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Misc 19


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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

  1. Chapter 8 Class 12 Application of Integrals
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About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo