Problem 212 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(I=\int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x\) and \(J=\int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x .\) Then which one of the following is true? \([2008]\) (A) \(I>\frac{2}{3}\) and \(J>2\) (B) \(I<\frac{2}{3}\) and \(J<2\) (C) \(I<\frac{2}{3}\) and \(J>2\) (D) \(I>\frac{2}{3}\) and \(J<2\)

Step-by-Step Solution

Answer

The correct answer is (B): \( I < \frac{2}{3} \) and \( J < 2 \).

Step 1: Consider Integral I

The integral is given by \( I = \int_{0}^{1} \frac{\sin x}{\sqrt{x}} \ dx \). To analyze its value, we consider the inequality \( \sin x \leq x \) for \( x > 0 \), which implies \( \frac{\sin x}{\sqrt{x}} \leq \frac{x}{\sqrt{x}} = \sqrt{x} \). Therefore, the integral \( I \) can be bounded as follows: \(\begin{align*}I & = \int_{0}^{1} \frac{\sin x}{\sqrt{x}} \, dx \leq \int_{0}^{1} \sqrt{x} \, dx \ & = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{2}{3}.\end{align*} \)Thus, \( I \leq \frac{2}{3} \).

Step 2: Consider Integral J

The integral is \( J = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} \ dx \). To approximate, note that \( \cos x \approx 1 \) for small \( x \). Therefore, \( \frac{\cos x}{\sqrt{x}} \approx \frac{1}{\sqrt{x}} \), which results in a comparison integral: \(\begin{align*}J & \approx \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx \ & = [2\sqrt{x}]_{0}^{1} = 2.\end{align*}\) However, because \( \cos x \leq 1 \), we achieve further refinement: \(\begin{align*}J & \leq \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx = 2.\end{align*}\)Thus, \( J < 2 \) since \( \cos x < 1 \) for \( x > 0 \).

Step 3: Verify Results Against Options

We found \( I \leq \frac{2}{3} \) and \( J < 2 \). This means that \( I \) can be less than or equal to \( \frac{2}{3} \) and \( J \) is strictly less than \( 2 \). The only option consistent with these results is (B): \( I < \frac{2}{3} \) and \( J < 2 \).

Key Concepts

Inequalities in Calculus

When dealing with the integrals of functions, particularly when some elements are not easily integrable directly, inequalities can simplify the process by providing bounds. In this problem, we're given the integral \( I = \int_{0}^{1} \frac{\sin x}{\sqrt{x}} \, dx \).
By acknowledging the inequality \( \sin x \leq x \) for small values of \( x \), we can simplify the approach to solving this integral. By replacing \( \sin x \) with \( x \), the integral simplifies to \( \int_{0}^{1} \sqrt{x} \, dx \), which is much easier to solve.
This gives a way to approximate the upper limit of the integral \( I \) indicating that \( I \leq \frac{2}{3} \). The use of inequalities allows us to reason about the value of an integral without directly calculating it.

Approximation Techniques

In calculus, approximation techniques are crucial when dealing with complex integrals or where exact values are hard to compute. For the integral \( J = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} \, dx \), direct computation is cumbersome, so approximation offers a simpler path.
By considering \( \cos x \approx 1 \) for small values of \( x \), we can approximate the function \( \frac{\cos x}{\sqrt{x}} \) as \( \frac{1}{\sqrt{x}} \). This is particularly useful because \( \frac{1}{\sqrt{x}} \) is a familiar form, and its integral from 0 to 1 is straightforward to compute, resulting in a value of 2.
While this initial approximation suggests \( J \approx 2 \), acknowledging that \( \cos x < 1 \) for most of this range ensures that \( J < 2 \), thus refining our approximation further and giving a realistic bound to work with.

Trigonometric Integrals

Evaluating integrals that incorporate trigonometric functions can often involve special strategies due to their unique properties and periodicity. In our example with integrals involving \( \sin x \) and \( \cos x \), these functions behave predictably for small values of \( x \).
For \( \sin x \), the fact that it equals or is less than \( x \) itself for values nearing zero helps us to establish an upper bound integral. This means we can confidently say \( \sin x \) effectively provides its own natural approximation when bounded with functions of \( x \).
Trigonometric integrals often depend on known identities or inequalities like \( \sin x \leq x \) and \( \cos x \leq 1 \). Recognizing these properties helps simplify calculations or allows approximations that are near to the actual values, often yielding valuable insights without necessitating complete integration.