JEE Exam > JEE Notes > Mathematics (Maths) for JEE Main & Advanced > Solved Example: Application of Integral
Solved Example: Application of Integral | Mathematics (Maths) for JEE Main & Advanced PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1 JEE Solved Example on Application of Integral JEE Mains Q1: The area bounded by the ?? -axis, the curve ?? = ?? ( ?? ) and the lines ?? = ?? , ?? = ?? is equal to v?? ?? + ?? - v ?? for all ?? > ?? , then ?? ( ?? ) is (a) v ?? - ?? (b) v ?? + ?? (c) v ?? ?? - ?? (d) ?? v ?? +?? ?? Ans: (d) ? 1 ?? ??? ( ?? ) ???? = v ?? 2 + 1 - v 2 = [v ?? 2 + 1] 1 ?? ? ?? ( ?? ) = 1 2 · 2?? v?? 2 + 1 Hence ?? ( ?? ) = ?? v1+?? 2 . Q2: The area of the region bounded by the curve ?? = ?? - ?? ?? between ?? = ?? and ?? = ?? is (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) ?? ?? Ans: (a) Required Area = ? 0 1 ?( ?? - ?? 2 ) ???? = [ ?? 2 2 - ?? 3 3 ] 0 1 = 1 2 - 1 3 = 1 6 . Q3: Find the area bounded between the curve ?? ?? = ?? ?? - ?? and ?? -axis . (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) 5 Ans: (a) The area between the given curve ?? = 2?? - ?? 2 and ?? -axis will be as shown Page 2 JEE Solved Example on Application of Integral JEE Mains Q1: The area bounded by the ?? -axis, the curve ?? = ?? ( ?? ) and the lines ?? = ?? , ?? = ?? is equal to v?? ?? + ?? - v ?? for all ?? > ?? , then ?? ( ?? ) is (a) v ?? - ?? (b) v ?? + ?? (c) v ?? ?? - ?? (d) ?? v ?? +?? ?? Ans: (d) ? 1 ?? ??? ( ?? ) ???? = v ?? 2 + 1 - v 2 = [v ?? 2 + 1] 1 ?? ? ?? ( ?? ) = 1 2 · 2?? v?? 2 + 1 Hence ?? ( ?? ) = ?? v1+?? 2 . Q2: The area of the region bounded by the curve ?? = ?? - ?? ?? between ?? = ?? and ?? = ?? is (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) ?? ?? Ans: (a) Required Area = ? 0 1 ?( ?? - ?? 2 ) ???? = [ ?? 2 2 - ?? 3 3 ] 0 1 = 1 2 - 1 3 = 1 6 . Q3: Find the area bounded between the curve ?? ?? = ?? ?? - ?? and ?? -axis . (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) 5 Ans: (a) The area between the given curve ?? = 2?? - ?? 2 and ?? -axis will be as shown ? Required Area = ? ? 2 0 ?( 2?? - ?? 2 ) ???? = [?? 2 - ?? 3 3 ] 0 2 = 4 3 Q4: Find the area bounded by the curves ?? = ?? ?????? ?? , ?? = ?? ?????? ?? in the first quadrant (a) ?????? ?? (b) ?? ?? ?? ?? ?? (c) ???? ?? ?? ?? (d) None of these Ans: (a) Clearly the given equation are the parametric equation of ellipse ?? 2 ?? 2 + ?? 2 ?? 2 = 1. Curve meet the ?? -axis in the first quadrant at ( ?? , ?? ) ? Required area ? 0 ?? ??????? = ? ?? 2 0 ?( ?? sin ?? ) ( -?? cos ?? ) ???? = ???? ? 0 ?? /2 ?sin 2 ?????? = ( ?????? 4 ) ( ? At ?? = 0, ?? = ?? /2 and ?? = ?? , ?? = 0) Q5: Find the whole area of circle ?? ?? + ?? ?? = ?? ?? (a) ?? (b) ?? ?? ?? (c) ?? ?? ?? (d) ?? ?? Ans: (b) The required area is symmetric about both the axis as shown in figure Page 3 JEE Solved Example on Application of Integral JEE Mains Q1: The area bounded by the ?? -axis, the curve ?? = ?? ( ?? ) and the lines ?? = ?? , ?? = ?? is equal to v?? ?? + ?? - v ?? for all ?? > ?? , then ?? ( ?? ) is (a) v ?? - ?? (b) v ?? + ?? (c) v ?? ?? - ?? (d) ?? v ?? +?? ?? Ans: (d) ? 