Problem 103 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

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) \(I_{2}=a I_{1}\)

Step-by-Step Solution

Answer

Option (C) \(I_1 = a I_2\).

Step 1: Understand the Definitions

The integral part of a number \( x \) is denoted by \([x]\) and represents the largest integer less than or equal to \( x \). The fractional part of \( x \) is \(\{x\} = x - [x]\). We are given that \( a \) is a positive integer.

Step 2: Evaluate Integral for the Integral Part, \( I_1 \)

We know that the integral of \([x]\) from \(0\) to \(a\) can be separated over each integer. So, \( I_1 = \int_0^a [x] \, dx = \sum_{k=0}^{a-1} \int_k^{k+1} k \, dx = \sum_{k=0}^{a-1} k = 0 + 1 + 2 + \ldots + (a-1)\). This is an arithmetic series.

Step 3: Sum for Arithmetic Series \( I_1 \)

The sum for the arithmetic series \( 0 + 1 + 2 + \ldots + (a-1) \) is given by the formula \( \frac{(a-1)\cdot a}{2} \). Therefore, \( I_1 = \frac{a(a-1)}{2} \).

Step 4: Evaluate Integral for the Fractional Part, \( I_2 \)

The integral of the fractional part \( \{x\} \) from \(0\) to \(a\) is \( I_2 = \int_0^a (x - [x]) \, dx = \int_0^a x \, dx - \int_0^a [x] \, dx \).

Step 5: Simplifying \( I_2 \)

First, calculate \( \int_0^a x \, dx = \frac{a^2}{2} \). As calculated earlier, \( \int_0^a [x] \, dx = I_1 \). Thus, \( I_2 = \frac{a^2}{2} - I_1 = \frac{a^2}{2} - \frac{a(a-1)}{2} \).

Step 6: Final Calculation for \( I_2 \)

Substitute \( I_1 \) from Step 3 into the expression for \( I_2 \): \( I_2 = \frac{a^2}{2} - \frac{a(a-1)}{2} = \frac{a(a - (a-1))}{2} = \frac{a}{2} \).

Step 7: Relate \( I_1 \) and \( I_2 \)

From Step 3, \( I_1 = \frac{a(a-1)}{2} \) and from Step 6, \( I_2 = \frac{a}{2} \). Thus, \( I_1 = a \, I_2 \).

Step 8: Choose the Correct Option

By comparing derived expressions with options, we find that option (C) \( I_1 = a I_2 \) matches what we have derived.

Key Concepts

Fractional Part of a Number

When we talk about the **fractional part** of a number, we are discussing the portion of the number that comes after the decimal point. This is an important concept in mathematics, particularly when working with integrals and series.
The fractional part of a real number \( x \), denoted by \( \{ x \} \), is given by the formula:

• \( \{ x \} = x - [x] \)

Here, \( [x] \) represents the integral part of \( x \), which we'll cover in the next section.
Understanding the fractional part helps us dissect numbers into simpler forms for summation and integration.
In the context of definite integrals, the fractional part function helps identify how the portion of a number that isn't a whole number behaves within an integral.
This concept is crucial when evaluating the separation of number components across integrals, as seen in problems like the one we explored above.

Integral Part of a Number

The **integral part** of a number \( x \) refers to the integer component when a real number is expressed in decimal form. Often denoted as \( [x] \), it represents the largest integer that is less than or equal to \( x \).
Here's a simple way to remember:

• For any positive number like \( x = 3.7 \), the integral part is \( [3.7] = 3 \).
• For \( x = -2.9 \), the integral part would be \( [-2.9] = -3 \) because -3 is less than -2.9 but still the closest integer to it from the left on a number line.

This concept becomes invaluable when we split an integral range into discrete segments, particularly when evaluating the integral of functions defined in terms of \([x]\). In our example, \( I_1 \) was calculated by summing the integral parts over a defined interval, leading to an arithmetic series.
Integrals of integral parts, like \( I_1 = \int_0^a [x] \, dx \), can thus be broken down into simpler arithmetic operations that make complex calculus problems more manageable.

Arithmetic Series

An **arithmetic series** is a sequence of numbers in which the difference between consecutive terms is constant. This difference is known as the "common difference." Arithmetic series are fundamental in calculus and algebra, especially when dealing with integrals that involve summation over integer intervals.
For instance, if we have a sequence like \( 0, 1, 2, \ldots, (a-1) \), the sum of this series can be computed using the formula:

• \( S_n = \frac{n(n+1)}{2} \), where \( n \) is the last number in the series.

In our problem, the integral \( I_1 \) was evaluated as such a series from \( 0 \) to \( (a-1) \).
The sum \( 0 + 1 + 2 + \ldots + (a-1) \) can be simplified to \( \frac{(a-1)\cdot a}{2} \).
This approach shows how arithmetic series facilitate the calculation of integrals over discrete sums, converting a potential infinite sum into a simple algebraic expression, thus simplifying the problem significantly.