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Misc 19 - Area bounded by y-axis, y = cos x, y = sin x - Miscellaneous

Misc 19 - Chapter 8 Class 12 Application of Integrals - Part 2
Misc 19 - Chapter 8 Class 12 Application of Integrals - Part 3
Misc 19 - Chapter 8 Class 12 Application of Integrals - Part 4
Misc 19 - Chapter 8 Class 12 Application of Integrals - Part 5
Misc 19

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Transcript

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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.