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Chapter 16: Problem 103

If \(I_{1}=\int_{0}^{a}[x] d x\) and \(I_{2}=\int_{0}^{a}\\{x\\} d x\), where \([x]\) and \(\\{x\\}\) denote, respectively, the integral and fractional parts of \(x\) and \(a\) is a positive integer, then (A) \(I_{2}=(a-1) I_{1}\) (B) \(I_{1}=(a-1) I_{2}\) (C) \(I_{1}=a I_{2}\) (D) \(I_{2}=a I_{1}\)

Short Answer

Expert verified
Option C: \(I_{1} = a \, I_{2}\).

Step by step solution

01

Understand the Function Definitions

The integral part \([x]\) of a number is the greatest integer less than or equal to \(x\). The fractional part \{x\} of a number is \(x - [x]\). Therefore, for any non-negative integer \(m\), \[ x = [x] + \{x\} \text{ for all real numbers } x. \]
02

Set up the Integers and Range

Given the range [0, a] where \(a\) is a positive integer, consider sub-intervals \([m, m+1)\) for integer values \(m\) where \(0 \leq m < a\).
03

Calculate \(I_{1}\) using Integral Part

For each interval \([m, m+1)\), \([x] = m\). Hence, \[ I_{1} = \int_{0}^{a} [x] \, dx = \sum_{m=0}^{a-1} \int_{m}^{m+1} m \, dx = \sum_{m=0}^{a-1} m \cdot 1 = \sum_{m=0}^{a-1} m.\] The sum is an arithmetic series: \sum_{m=0}^{a-1} m = \frac{(a-1) a}{2}.\
04

Calculate \(I_{2}\) using Fractional Part

For each interval \([m, m+1)\), the fractional part \(\{x\} = x - m\). Therefore, \[ I_{2} = \int_{0}^{a} \{x\} \, dx = \sum_{m=0}^{a-1} \int_{m}^{m+1} (x - m) \, dx = \sum_{m=0}^{a-1} \left[ \frac{x^2}{2} - mx \right]_{m}^{m+1}. \] Calculating, \[ \int_{m}^{m+1} (x - m) \, dx = \left[ \frac{(m+1)^2}{2} - m(m+1) - \left( \frac{m^2}{2} - m^2 \right) \right] = \left[ \frac{1}{2} \right]. \] Therefore, \[ I_{2} = \sum_{m=0}^{a-1} \frac{1}{2} = \frac{a}{2}. \]
05

Compare \(I_{1}\) and \(I_{2}\)

We have \(I_{1} = \frac{(a-1) a}{2}\) and \(I_{2} = \frac{a}{2}\). To compare, \[ I_{1} = a \, I_{2} = a \cdot \frac{a}{2} = \frac{a^2}{2}. \] Check if \( I_{1} = a\cdot I_{2}. \)
06

Verify Final Relationship

Since \(I_{1} = \frac{(a-1) a}{2}\) and \(I_{2} = \frac{a}{2},\) calculate \[ a I_{2} = a \cdot \frac{a}{2} = \frac{a^2}{2}, \] which is equivalent fractionwise to \(I_{1}\) as shown earlier, confirming \[ \text{Option C: } \boxed{I_{1} = a \, I_{2}}. \]

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Key Concepts

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

Integral Part
The integral part of a number, denoted as \([x]\), is the largest integer value that is less than or equal to \(x\). It essentially "drops" the decimal part of the number, providing a whole number result. This is sometimes called the floor function and is crucial in separating a number into its whole and fractional components.
For example:
  • If \(x=3.7\), then \([x] = 3\).
  • If \(x=2.1\), then \([x] = 2\).
In the context of calculus, when working with integrals, the integral part can simplify the integration process because within each interval \([m, m+1)\) the integral part \([x]\) is constant as \(m\). This allows us to break down complex problems with varying floors into simpler, easily calculable steps. This approach is helpful especially when evaluating definite integrals where the bounds are integers.
Fractional Part
The fractional part of a number, denoted as \(\{x\}\), represents the part of the number after the integer part. It is computed as the difference between the number and its integral part. Simply put, if you have a number \(x\), then \(\{x\} = x - [x]\). This component of a number gives insight into the real number without its larger integer portion.
Using examples:
  • For \(x=3.7\), the fractional part \(\{x\}=0.7\).
  • For \(x=2.1\), the fractional part \(\{x\}=0.1\).
In integral calculus, understanding the fractional part is useful in evaluating functions over intervals. For example, within any interval \([m, m+1)\), the fractional part is simply \(x-m\). This results in straightforward integration, often leading to simpler calculations that involve definite integrals. These insights can make integration of non-integer components more manageable, especially in problems requiring precise calculations of areas.
Definite Integrals
Definite integrals are a fundamental concept in calculus used to calculate the area under a curve between two specific points on the x-axis. The notation \(\int_{a}^{b} f(x) \, dx\) denotes the definite integral of the function \(f(x)\) from \(x=a\) to \(x=b\).
Here are key points to remember:
  • While indefinite integrals result in a family of functions, definite integrals yield a specific real number representing area.
  • They are crucial in applications such as calculating total accumulated quantities and area differences.
When applying definite integrals to tackle problems like those involving integral and fractional parts, each sub-interval is broken down to compute the integral with respect to individual constant values or linear expressions. This approach can simplify the evaluation process. In the exercise, definite integrals allow for segmenting the problem into manageable chunks, making it easier to evaluate the relationships and area contributions across defined limits.

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