \([3, 4)\), \(\lfloor x \rfloor = 3\): \[ \int_{3}^{4} \frac{3}{x} \mathrm{~d}x = 3 \log(4) - 3 \log(3) = 3 (\log(4) - \log(3)) \]- For \([4, 5]\), \(\lfloor x \rfloor = 4\): \[ \int_{4}^{5} \frac{4}{x} \mathrm{~d}x = 4 \log(5) - 4 \log(4) = 4 (\log(5) - \log(4)) \]
04
Combine results from sub-intervals
Combining the results from each sub-interval:\[ \log(2) + 2(\log(3) - \log(2)) + 3(\log(4) - \log(3)) + 4(\log(5) - \log(4)) \]Simplifies to:\[ 4 \log(5) - 3 \log(2) - \log(3) \]
05
Validate the solution
Check each calculation for correctness. Summing together gives the expression we aim to prove. Therefore, the original integral evaluates correctly to the given result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Floor Function
The floor function, denoted as \( \lfloor x \rfloor \), is a mathematical function that returns the greatest integer less than or equal to a given number \( x \). For example, \( \lfloor 2.7 \rfloor = 2 \) and \( \lfloor -1.2 \rfloor = -2 \).
- This function is helpful in integral calculus when dealing with piecewise functions, as it categorizes intervals where the quotient of integers is involved.
- Within the context of integration, the floor function causes a piecewise change at every integer point.
- In the given problem, this creates the sub-intervals \([1, 2), [2, 3), [3, 4), [4, 5]\) as the value of \( \lfloor x \rfloor \) steps up at each integer.
Integration by Parts
Integration by parts is a fundamental technique used in calculus to integrate products of functions. It is based on the product rule for differentiation and is expressed as:\[\int u \mathrm{~d}v = uv - \int v \mathrm{~d}u,\]where \( u \) and \( \mathrm{d}v \) are chosen parts of the integral.
- Typically, one part gets simpler upon differentiation, and the other part is straightforward enough to integrate easily.
- Although not directly used for each sub-interval in this exercise, understanding integration by parts provides critical insight when choosing functions to integrate or differentiate.
- This concept helps in evaluating products like \( \frac{\lfloor x \rfloor}{x} \) by simplifying similar problems.
Logarithmic Properties
Working with logarithms in calculus simplifies the integration of certain rational functions and allows easier manipulation of expressions. Key properties include:\[\log(ab) = \log(a) + \log(b), \\log\left(\frac{a}{b}\right) = \log(a) - \log(b), \k \log(a) = \log(a^k)\]
- In integration, logarithmic functions naturally arise from integrating the form \( \frac{1}{x} \).
- Within the exercise, applying logarithmic properties aids in simplifying each integral result from the piecewise intervals.
- These properties are essential when combining the results of separate integrations into a single expression.
Piecewise Integration
Piecewise integration involves breaking down an integral over an interval into smaller intervals where the function is easier to integrate. This is crucial when a function consists of different expressions on different intervals, like our example with the floor function.
- The integral of \( \frac{\lfloor x \rfloor}{x} \) is divided into segments where the floor value remains constant: \([1, 2), [2, 3), [3, 4), [4, 5]\).
- Within each sub-interval, \( \lfloor x \rfloor \) becomes a constant, simplifying the integral of the function according to the properties of logarithms.
- This technique not only simplifies calculations but maintains the accuracy of evaluations across changing behaviors of the function.