1 ?? ??? ( ?? ) ???? = v ?? 2 + 1 - v 2 = [v ?? 2 + 1] 1 ?? ? ?? ( ?? ) = 1 2 · 2?? v?? 2 + 1 Hence ?? ( ?? ) = ?? v1+?? 2 . Q2: The area of the region bounded by the curve ?? = ?? - ?? ?? between ?? = ?? and ?? = ?? is (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) ?? ?? Ans: (a) Required Area = ? 0 1 ?( ?? - ?? 2 ) ???? = [ ?? 2 2 - ?? 3 3 ] 0 1 = 1 2 - 1 3 = 1 6 . Q3: Find the area bounded between the curve ?? ?? = ?? ?? - ?? and ?? -axis . (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) 5 Ans: (a) The area between the given curve ?? = 2?? - ?? 2 and ?? -axis will be as shown ? Required Area = ? ? 2 0 ?( 2?? - ?? 2 ) ???? = [?? 2 - ?? 3 3 ] 0 2 = 4 3 Q4: Find the area bounded by the curves ?? = ?? ?????? ?? , ?? = ?? ?????? ?? in the first quadrant (a) ?????? ?? (b) ?? ?? ?? ?? ?? (c) ???? ?? ?? ?? (d) None of these Ans: (a) Clearly the given equation are the parametric equation of ellipse ?? 2 ?? 2 + ?? 2 ?? 2 = 1. Curve meet the ?? -axis in the first quadrant at ( ?? , ?? ) ? Required area ? 0 ?? ??????? = ? ?? 2 0 ?( ?? sin ?? ) ( -?? cos ?? ) ???? = ???? ? 0 ?? /2 ?sin 2 ?????? = ( ?????? 4 ) ( ? At ?? = 0, ?? = ?? /2 and ?? = ?? , ?? = 0) Q5: Find the whole area of circle ?? ?? + ?? ?? = ?? ?? (a) ?? (b) ?? ?? ?? (c) ?? ?? ?? (d) ?? ?? Ans: (b) The required area is symmetric about both the axis as shown in figure ? Required area = 4 ? ? ?? 0 ? v?? 2 - ?? 2 ???? = 4 [ ?? 2 v?? 2 - ?? 2 + ?? 2 2 sin -1 ?? ?? ] 0 ?? = 4 [ ?? 2 × ?? 2 2 ] = ?? ?? 2 Q6: Find the area bounded by the parabola ?? ?? = ?? ?? and its latus rectum [ (a) ?? ?? (b) ?? ?? (c) ???? ?? (d) None of these Ans: (a) Since the curve is symmetrical about ?? -axis, therefore the required area = 2 ? ? 1 0 ??????? = 2 ? ? 1 0 ?v 4?? ???? = 4 · 2 3 [?? 3 2 ] 0 1 = 8 3 Page 4 JEE Solved Example on Application of Integral JEE Mains Q1: The area bounded by the ?? -axis, the curve ?? = ?? ( ?? ) and the lines ?? = ?? , ?? = ?? is equal to v?? ?? + ?? - v ?? for all ?? > ?? , then ?? ( ?? ) is (a) v ?? - ?? (b) v ?? + ?? (c) v ?? ?? - ?? (d) ?? v ?? +?? ?? Ans: (d) ? 1 ?? ??? ( ?? ) ???? = v ?? 2 + 1 - v 2 = [v ?? 2 + 1] 1 ?? ? ?? ( ?? ) = 1 2 · 2?? v?? 2 + 1 Hence ?? ( ?? ) = ?? v1+?? 2 . Q2: The area of the region bounded by the curve ?? = ?? - ?? ?? between ?? = ?? and ?? = ?? is (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) ?? ?? Ans: (a) Required Area = ? 0 1 ?( ?? - ?? 2 ) ???? = [ ?? 2 2 - ?? 3 3 ] 0 1 = 1 2 - 1 3 = 1 6 . Q3: Find the area bounded between the curve ?? ?? = ?? ?? - ?? and ?? -axis . (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) 5 Ans: (a) The area between the given curve ?? = 2?? - ?? 2 and ?? -axis will be as shown ? Required Area = ? ? 2 0 ?( 2?? - ?? 2 ) ???? = [?? 2 - ?? 