Slide41.JPG Slide42.JPG Slide43.JPG Misc 19

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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise


Misc 19 The area bounded by the 𝑦-axis, 𝑦=cos⁑π‘₯ and 𝑦=sin⁑π‘₯ when 0≀π‘₯β‰€πœ‹/2 is (A) 2 ( √("2 βˆ’1" )) (B) √("2 βˆ’1" ) (C) √("2 " )+1 (D) √("2 " ) Finding point of intersection B Solving 𝑦=cos⁑π‘₯ and 𝑦=s𝑖𝑛⁑π‘₯ cos⁑π‘₯=s𝑖𝑛⁑π‘₯ At π‘₯=πœ‹/4 , both are equal Also, 𝑦=cos⁑π‘₯ = cos πœ‹/4 = 1/√2 So, B =((πœ‹ )/4 , 1/√2) Area Required Area Required = Area ABCO – Area BCO Area ABCO Area ABCO = ∫_0^(πœ‹/4)▒〖𝑦 𝑑π‘₯γ€— Here, 𝑦=cos⁑π‘₯ Thus, Area ABCO =∫_0^(πœ‹/4)β–’γ€–cos⁑π‘₯ 𝑑π‘₯γ€— =[sin⁑π‘₯ ]_0^(πœ‹/4) =[sinβ‘γ€–πœ‹/4βˆ’sin⁑0 γ€— ] =1/√2βˆ’0 =1/√2 Area BCO Area BCO = ∫_0^(πœ‹/4)▒〖𝑦 𝑑π‘₯γ€— Here, 𝑦=sin⁑π‘₯ Thus, Area BCO =∫_0^(πœ‹/4)β–’γ€–sin⁑π‘₯ 𝑑π‘₯γ€— =[γ€–βˆ’cπ‘œπ‘ γ€—β‘π‘₯ ]_0^(πœ‹/4) =βˆ’[cosβ‘γ€–πœ‹/4βˆ’cos⁑(0) γ€— ] =βˆ’[1/√2βˆ’1] =1βˆ’1/√2 Therefore Area Required = Area ABCO – Area BCO =1/√2βˆ’[1βˆ’1/√2] =1/√2+1/√2βˆ’1 =2/√2βˆ’1 =βˆšπŸβˆ’πŸ ∴ Option B is Correct

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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.