Problem 110 - A Complete Resource Book in Mathematics for JEE Main

Question

If \([x]\) and \(\\{x\\}\) denote the integral and fractional parts of \(x\), respectively, then \(\int_{0}^{x}\left(x-[x]-\frac{1}{2}\right) d x\) is equal to (A) \(\frac{1}{2}\\{x\\}(\\{x\\}-1)\) (B) \(\frac{1}{2}\\{x\\}(\\{x\\}+1)\) (C) \(\\{x\\}(\\{x\\}-1)\) (D) None of these

Step-by-Step Solution

Answer

(A) \(\frac{1}{2}\{x\}(\{x\} - 1)\).

Step 1: Understanding the Problem

The integral we need to evaluate is \( \int_{0}^{x} \left(x-[x]-\frac{1}{2}\right) dx \). Here, \([x]\) is the greatest integer less than or equal to \(x\), known as the integral part, and \(\{x\} = x - [x]\), the fractional part of \(x\).

Step 2: Simplifying the Expression within the Integral

Since \(x\) can be expressed as \([x] + \{x\}\), the expression inside the integral \(x - [x] - \frac{1}{2}\) simplifies to \(\{x\} - \frac{1}{2}\). This means we need to integrate \(\{x\} - \frac{1}{2}\) over the interval \([0, x]\).

Step 3: Change of Variable Based on \(x\)'s Interval

Consider the integral \(\int_{0}^{x} \left(\{t\} - \frac{1}{2}\right) dt\). Recognizing that the fractional part function \(\{t\}\) resets after each integer, we need to integrate piecewise over each interval \([n, n+1]\), where \(n\) is an integer.

Step 4: Evaluate the Definite Integral over Sub-Intervals

Within each interval \([n, n+1]\), the function \(\{t\} = t - n\). The evaluation of \(\int_{n}^{n+1} (t - n - \frac{1}{2}) dt\) yields: 1. \(\int_{n}^{n+1} (t - n) dt = \frac{1}{2}\). 2. \(\int_{n}^{n+1} -\frac{1}{2} dt = -\frac{1}{2}\).Thus, the integral over each segment \([n, n+1]\) is \(\frac{1}{2} - \frac{1}{2} = 0\).

Step 5: Summation Result Over Interval Segments

Because each complete interval \([n, n+1]\) contributes zero to the sum, only consider the interval from \([x]\) to \(x\) (where \([x]\) is the integral part). For this interval, use \(\int_{[x]}^{x} ((x- n) - \frac{1}{2}) dx\).

Step 6: Compute the Integral from Integral Part to \(x\)

For the remaining part from \([x]\) to \(x\), integrate \((x - [x] - \frac{1}{2}) = (\{x\} - \frac{1}{2})\). Thus: \[ \int_{[x]}^{x} \left(\{t\} - \frac{1}{2}\right) dt = \int_{0}^{\{x\}} \left(t - \frac{1}{2}\right) dt. \] Evaluating this, we have:- \(\int_{0}^{\{x\}} t dt = \frac{1}{2} \{x\}^2\).- \(\int_{0}^{\{x\}} \left(-\frac{1}{2}\right) dt = -\frac{1}{2} \{x\}\).Final part: \(\frac{1}{2} \{x\}^2 - \frac{1}{2} \{x\} = \frac{1}{2}\{x\}(\{x\} - 1)\).

Step 7: Final Evaluation and Answer

Thus the evaluated integral over \([0, x]\) is \(\frac{1}{2}\{x\}(\{x\} - 1)\). Comparing this result with the given choices, the correct answer is Option (A).

Key Concepts

Integral Part

The integral part of a real number, denoted by \([x]\), is the greatest integer that is less than or equal to that number. To put it simply, it's the "whole" number part of a real number that you're used to seeing before a decimal.

• If the number is a whole number, like 3, the integral part is also 3.
• However, if you take a number like 3.7, the integral part will still be 3, because it's the largest integer that doesn't exceed 3.7.

This concept comes up frequently when working with definite integrals that involve step changes, such as the floor function or greatest integer function.
In the context of the given problem, the integral part helps us understand the basic structure of how number behavior changes at each integer crossover. This understanding is essential for calculating the definite integral, especially when you have to consider intervals that are defined between integers.

Fractional Part

The fractional part of a real number, written as \( \{x\} \), is simply the portion of the number that follows the decimal point. This can be calculated by subtracting the integral part from the original number.

• For instance, if you have the number 3.7, the fractional part is calculated as 3.7 - 3, which equals 0.7.
• In mathematical terms, if \( x = 5.25 \), then \([x] = 5\) and \( \{x\} = 0.25\).

The fractional part becomes extremely relevant when you're breaking down an interval for evaluation, as it helps in understanding how the value behaves in sections spanning an integer. The problem at hand uses this concept to break the integration task down into manageable parts.
It's the part that varies continuously as you move between integers, while the integral part stays fixed until the next integer is reached.

Piecewise Integration

Piecewise integration is a technique used to evaluate an integral by breaking it down into smaller sub-integrals over set intervals. This is particularly useful when dealing with functions that have different behaviors over different segments of their domain, like the fractional part function \( \{x\} \).

• Imagine you are trying to integrate a function over a range that spans multiple integer intervals, such as from 0.5 to 3.6. Here, the behavior of \( \{x\} \) changes at each integer, resetting to zero.
• This means you'll need to evaluate the integral separately over each interval from one integer to the next, essentially compartmentalizing your integration efforts.

In the original problem, this method allows us to deal precisely with the way that \( \{x\} \) resets at each integer point, allowing us to calculate the integral accurately over each of these intervals. By understanding how the function behaves within these slices, you can obtain a complete and exact understanding of the area under the curve of the function being integrated.