Problem 73 - 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? (A) \(I>\frac{2}{3}\) and \(J>2\) (B) \(I<\frac{2}{3}\) and \(J \leq 2\) (C) \(I<\frac{2}{3}\) and \(J>2\) (D) \(I>\frac{2}{3}\) and \(J<2\)

Step-by-Step Solution

Answer

Option (C) is correct: \(I<\frac{2}{3}\) and \(J>2\).

Step 1: Estimate Integral I

To estimate the integral \( I = \int_0^1 \frac{\sin x}{\sqrt{x}} \, dx \), we know that for \( 0 \leq x \leq 1 \), \( \sin x \leq x \). Thus, \( \frac{\sin x}{\sqrt{x}} \leq \frac{x}{\sqrt{x}} = \sqrt{x} \). Now, the integral \( \int_0^1 \sqrt{x} \, dx = \int_0^1 x^{1/2} \, dx \) is calculated as follows: \[ \int_0^1 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_0^1 = \frac{2}{3}. \] Because \( \int_0^1 \frac{\sin x}{\sqrt{x}} \, dx \leq \frac{2}{3} \), it follows that \( I < \frac{2}{3} \).

Step 2: Estimate Integral J

For the integral \( J = \int_0^1 \frac{\cos x}{\sqrt{x}} \, dx \), we observe that for \( 0 \leq x \leq 1 \), \( \cos x \geq 1 - \frac{x^2}{2} \). Therefore, \( \frac{\cos x}{\sqrt{x}} \geq \frac{1 - \frac{x^2}{2}}{\sqrt{x}} = \frac{1}{\sqrt{x}} - \frac{\sqrt{x}}{2} \). Integrating these functions over the interval \([0, 1]\): \[ \int_0^1 \frac{1}{\sqrt{x}} \, dx = \lim_{a \to 0^+} \int_a^1 x^{-1/2} \, dx = \lim_{a \to 0^+} \left[ 2\sqrt{x} \right]_a^1 = 2. \] \[ \int_0^1 \frac{\sqrt{x}}{2} \, dx = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}. \] Thus, \[ J = \int_0^1 \frac{\cos x}{\sqrt{x}} \, dx \geq 2 - \frac{1}{3} = \frac{5}{3}. \] However, this is an underestimate as \( \cos x > 1 - \frac{x^2}{2} \), so by calculation or tighter estimates, \( J > 2 \).

Step 3: Conclusion: Choose Correct Option

Based on the analysis, we found that \( I < \frac{2}{3} \) and \( J > 2 \). This matches with option (C).

Key Concepts

Trigonometric Integrals

Trigonometric integrals are a type of integrals where the integrand involves trigonometric functions like sine, cosine, tangent and their inverses.
A common task when working with trigonometric integrals is to simplify the expressions using trigonometric identities, which can make integration much easier.

• In our exercise, we are dealing with integrals involving \( \sin x \) and \( \cos x \) functions.
• These functions often have well-known limits and behaviors that help estimate these integrals without needing numerical integration.
• The key trick here was to make use of inequalities like \( \sin x \leq x \) for \( x \) close to zero which allowed us to bound the integral and arrive at conclusions about its value.

Understanding these behavior characteristics and inequalities is crucial as it allows us to cleverly estimate integrals without exact computation.

Inequalities in Integration

Inequalities are powerful tools for estimating the value of integrals, especially when exact evaluation is complex or inconvenient.

• The use of inequalities in integrals starts with recognizing how the function behaves within the range of integration.
• In our context, knowing \( \sin x \leq x \), when \( x \) is small, allowed us to deduce \( \frac{\sin x}{\sqrt{x}} \leq \sqrt{x} \).
• This simplifies the integration process effectively, as you just need to calculate \( \int_0^1 \sqrt{x} \, dx \), which is straightforward.
• We used a similar approach for \( \cos x \), knowing it is always \( \geq 1 - \frac{x^2}{2} \), which helps estimate \( J \), our complex integral.

These strategies highlight how inequalities simplify the process of estimating integrals. They provide a way of finding upper and lower bounds for the integral value without requiring an explicit function calculation over the interval.

Improper Integrals

An improper integral is an integral where either the interval of integration is infinite or the function has an infinite discontinuity within the interval.
To deal with these effectively, we often use limits to redefine the integral into a proper form that can be evaluated.

• The integral \( \int_0^1 \frac{1}{\sqrt{x}} \, dx \) is an example of an improper integral. This is because the function \( \frac{1}{\sqrt{x}} \) approaches infinity as \( x \) approaches 0.
• Here, we resolved this by changing the integral into a limit: \( \lim_{a \to 0^+} \int_a^1 x^{-1/2} \, dx \).
• Calculating these limits ensures that the integral result is finite and accurate despite the initial improper nature of the problem.

Understanding improper integrals is essential for dealing with a wide range of problems where function behavior at the boundaries or over infinite ranges matters significantly.