3 3 ] 0 2 = 4 3 Q4: Find the area bounded by the curves ?? = ?? ?????? ?? , ?? = ?? ?????? ?? in the first quadrant (a) ?????? ?? (b) ?? ?? ?? ?? ?? (c) ???? ?? ?? ?? (d) None of these Ans: (a) Clearly the given equation are the parametric equation of ellipse ?? 2 ?? 2 + ?? 2 ?? 2 = 1. Curve meet the ?? -axis in the first quadrant at ( ?? , ?? ) ? Required area ? 0 ?? ??????? = ? ?? 2 0 ?( ?? sin ?? ) ( -?? cos ?? ) ???? = ???? ? 0 ?? /2 ?sin 2 ?????? = ( ?????? 4 ) ( ? At ?? = 0, ?? = ?? /2 and ?? = ?? , ?? = 0) Q5: Find the whole area of circle ?? ?? + ?? ?? = ?? ?? (a) ?? (b) ?? ?? ?? (c) ?? ?? ?? (d) ?? ?? Ans: (b) The required area is symmetric about both the axis as shown in figure ? Required area = 4 ? ? ?? 0 ? v?? 2 - ?? 2 ???? = 4 [ ?? 2 v?? 2 - ?? 2 + ?? 2 2 sin -1 ?? ?? ] 0 ?? = 4 [ ?? 2 × ?? 2 2 ] = ?? ?? 2 Q6: Find the area bounded by the parabola ?? ?? = ?? ?? and its latus rectum [ (a) ?? ?? (b) ?? ?? (c) ???? ?? (d) None of these Ans: (a) Since the curve is symmetrical about ?? -axis, therefore the required area = 2 ? ? 1 0 ??????? = 2 ? ? 1 0 ?v 4?? ???? = 4 · 2 3 [?? 3 2 ] 0 1 = 8 3 Q7: The area bounded by the curve ?? ?? = ?? ?? and ?? ?? = ?? ?? is (a) ???? ?? sq. units (b) ?? ???? sq. units (c) ???? ?? sq. units (d) ?? ???? sq. units Ans: (a) Required area = ? 0 4 ?( ???????? - ???????? ) Region = ? 0 4 ?(v 4?? - ?? 2 4 )???? = 16 3 square unit. Trick : From Important Tips' the area of the region bounded by ?? 2 = 4???? and ?? 2 = 4???? is 16???? 3 square unit. Here ?? 2 = 4?? and ?? 2 = 4?? , so ?? = 1 and ?? = 1 Required area = 16 3 ( 1) ( 1) = 16 3 square unit. Q8: The area of the bounded region by the curve ?? = ?????? ?? , the ?? -axis and the line ?? = ?? and ?? = ?? is (a) 4 (b) 2 (c) ?? (d) None of these Ans: (b) Required area = ? ? ?? 0 ?sin ?????? = 2 ? ? ?? /2 0 ?sin ?????? = 2[-cos ?? ] 0 ?? /2 = 2[( -cos ?? /2)- ( -cos 0) ] = 2( 1) = 2 square unit. Page 5 JEE Solved Example on Application of Integral JEE Mains Q1: The area bounded by the ?? -axis, the curve ?? = ?? ( ?? ) and the lines ?? = ?? , ?? = ?? is equal to v?? ?? + ?? - v ?? for all ?? > ?? , then ?? ( ?? ) is (a) v ?? - ?? (b) v ?? + ?? (c) v ?? ?? - ?? (d) ?? v ?? +?? ?? Ans: (d) ? 1 ?? ??? ( ?? ) ???? = v ?? 2 + 1 - v 2 = [v ?? 2 + 1] 1 ?? ? ?? ( ?? ) = 1 2 · 2?? v?? 2 + 1 Hence ?? ( ?? ) = ?? v1+?? 2 . Q2: The area of the region bounded by the curve ?? = ?? - ?? ?? between ?? = ?? and ?? = ?? is (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) ?? ?? Ans: (a) Required Area = ? 0 1 ?( ?? - ?? 2 ) ???? = [ ?? 2 2 - ?? 3 3 ] 0 1 = 1 2 - 1 3 = 1 6 . Q3: Find the area bounded between the curve ?? ?? = ?? ?? - ?? and ?? -axis . (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) 5 Ans: (a) The area between the given curve ?? = 2?? - ?? 2 and ?? -axis will be as shown ? Required Area = ? ? 2 0 ?( 2?? - ?? 2 ) ???? = [?? 2 - ?? 3 3 ] 0 2 = 4 3 Q4: Find the area bounded by the curves ?? = ?? ?????? ?? , ?? = ?? ?????? ?? in the first quadrant (a) ?????? ?? (b) ?? ?? ?? ?? ?? (c) ???? ?? ?? ?? (d) None of these Ans: (a) Clearly the given equation are the parametric equation of ellipse ?? 