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Area between curve and curve
Misc 13
Ex 8.2, 4 Important Deleted for CBSE Board 2023 Exams
Ex 8.2, 5 Important Deleted for CBSE Board 2023 Exams
Misc 14 Important
Misc 4 Important
Misc 11 Important
Ex 8.2,1 Important Deleted for CBSE Board 2023 Exams
Misc 15 Deleted for CBSE Board 2023 Exams
Example 7 Important Deleted for CBSE Board 2023 Exams
Misc 18 (MCQ) Deleted for CBSE Board 2023 Exams
Example 15 Important
Ex 8.2, 2 Deleted for CBSE Board 2023 Exams
Example 10 Important Deleted for CBSE Board 2023 Exams
Example 6 Important Deleted for CBSE Board 2023 Exams
Misc 19 (MCQ) Important Deleted for CBSE Board 2023 Exams You are here
Area between curve and curve
Last updated at Aug. 11, 2021 by Teachoo
Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month
Misc 19 The area bounded by the π¦-axis, π¦=cosβ‘π₯ and π¦=sinβ‘π₯ when 0β€π₯β€π/2 is (A) 2 ( β("2 β1" )) (B) β("2 β1" ) (C) β("2 " )+1 (D) β("2 " ) Finding point of intersection B Solving π¦=cosβ‘π₯ and π¦=sππβ‘π₯ cosβ‘π₯=sππβ‘π₯ At π₯=π/4 , both are equal Also, π¦=cosβ‘π₯ = cos π/4 = 1/β2 So, B =((π )/4 , 1/β2) Area Required Area Required = Area ABCO β Area BCO Area ABCO Area ABCO = β«_0^(π/4)βγπ¦ ππ₯γ Here, π¦=cosβ‘π₯ Thus, Area ABCO =β«_0^(π/4)βγcosβ‘π₯ ππ₯γ =[sinβ‘π₯ ]_0^(π/4) =[sinβ‘γπ/4βsinβ‘0 γ ] =1/β2β0 =1/β2 Area BCO Area BCO = β«_0^(π/4)βγπ¦ ππ₯γ Here, π¦=sinβ‘π₯ Thus, Area BCO =β«_0^(π/4)βγsinβ‘π₯ ππ₯γ =[γβcππ γβ‘π₯ ]_0^(π/4) =β[cosβ‘γπ/4βcosβ‘(0) γ ] =β[1/β2β1] =1β1/β2 Therefore Area Required = Area ABCO β Area BCO =1/β2β[1β1/β2] =1/β2+1/β2β1 =2/β2β1 =βπβπ β΄ Option B is Correct