/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 Let \(I=\int_{0}^{1} \frac{\sin ... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(J<2\)

Short Answer

Expert verified
Option (B): \( I < \frac{2}{3} \) and \( J \leq 2 \).

Step by step solution

01

Analyze the Integrals I and J

Given the integrals \( I = \int_{0}^{1} \frac{\sin x}{\sqrt{x}} \, dx \) and \( J = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} \, dx \), we aim to determine which relationships between \( I \) and \( J \) hold. Recognizing that the integrands \( \frac{\sin x}{\sqrt{x}} \) and \( \frac{\cos x}{\sqrt{x}} \) involve oscillatory trigonometric functions weighted by \( \frac{1}{\sqrt{x}} \), we need to estimate these integrals.
02

Use Integration Techniques – I

Examine \( \int_{0}^{1} \sin x \, dx \) and \( \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx \) separately: \( \int_{0}^{1} \sin x \, dx = 1 - \cos 1 \), and \( \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx = 2 \). Consider that the integral of \( \sin x \) over this range shows positive behavior, but is diminished by \( \frac{1}{\sqrt{x}} \) at \( x = 0 \). The integral \( I \) is likely less than an estimate built from these two separate behaviors: \( \int_0^1 \frac{\sin x}{\sqrt{x}} \, dx \leq \int_0^1 \sin x \, dx \cdot \int_0^1 \frac{1}{\sqrt{x}} \, dx = (1-\cos 1) \cdot 2 \approx 1.08 \).
03

Estimate I

To approximate \( I \), compare \( \sin x \) with simple bounds. Noting \( 0 \leq \sin x \leq x \) for \( x \in [0,1] \), simplify: \( 0 \leq \frac{\sin x}{\sqrt{x}} \leq \sqrt{x} \). So, \( I \approx \int_{0}^{1} \sqrt{x} \, dx = \frac{2}{3} \). Hence, \( I < \frac{2}{3} \).
04

Use Integration Techniques – J

Focus on \( J = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} \, dx \). Like \( I \), handle \( \int_{0}^{1} \cos x \, dx \) and \( \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx \) separately to get upper potential bounds. Since \( (\sin 1) \cdot 2 \approx 1.68 \), this shows \( \cos x \) does not serve to increases significantly beyond higher order behavior at infinity, implying \( J \leq 2 \).
05

Estimate J

By examining the asymptotic behavior near \( x=0 \) and more rigorously comparing with approximations, bounding functions such as polynomials are used: set \( 0 \leq \cos x \leq 1 \) so that \( J \approx \int_0^1 \frac{1}{\sqrt{x}} \, dx \cdot \int_0^1 1 \, dx = 2 \). The nature of behavior confirms \( J \leq 2 \).
06

Conclusion

From these analysis steps, we find the conditions that satisfy \( I < \frac{2}{3} \) and \( J \leq 2 \). Therefore, the correct option corresponding to these evaluations is choice (B).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Integrals
Trigonometric integrals involve the integration of functions that include trigonometric expressions, either sine, cosine, or their combinations. In this exercise, we encounter the integrals\[I = \int_{0}^{1} \frac{\sin x}{\sqrt{x}} \, dx \quad \text{and} \quad J = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} \, dx.\]By integrating these expressions, we explore how oscillatory behavior, characteristic of sine and cosine, interacts with other functions—in this case, the reciprocal square root function.
One intuitive approach in evaluating such integrals is to consider bounds for the trigonometric functions. Given that \(\sin x\) ranges between 0 and 1 and \(\cos x\) ranges between -1 and 1 over their domain \([0, \pi/2]\), applying these considerations helps approximate or bound the integrals.
This theory is crucial for students because understanding these integrals involves recognizing the interplay between the shape and periodic nature of trigonometric functions and other components in the integrand, especially when limits of integration begin or end where the sine or cosine value is zero or one. This understanding also helps in formulating approaches to evaluate or estimate more complex trigonometric integrals.
Definite Integrals
Definite integrals, such as those we are solving here, are integrals with upper and lower limits. In this case, both integrals are evaluated between 0 and 1. Definite integrals are useful as they provide the exact area under a curve between two specified points.
For our exercise, the definite integrals are:\[I = \int_{0}^{1} \frac{\sin x}{\sqrt{x}} \, dx \quad \text{and} \quad J = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} \, dx.\]Evaluating these involves calculating the areas taking into account both the oscillatory nature of the trig functions and the inverse square root that creates a singularity at 0.
Employing definite integrals not only finds the specific area but also helps in interpreting the behavior of complex functions over an interval, making them essential in practical applications such as physics and engineering, where understanding total quantities or behaviors is necessary.
Integration Techniques
The solution of our problem leverages various integration techniques. Understanding and applying the right techniques is fundamental for approximating or solving integrals that may not have elementary antiderivatives.
In analyzing our given integrals, methods such as 'integral comparison tests' or 'upper and lower bounding functions' play a fundamental role.- **Comparative Analysis**: Realize that \(\sin x\) can be approximated by simpler functions like \(x\) for small intervals, and \(\cos x\) by 1. This helps in setting upper bounds when finding exact integrals analytically is challenging.- **Substitution and Standard Forms**: Breaking complex parts of the integral into standard integration problems, or using substitutions can simplify the integral into a more familiar form.
The gained insight from such techniques builds the foundation necessary for tackling more complex integrals and diverse mathematical scenarios students may encounter later.

