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


Example 6 Find the area of the region bounded by the two parabolas ๐‘ฆ=๐‘ฅ2 and ๐‘ฆ2 = ๐‘ฅ Drawing figure Here, we have parabolas ๐‘ฆ^2=๐‘ฅ ๐‘ฅ^2=๐‘ฆ Area required = Area OABC Finding Point of intersection B Solving ๐‘ฆ2 = ๐‘ฅ ๐‘ฅ2 =๐‘ฆ Put (2) in (1) ๐‘ฆ2 = ๐‘ฅ (๐‘ฅ^2 )^2=๐‘ฅ ๐‘ฅ^4โˆ’๐‘ฅ=0 ๐‘ฅ(๐‘ฅ^3โˆ’1)=0 Finding y โ€“ coordinate For ๐’™=๐ŸŽ ๐‘ฆ=๐‘ฅ^2=0^2= 0 So, coordinates are (0 , 0) For ๐’™=๐Ÿ ๐‘ฆ=๐‘ฅ^2=1^2=1 So, coordinates are (1 , 1) Since point B lies in 1st quadrant So, co-ordinate of B is (1 , 1) Finding Area Area OABC = Area OABD โ€“ Area OCBD Finding Area OABD Area OABD =โˆซ_0^1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— Here, ๐‘ฆ^2=๐‘ฅ ๐‘ฆ=ยฑโˆš๐‘ฅ As OABD is in 1st quadrant, value of y is positive โˆด ๐‘ฆ=โˆš๐‘ฅ Area OBQP =โˆซ_0^1โ–’ใ€–โˆš๐‘ฅ ๐‘‘๐‘ฅใ€— =โˆซ_0^1โ–’ใ€–โˆš๐‘ฅ ๐‘‘๐‘ฅใ€— =โˆซ_0^1โ–’ใ€–๐‘ฅ^(1/2) ๐‘‘๐‘ฅใ€— = [๐‘ฅ^(1/2 + 1)/(1/2 + 1)]_0^1 = [๐‘ฅ^(3/2)/(3/2)]_0^1 = 2/3 [๐‘ฅ^(3/2) ]_0^1 =2/3 [(1)^(3/2)โˆ’(0)^(3/2) ] =2/3 [1โˆ’0] =2/3 Area OCBD Area OCBD =โˆซ_0^1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— Here, ๐‘ฅ^2=๐‘ฆ ๐‘ฆ=๐‘ฅ^2 Area OAQP =โˆซ_0^1โ–’ใ€–๐‘ฅ^2 ๐‘‘๐‘ฅใ€— =[๐‘ฅ^(2 + 1)/(2 + 1)]_0^1 =1/3 [๐‘ฅ^3 ]_0^1 =1/3 [1^3โˆ’0^3 ] =๐Ÿ/๐Ÿ‘ Therefore, Area OABC = Area OABD โ€“ Area OCBD = 2/3โˆ’1/3 = ๐Ÿ/๐Ÿ‘ square units

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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.