# Misc 19

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 19 The area bounded by the 𝑦-axis, 𝑦=cos𝑥 and 𝑦=sin𝑥 when 0≤𝑥≤ 𝜋2 is (A) 2 ( 2 −1) (B) 2 −1 (C) 2 +1 (D) 2 Finding point of intersection B Solving 𝑦=cos𝑥 and 𝑦=s𝑖𝑛𝑥 cos𝑥=s𝑖𝑛𝑥 At 𝑥= 𝜋4 , both are equal Also, 𝑦=cos𝑥 = cos 𝜋4 = 1 2 So, B = 𝜋 4 , 1 2 Step 3: Finding Area Area Required = Area ABCO – Area BCO Area ABCO Area ABCO = 0 𝜋4𝑦 𝑑𝑥 Here, 𝑦= cos𝑥 Thus, Area ABCO = 0 𝜋4 cos𝑥 𝑑𝑥 = sin𝑥0 𝜋4 = sin 𝜋4− sin0 = 1 2−0 = 1 2 Area BCO Area BCO = 0 𝜋4𝑦 𝑑𝑥 Here, 𝑦= sin𝑥 Thus, Area BCO = 0 𝜋4 sin𝑥 𝑑𝑥 = −c𝑜𝑠𝑥0 𝜋4 =− cos 𝜋4− cos 0 =− 1 2−1 =1− 1 2 Therefore Area Required = Area ABCO – Area BCO = 1 2− 1− 1 2 = 1 2+ 1 2−1 = 2 2−1 = 2−1 ∴ Option B is Correct

Chapter 8 Class 12 Application of Integrals

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.