Problem 48 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The value of \(\int_{-\pi / 2}^{\pi / 2}\left(\left[\frac{x}{\pi}\right]+0.5\right) d x\) is [where [.] denotes the greates integer function) (A) \(\pi\) (B) \(\pi 2\) (C) 0 (D) \(-\pi / 2\)

Step-by-Step Solution

Answer

The value of the integral is 0.

Step 1: Understand the Greatest Integer Function

The greatest integer function, also known as the floor function, \([\cdot]\) returns the greatest integer less than or equal to the given number. For instance, \[\left[\frac{x}{\pi}\right]\text{ for } x=-\frac{\pi}{2} ext{ to } x=\frac{\pi}{2}\] results in -1 for negative values of \(x \) and 0 for non-negative values up to 0.

Step 2: Define the Integral Over Given Ranges

Split the integral into two parts based on the change in the greatest integer function. The ranges determined are \(-\frac{\pi}{2} \leq x < 0\) and \(0 \leq x \leq \frac{\pi}{2}\). This gives us two separate integrals to solve based on these ranges.

Step 3: Solve the Integral for Negative Range

For \(-\frac{\pi}{2} \leq x < 0\), the function becomes \([\frac{x}{\pi}] + 0.5 = -1 + 0.5 = -0.5\). Thus, the integral is \[\int_{-\pi/2}^{0} -0.5 \, dx = -0.5 \times (0 - (-\pi/2)) = -0.5 \times \frac{\pi}{2} = -\frac{\pi}{4}.\]

Step 4: Solve the Integral for Non-Negative Range

For \(0 \leq x \leq \frac{\pi}{2}\), the function becomes \([\frac{x}{\pi}] + 0.5= 0 + 0.5 = 0.5\). Thus, the integral is \[\int_{0}^{\pi/2} 0.5 \, dx = 0.5 \times (\frac{\pi}{2} - 0) = 0.5 \times \frac{\pi}{2} = \frac{\pi}{4}.\]

Step 5: Combine Results from Both Ranges

Add the results from both ranges to find the value of the integral: \[-\frac{\pi}{4} + \frac{\pi}{4} = 0.\]

Key Concepts

Greatest Integer Function

The greatest integer function, often denoted as \([ ext{floor}(x)]\), plays a key role in understanding how specific values transform when integrated. Here's the core idea: given any real number \(x\), the function returns the largest integer less than or equal to \(x\). For example:

• \([rac{3}{2}] = 1\) because 1 is the largest integer less than 1.5.
• \([3] = 3\) since 3 is already an integer.
• \([-1.2] = -2\) as -2 is the largest integer less than -1.2.

By understanding the nature of this function, we can see how it modifies the value it operates on, particularly when combined with other mathematical operations. In definite integrals, the greatest integer function tends to segment the function into distinct parts, each yielding a constant value over integer-defined intervals.

Floor Function

The floor function, not surprisingly, overlaps significantly with the greatest integer function. This simply emphasizes that these terms refer to the same concept. Whenever you read about returning the greatest integer less than or equal to a number, think of the floor function, denoted as \(\lfloor x \rfloor\).

This concept might seem purely theoretical until you encounter integrals involving the floor function where:

• The integral is split at the points \(x\) where \(\lfloor \frac{x}{\pi} \rfloor\) changes its integer value.
• Each segment between these transitions results in a constant value which makes definite integration easier.

In the context of the exercise example, understanding the floor function helps in breaking down complex integrals into manageable components, such as adjusting the range to accommodate changes.

Integral Calculus

Integral calculus forms the backbone of finding areas under curves, with definite integrals being particularly important. Here’s a quick refresher on topical concepts:

• Definite integral evaluates the net area between the curve and the x-axis from one point to another.
• Symbolically, it is expressed as \(\int_{a}^{b} f(x) \, dx\).
• The result yields a numerical value representing accumulated change over \([a, b]\).

In combination with functions like the greatest integer function, integrals can be segmented to simplify calculations. As seen in the exercise, the integral was divided into two parts over two ranges because the behavior of the greatest integer function changed. This strategic segmentation results in easier math work, providing key insights necessary for assessing scenarios under differing conditions.

Mastering integral calculus with these components is essential for tackling more advanced calculus challenges that you'll meet along your learning journey.