Problem 19 - A Complete Resource Book in Mathematics for JEE Main

Question

The area bounded by the \(y=|\sin x|, x\)-axis and the lines \(|x|=\pi\) is (A) 2 (B) 1 (C) 4 (D) None of these

Step-by-Step Solution

Answer

The area is 4, option (C).

Step 1: Understanding the Problem

The problem asks us to find the area enclosed by the graph of the function \(y = |\sin x|\), the x-axis, and the vertical lines \(x = -\pi\) and \(x = \pi\). This means we need to determine the integral of \(|\sin x|\) over the interval \([-\pi, \pi]\).

Step 2: Analyzing the Function

The function \(y = |\sin x|\) describes the absolute value of the sine function. Since the sine function is periodic and symmetric about the y-axis, we can simplify our work by focusing on the interval from \(0\) to \(\pi\) and then doubling the result. The function will have the same area from \(-\pi\) to \(0\).

Step 3: Setting Up the Integral

We need to calculate two identical integrals from \(0\) to \(\pi\) because the function is symmetric. So, consider the integral: \[ \int_0^{\pi} |\sin x| \, dx \]. Since \(\sin x\) is non-negative in this interval, \[ \int_0^{\pi} \sin x \, dx \] will do just as well.

Step 4: Calculating the Integral

To find the integral \( \int_0^{\pi} \sin x \, dx \), we use the antiderivative of \(\sin x\), which is \(-\cos x\). Thus,\[ \int_0^{\pi} \sin x \, dx = [-\cos x]^\pi_0 = -\cos(\pi) + \cos(0) = -(-1) + 1 = 2. \]

Step 5: Doubling for Symmetry

Since \(|\sin x|\) is symmetric, we double the area from \(0\) to \(\pi\) to account for the interval from \(-\pi\) to \(\pi\): \[2 \times 2 = 4.\]

Step 6: Identifying the Correct Answer

The area bounded by \(y = |\sin x|\), the x-axis, and the lines \(|x| = \pi\) is 4, matching option (C) in the multiple-choice options.

Key Concepts

Periodic Functions

Periodic functions are fundamental concepts in mathematics, characterized by their repeating patterns over fixed intervals. The sine function, \(\sin x\), for example, is one of the most well-known periodic functions. It repeats its values every \(2\pi\) radians.

For any function \(f(x)\) to be periodic, there must exist a positive number \(T\) such that \(f(x + T) = f(x)\) for all \(x\). In the case of the sine function, \(T = 2\pi\), meaning the graph of \(\sin x\) repeats every \(2\pi\).

Understanding periodic functions helps us solve problems like finding areas under curves with known repeating patterns. When such a function is symmetric concerning the y-axis, calculations can often be simplified by focusing on half the period and doubling the result. This is exactly what we do when evaluating integrals involving sine-related functions, as we only need to compute the integral over one complete period to understand the entire behavior.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that relate to trigonometric functions such as sine, cosine, and tangent. These integrals are ubiquitous in calculus, and they often involve using known antiderivatives or basic identities to find their values.

In our specific example, the integral of \(\sin x\), in the range from \(0\) to \(\pi\), exploits the relationship of \(\sin x\) and its antiderivative, \(-\cos x\). The calculation \[ \int_0^{\pi} \sin x \, dx = [-\cos x]^\pi_0 = -\cos(\pi) + \cos(0) = -(-1) + 1 = 2. \] lays the groundwork for understanding how areas under trigonometric curves can be calculated.

Overall, mastering trigonometric integrals is essential for solving a myriad of problems involving oscillations, waves, and harmonic motions in various fields of science and engineering.

Symmetry in Mathematics

Symmetry is a powerful tool in mathematics that can streamline the process of solving complex problems. It occurs when a figure or function remains unchanged under certain operations, like reflection or rotation. In the context of the sine function, symmetry around the y-axis, known as even symmetry, plays a crucial role.

When analyzing \(y = |\sin x|\), the graph is symmetric about the y-axis. This means that the area under the curve from \(0\) to \(\pi\) is equal to the area from \(-\pi\) to \(0\). Because of this symmetrical property, to find the total area over an entire period, we only compute the integral once for half the range and then double it to account for the full range. This kind of symmetry vastly simplifies the calculation process.

Understanding symmetry helps recognize patterns and simplify calculations, making otherwise daunting integrals manageable and ensuring accuracy in results.