2 ?? 2 + ?? 2 ?? 2 = 1. Curve meet the ?? -axis in the first quadrant at ( ?? , ?? ) ? Required area ? 0 ?? ??????? = ? ?? 2 0 ?( ?? sin ?? ) ( -?? cos ?? ) ???? = ???? ? 0 ?? /2 ?sin 2 ?????? = ( ?????? 4 ) ( ? At ?? = 0, ?? = ?? /2 and ?? = ?? , ?? = 0) Q5: Find the whole area of circle ?? ?? + ?? ?? = ?? ?? (a) ?? (b) ?? ?? ?? (c) ?? ?? ?? (d) ?? ?? Ans: (b) The required area is symmetric about both the axis as shown in figure ? Required area = 4 ? ? ?? 0 ? v?? 2 - ?? 2 ???? = 4 [ ?? 2 v?? 2 - ?? 2 + ?? 2 2 sin -1 ?? ?? ] 0 ?? = 4 [ ?? 2 × ?? 2 2 ] = ?? ?? 2 Q6: Find the area bounded by the parabola ?? ?? = ?? ?? and its latus rectum [ (a) ?? ?? (b) ?? ?? (c) ???? ?? (d) None of these Ans: (a) Since the curve is symmetrical about ?? -axis, therefore the required area = 2 ? ? 1 0 ??????? = 2 ? ? 1 0 ?v 4?? ???? = 4 · 2 3 [?? 3 2 ] 0 1 = 8 3 Q7: The area bounded by the curve ?? ?? = ?? ?? and ?? ?? = ?? ?? is (a) ???? ?? sq. units (b) ?? ???? sq. units (c) ???? ?? sq. units (d) ?? ???? sq. units Ans: (a) Required area = ? 0 4 ?( ???????? - ???????? ) Region = ? 0 4 ?(v 4?? - ?? 2 4 )???? = 16 3 square unit. Trick : From Important Tips' the area of the region bounded by ?? 2 = 4???? and ?? 2 = 4???? is 16???? 3 square unit. Here ?? 2 = 4?? and ?? 2 = 4?? , so ?? = 1 and ?? = 1 Required area = 16 3 ( 1) ( 1) = 16 3 square unit. Q8: The area of the bounded region by the curve ?? = ?????? ?? , the ?? -axis and the line ?? = ?? and ?? = ?? is (a) 4 (b) 2 (c) ?? (d) None of these Ans: (b) Required area = ? ? ?? 0 ?sin ?????? = 2 ? ? ?? /2 0 ?sin ?????? = 2[-cos ?? ] 0 ?? /2 = 2[( -cos ?? /2)- ( -cos 0) ] = 2( 1) = 2 square unit. Trick : For the curve ?? = sin ?? or cos ?? , the area of ? 0 ?? /2 ?sin ?????? = 1, ? 0 ?? ?sin ?????? = 2, ? 0 3?? /2 ?sin ?????? = 3, ? 0 2?? ?sin ?????? = 4 and so on. Q9: The area enclosed by the parabola ?? ?? = ?? ?? and the line ?? = ?? ?? is (a) ?? ?? (b) ?? ?? (c) ?? ?? (d) ?? ?? Ans: (a) Solve the equation ?? 2 = 8?? and the line ?? = 2?? , we get the point of intersection. Then find the required area bounded by this region. It is 4 3 . Trick : Required area = 8( 2) 2 3( 2) 3 = 32 24 = 4 3 [? Area bounded by ?? 2 = 4???? and ?? = ???? is 8?? 2 3?? 3 . Here ?? = 2, ?? = 2] Q10: If the area bounded by ?? = ?? ?? ?? and ?? = ?? ?? ?? , ?? > ?? , is 1 , then ?? = (a) 1 (b) ?? v ?? (c) ?? ?? (d) - ?? v ?? Ans: (b) The ?? coordinate of A is 1 ?? According to the given conditionRead More
|
209 videos|443 docs|143 tests
|
FAQs on Solved Example: Application of Integral - Mathematics (Maths) for JEE Main & Advanced
| 1. What is the application of integrals in JEE? |  |
Ans. In JEE, the application of integrals is used to solve problems related to finding areas, volumes, and even calculating work done in physics problems. It is an essential concept in calculus that helps in solving real-life problems.
