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Last updated at Dec. 12, 2019 by Teachoo

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Misc 11 Using the method of integration find the area bounded by the curve |๐ฅ|+|๐ฆ|=1 [Hint: The required region is bounded by lines ๐ฅ+๐ฆ= 1, ๐ฅ โ๐ฆ=1, โ๐ฅ+๐ฆ =1 and โ๐ฅ โ๐ฆ=1 ] We know that "โ" ๐ฅ"โ"={โ(๐ฅ, ๐ฅโฅ0@&โ๐ฅ, ๐ฅ<0)โค & "โ" ๐ฆ"โ"={โ(๐ฆ, ๐ฆโฅ0@&โ๐ฆ, ๐ฆ<0)โค So, we can write โ๐ฅ"โ+โ" ๐ฆ"โ"=1 as {โ(โ(โ( ๐ฅ+๐ฆ=1 ๐๐๐ ๐ฅ>0 , ๐ฆ>0@โ๐ฅ+๐ฆ=1 ๐๐๐ ๐ฅ<0 ๐ฆ>0)@โ( ๐ฅโ๐ฆ =1 ๐๐๐ ๐ฅ>0 , ๐ฆ<0@โ๐ฅโ๐ฆ=1 ๐๐๐ ๐ฅ<0 ๐ฆ<0)))โค For ๐+๐=๐ For โ๐+๐=๐ For โ๐โ๐=๐ For ๐โ๐=๐ Joining them, we get our diagram Since the Curve symmetrical about ๐ฅ & ๐ฆโ๐๐ฅ๐๐ Required Area = 4 ร Area AOB Area AOB Area AOB = โซ_0^1โใ๐ฆ ๐๐ฅใ where ๐ฅ+๐ฆ=1 ๐ฆ=1โ๐ฅ Therefore, Area AOB = โซ_0^1โใ(1โ๐ฅ) ๐๐ฅใ = [๐ฅโ๐ฅ^2/2]_0^1 =1โใ 1ใ^2/2โ(0โ0^2/2)^2 =1โ1/2 =1/2 Hence, Required Area = 4 ร Area AOB = 4 ร 1/2 = 2 square units

Chapter 8 Class 12 Application of Integrals

Class 12

Important Questions for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.