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

Transcript

Ex 8.1, 8 The area between ๐‘ฅ=๐‘ฆ2 and ๐‘ฅ = 4 is divided into two equal parts by the line ๐‘ฅ=๐‘Ž, find the value of a. Given curve ๐‘ฆ^2=๐‘ฅ Let AB represent the line ๐‘ฅ=๐‘Ž CD represent the line ๐‘ฅ=4 Since the line ๐‘ฅ=๐‘Ž divides the region into two equal parts โˆด Area of OBA = Area of ABCD 2 ร— โˆซ_0^๐‘Žโ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€—="2 ร—" โˆซ_๐‘Ž^4โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— โˆซ_๐ŸŽ^๐’‚โ–’ใ€–๐’š ๐’…๐’™ใ€—=โˆซ_๐’‚^๐Ÿ’โ–’ใ€–๐’š ๐’…๐’™ใ€— Now, y2 = x y = ยฑ โˆš๐‘ฅ Since, the curve is symmetric about x-axis we can take either positive or negative value of ๐‘ฆ So, lets take ๐‘ฆ=โˆš๐‘ฅ Now, From (1) โˆซ_0^๐‘Žโ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€—=โˆซ_๐‘Ž^4โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— โˆซ_0^๐‘Žโ–’โˆš๐‘ฅ ๐‘‘๐‘ฅ=โˆซ_๐‘Ž^4โ–’โˆš๐‘ฅ ๐‘‘๐‘ฅ [๐‘ฅ^(1/2 + 1)/(1/2 + 1)]_0^๐‘Ž=[๐‘ฅ^(1/2+1)/(1/2+1)]_๐‘Ž^4 [๐‘ฅ^((1+2)/2) ]_0^๐‘Ž=[๐‘ฅ^((1+2)/2) ]_๐‘Ž^4 [๐‘ฅ^(3/2) ]_0^๐‘Ž=[๐‘ฅ^(3/2) ]_๐‘Ž^4 (๐‘Ž)^(3/2)โˆ’0=(4)^(3/2)โˆ’(๐‘Ž)^(3/2) 2(๐‘Ž)^(3/2)=(4)^(3/2) Taking ใ€–2/3ใ€—^๐‘กโ„Ž root on both sides (2)^(2/3) ๐‘Ž^(3/2 ร— 2/3)=4^(3/2 ร— 2/3) (2)^(2/3) ๐‘Ž=4 ๐‘Ž=(2)^2/(2)^(2/3) ๐‘Ž=(2)^(2 โˆ’ 2/3) ๐‘Ž=(2)^((6 โˆ’ 2)/3) ๐‘Ž=(2)^(4/3) ๐‘Ž=(2)^(2 ร— 2/3) ๐‘Ž=[2^2 ]^(2/3) ๐’‚=(๐Ÿ’)^(๐Ÿ/๐Ÿ‘) So, value of a is (4)^(2/3)

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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.