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Most popular questions from this chapter

The value of the integral \(\int_{0}^{1} \frac{8 \log (1+x)}{1+x^{2}} d x\) is (A) \(\frac{\pi}{8} \log 2\) (B) \(\frac{\pi}{2} \log 2\) (C) \(\log 2\) (D) \(\pi \log 2\)

A function \(f\) defined on \([a, b]\) is bounded above if there exists a real number \(K\) such that \(|f(x)| \leq K, \forall x \in[a, b]\). The real number \(K\) is called an upper bound of \(f . f\) is called bounded below if there exists a real number \(k\) such that \(|f(x)| \geq k, \forall x \in[a, b]\). The real number \(k\) is called a lower bound of \(f\). It may be observed that if \(K\) is an upper bound of \(f\), then every real number greater than or equal to \(K\) is also an upper bound of \(f\). The smallest of all the upper bounds of \(f\) is called the least upper bound or the supremum (sup.) of \(f\). Similarly, the greatest of all the lower bounds of \(f\) is called the greatest lower bound or the infimum (Inf.) of \(f\). If the functions \(f\) and \(g\) are bounded and continuous on \([a, b]\) and \(g\) keeps the same sign on \([a, b]\), then there exists a number \(\mu\) between the infimum and supremum of \(f\) on \([a, b]\) such that $$ \int_{a}^{b} f(x) g(x) d x=\mu \int_{a}^{b} g(x) d x $$. \(\int_{0}^{1} \frac{\sin \pi x}{1+x^{2}} d x\) lies between (A) \(\frac{1}{\pi}\) and \(\frac{2}{\pi}\) (B) \(\frac{\pi}{2}\) and \(\pi\) (C) \(\pi\) and \(\frac{3 \pi}{2}\) (D) None of these

The area of the region bounded by the parabola \((y-2)^{2}=x-1\), the tangent to the parabola at the point \((2,3)\) and the \(x\)-axis is (A) 3 (B) 6 (C) 9 (D) 12

If \(\int_{a}^{b}|\sin x| d x=8\) and \(\int_{a}^{a+b}|\cos x| d x=\frac{9}{2}\), then (A) \(a=\frac{\pi}{2}\) (B) \(b=\frac{11 \pi}{4}\) (C) \(a=\frac{\pi}{4}\) (D) \(b=\frac{17 \pi}{4}\)

In the following questions an Assertion (A) is given followed by a Reason (R). Mark your responses from the following options: (A) Assertion (A) is True and Reason (R) is True; Reason \((\mathrm{R})\) is a correct explanation for Assertion (A) (B) Assertion (A) is True, Reason \((\mathrm{R})\) is True; Reason (R) is not a correct explanation for Assertion (A) (C) Assertion (A) is True, Reason (R) is False (D) Assertion (A) is False, Reason (R) is True Assertion: If \(f(x)\) is a non-negative continuous function such that \(f(x)+f\left(x+\frac{1}{2}\right)=1\), then \(\int_{0}^{2} f(x) d x=1\) Reason: \(f(x)\) is a periodic function having period \(1 .\)

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