# Misc 11

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 11 Using the method of integration find the area bounded by the curve 𝑥+ 𝑦=1 [Hint: The required region is bounded by lines 𝑥+𝑦= 1, 𝑥 –𝑦=1, –𝑥+𝑦 =1 and −𝑥 −𝑦=1 ] We know that │𝑥│= 𝑥, 𝑥≥0&−𝑥, 𝑥<0 & │𝑦│= 𝑦, 𝑦≥0&−𝑦, 𝑦<0 So, we can write │𝑥│+│𝑦│=1 as 𝑥+𝑦=1 𝑓𝑜𝑟 𝑥>0 , 𝑦>0−𝑥+𝑦=1 𝑓𝑜𝑟 𝑥<0 𝑦>0 𝑥−𝑦 =1 𝑓𝑜𝑟 𝑥>0 , 𝑦<0−𝑥−𝑦=1 𝑓𝑜𝑟 𝑥<0 𝑦<0 For 𝒙+𝒚=𝟏 For −𝒙+𝒚=𝟏 Hence the figure is Since the Curve symmetrical about 𝑥 & 𝑦−𝑎𝑥𝑖𝑠 Required Area = 4 × Area AOB Area AOB Area ABO = 01𝑦 𝑑𝑥 where 𝑥+𝑦=1 𝑦=1−𝑥 Therefore, Area ABO = 01 1−𝑥 𝑑𝑥 = 𝑥− 𝑥2201 =1− 122− 0− 0222 =1− 12 = 12 Hence, Required Area = 4 × Area AOB = 4 × 12 = 2 square units

Class 12

Important Question 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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.