Problem 110 - A Complete Resource Book in Mathematics for JEE Main 2017

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: Understand the integral

The given integral to solve is \( \int_{0}^{x} \left( x - [x] - \frac{1}{2} \right) \, dx \). Here, \([x]\) is the greatest integer less than or equal to \(x\), and \(\{x\} = x - [x]\) is the fractional part of \(x\). Thus, the expression becomes \( \int_{0}^{x} \left( \{x\} - \frac{1}{2} \right) \, dx \).

Step 2: Recognize the function within the interval

Notice that between any two integers \(n\) and \(n+1\), the fractional part \( \{ x \} \) is simply \(x - n\). Therefore, inside this range, the function reduces to \( \{ x \} - \frac{1}{2} = (x - n) - \frac{1}{2} = x - n - \frac{1}{2}\).

Step 3: Set the integration limit

Since the integral is calculated from 0 to \( x \), but since the expression depends on \( \{ x \} \), we separate this integral into intervals \( [n, n+1) \). Within such an interval, the fractional part \( \{ x \} = x - n \). For the upper bound of the specific interval for \(x\), say, \(n < x \leq n+1\), split the initial integral from 0 to \(n + \{x\}\).

Step 4: Calculate the integral over [n, n+1]

The integral now becomes \( \int_{n}^{n + \{x\}} (x - n - \frac{1}{2}) \, dx \). Evaluating this integral gives:\[ \int_{n}^{n + \{x\}} \left(x - n - \frac{1}{2} \right) \, dx = \left[ \frac{(x - n)^2}{2} - \frac{x - n}{2} \right]_{n}^{n + \{x\}} \].

Step 5: Apply the limits of integration

Apply the limits of integration:- When \(x = n + \{x\}\), the expression evaluates to \( \frac{(\{x\})^2}{2} - \frac{\{x\}}{2} = \frac{1}{2} \{x\}(\{x\} - 1) \).- When \(x = n\), the expression is 0.

Step 6: Combine the result

The result of the integral between any integer \(n \) to \( n + \{x\} \) is \( \frac{1}{2} \{x\}(\{x\} - 1) \). This matches option (A).

Key Concepts

Integral Function Analysis

Integral calculus revolves around finding the accumulation of quantities, which often means calculating areas under curves. In the given problem, we analyze the integral of a function involving the expression \(x - [x] - \frac{1}{2}\). This expression simplifies to integrations over the fractional part of \(x\), \(\{x\}\). Integrals provide insight into continuous accumulation, allowing for the analysis of functions over specific intervals.

• An integral function, such as the one in our problem, requires identifying the nature of its components—including constants and variable parts.
• It's crucial to set integration limits properly, such as from 0 to \(x\), to cover the function's domain.
• The integral transforms subtraction within the function, such as \([x]\) representing the integer part, to focus on continuous parts like \(\{x\}\).

Using integral function analysis, the evaluator understands the interaction between variable expressions within specific boundaries, which is essential for accurate computation.

Fractional Part Function

The fractional part function, denoted as \(\{x\}\), is crucial in distinguishing between the whole and fractional elements of a number. It is defined as \(\{x\} = x - [x]\), where \([x]\) is the greatest integer less than or equal to \(x\). This function finds frequent use in periodic functions and problems requiring modular arithmetic.

• In mathematics, \(\{x\}\) captures the non-integer part of \(x\), measured as \(x - \lfloor x \rfloor\).
• In our integral, the fractional part isolates the element that changes within each unit interval as \([x]\) increases stepwise.
• Understanding \(\{x\}\) helps in simplifying expressions or integrals where continuity and periodicity are involved.

When evaluating an integral with \(\{x\}\), the function alters slightly over the base integer grid, which adds a level of complexity to direct calculations or computations.

Integral Evaluation Techniques

Integral evaluation techniques are methods to simplify or calculate integrals accurately and efficiently. In the exercise, we used limits and parts of functions to find the answer easily. Proper evaluation of this integral involved breaking the problem into manageable intervals and components, such as within \([n, n+1]\).

• Beginning by recognizing function behavior within specific intervals aids in simplifying the evaluation process without oversized calculations.
• Establishing variable substitutions and transformations, like rewriting \(x - [x]\) as \(\{x\}\), is fundamental.
• Using integration laws, like the power rule, helps in evaluating the integrated function accurately, as seen when integrating \((x - n - \frac{1}{2})\).

These techniques enable solving complex-looking integrals by converting them into simpler forms, applying boundaries effectively, and nullifying redundant terms, which significantly enhance computational efficiency.