# Misc 9 - Chapter 8 Class 12 Application of Integrals (Important Question)

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Misc 9 Find the area of the smaller region bounded by the ellipse 2 2 + 2 2 =1 & + = 1 Step 1: Drawing figure + = is an which is a equation ellipse with as principle For + = 1 Points A(a, 0) and B(0, b) passes through both line and ellipse Required Area Required Area = Area OACB Area OAB Area OACB Area OACB = 0 Equation of ellipse 2 2 + 2 2 =1 2 2 =1 2 2 2 = 2 1 2 2 = 2 1 2 2 = 1 2 2 Therefore, Area OACB = 0 1 2 2 =b 0 2 2 2 = 0 2 2 = 1 2 2 2 + 2 2 sin 1 0 = 1 2 . 2 2 + 2 2 sin 1 1 2 0 2 0 2 + 2 2 sin 1 0 = 0+ 2 2 . 0 0 = 0+ 2 2 . = 2 2 2 = 4 Area OAB Area OAB = 0 Equation of line + =1 =1 = 1 Therefore, Area OAB = 0 1 = 2 2 0 = 2 2 0 0 2 2 = 2 0 = 2 Area Required = Area OACB Area OAB = 4 2 = 2 2 1 = 2 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

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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.