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Ex 8.1, 8 - Area between x = y2, x = 4 is divided into equal

Ex 8.1, 8 - Chapter 8 Class 12 Application of Integrals - Part 2
Ex 8.1, 8 - Chapter 8 Class 12 Application of Integrals - Part 3 Ex 8.1, 8 - Chapter 8 Class 12 Application of Integrals - Part 4

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Question 6 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.