| 2. How can integrals be used to find the area under a curve in JEE? | |
Ans. Integrals can be used to find the area under a curve by setting up definite integrals with appropriate limits of integration. By integrating the function over the interval, the area under the curve can be calculated accurately.
| 3. Can integrals be used to calculate the volume of a solid in JEE? | |
Ans. Yes, integrals can be used to calculate the volume of a solid by setting up triple integrals for three-dimensional shapes or using double integrals for two-dimensional shapes. By integrating the appropriate function, the volume of the solid can be determined.
| 4. How does the concept of integrals help in solving optimization problems in JEE? | |
Ans. The concept of integrals helps in solving optimization problems in JEE by setting up an equation to be optimized and using calculus techniques to find the maximum or minimum value. Integrals play a crucial role in determining the optimal solution in such problems.
| 5. What are some practical applications of integrals in JEE beyond calculus problems? | |
Ans. Some practical applications of integrals in JEE beyond calculus problems include using integrals in physics to calculate work done, in economics to find total revenue or profit, and in engineering to analyze stress distribution in structures. Integrals have diverse applications in various fields beyond mathematics.
Related Exams
About this Document
|
4.84/5 Rating |
|
Nov 23, 2024 Last updated |
Document Description: Solved Example: Application of Integral for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation.
The notes and questions for Solved Example: Application of Integral have been prepared according to the JEE exam syllabus. Information about Solved Example: Application of Integral covers topics
like and Solved Example: Application of Integral Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Solved Example: Application of Integral.
Introduction of Solved Example: Application of Integral in English is available as part of our Mathematics (Maths) for JEE Main & Advanced
for JEE & Solved Example: Application of Integral in Hindi for Mathematics (Maths) for JEE Main & Advanced course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: Solved Example: Application of Integral | Mathematics (Maths) for JEE Main & Advanced
Description
Full syllabus notes, lecture & questions for Solved Example: Application of Integral | Mathematics (Maths) for JEE Main & Advanced - JEE | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) for JEE Main & Advanced | Best notes, free PDF download
Information about Solved Example: Application of Integral
In this doc you can find the meaning of Solved Example: Application of Integral defined & explained in the simplest way possible. Besides explaining types of
Solved Example: Application of Integral theory, EduRev gives you an ample number of questions to practice Solved Example: Application of Integral tests, examples and also practice JEE
tests
|
209 videos|443 docs|143 tests
|
Download as PDF
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
past year papers
,mock tests for examination
,Solved Example: Application of Integral | Mathematics (Maths) for JEE Main & Advanced
,Previous Year Questions with Solutions
,Semester Notes
,Solved Example: Application of Integral | Mathematics (Maths) for JEE Main & Advanced
,MCQs
,study material
,video lectures
,Free
,Sample Paper
,shortcuts and tricks
,Viva Questions
,practice quizzes
,Objective type Questions
,ppt
,Extra Questions
,Exam
,Solved Example: Application of Integral | Mathematics (Maths) for JEE Main & Advanced
,Important questions
,Summary
;
Additional Information about Solved Example: Application of Integral for JEE Preparation
Solved Example: Application of Integral Free PDF Download
The Solved Example: Application of Integral is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Solved Example: Application of Integral now and kickstart your journey towards success in the JEE exam.
Importance of Solved Example: Application of Integral
The importance of Solved Example: Application of Integral cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Solved Example: Application of Integral Notes
Solved Example: Application of Integral Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Solved Example: Application of Integral.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Solved Example: Application of Integral Notes on EduRev are your ultimate resource for success.
Solved Example: Application of Integral JEE Questions
The "Solved Example: Application of Integral JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Solved Example: Application of Integral on the App
Students of JEE can study Solved Example: Application of Integral alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Solved Example: Application of Integral,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Solved Example: Application of Integral is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup