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

Transcript

Example 14 Prove that curves ๐‘ฆ2=4๐‘ฅ and ๐‘ฅ2=4๐‘ฆ divide the area of the square bounded by ๐‘ฅ=0, ๐‘ฅ=4, ๐‘ฆ=4 and ๐‘ฆ=0 into three equal parts Drawing figure Here, we have parabolas ๐‘ฆ^2=4๐‘ฅ ๐‘ฅ^2=4๐‘ฆ And, Square made by the lines x = 4, y = 4, x = 0, y = 0 We need to prove that area of square is divided into 3 parts by the curve So, we need to prove Area OPQA = Area OAQB = Area OBQR Area OPQA Area OPQA = โˆซ_0^4โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— Here, 4๐‘ฆ=๐‘ฅ^2 ๐‘ฆ=๐‘ฅ^2/4 So, Area OPQA =โˆซ_0^4โ–’ใ€–๐‘ฅ^2/4 ๐‘‘๐‘ฅใ€— = 1/4 [๐‘ฅ^3/3]_0^4 =1/12ร—[4^3โˆ’0^3 ] =1/12 ร— [64โˆ’0] =64/12=16/3 Area OPQA is the area by curve x2 = 4y in the x-axis from x = 0 to x = 4 Area OBQR Since Area is on ๐‘ฆโˆ’๐‘Ž๐‘ฅ๐‘–๐‘  , we use formula โˆซ1โ–’ใ€–๐‘ฅ ๐‘‘๐‘ฆใ€— Area OBQR = โˆซ_0^4โ–’ใ€–๐‘ฅ ๐‘‘๐‘ฆใ€— Here, ๐‘ฆ^2=4๐‘ฅ ๐‘ฅ=๐‘ฆ^2/4 So, Area OBQR =โˆซ_0^4โ–’ใ€–๐‘ฆ^2/4 ๐‘‘๐‘ฆใ€— = 1/4 [๐‘ฆ^3/3]_0^4=1/12ร—[4^3โˆ’0^3 ]=1/12 ร—[64โˆ’0]=16/3 Area OBQR is the area by curve y2 = 4x in the y-axis from y = 0 to y = 4 Area OAQB Area OAQB = Area OBQP โ€“ Area OAQP Finding Area OBQP Area OBQP =โˆซ_0^4โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— Here, ๐‘ฆ^2=4๐‘ฅ ๐‘ฆ=ยฑโˆš4๐‘ฅ As OBQP is in 1st quadrant, value of y is positive โˆด ๐‘ฆ=โˆš4๐‘ฅ Area OBQP =โˆซ_0^4โ–’ใ€–โˆš4๐‘ฅ ๐‘‘๐‘ฅใ€— =2โˆซ_0^4โ–’ใ€–โˆš๐‘ฅ ๐‘‘๐‘ฅใ€— =2โˆซ_0^4โ–’ใ€–๐‘ฅ^(1/2) ๐‘‘๐‘ฅใ€— =2 ร— [๐‘ฅ^(1/2+1)/(1/2+1)]_0^4 =2 ร— [๐‘ฅ^(3/2)/(3/2)]_0^4 =2 ร— 2/3 [๐‘ฅ^(3/2) ]_0^4 =4/3 [(4)^(3/2)โˆ’(0)^(3/2) ] =4/3 [8โˆ’0] =32/3 Area OAQP Area OAQP =โˆซ_0^4โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— Here, ๐‘ฅ^2=4๐‘ฆ ๐‘ฆ=๐‘ฅ^2/4 Area OAQP =โˆซ_0^4โ–’ใ€–๐‘ฅ^2/4 ๐‘‘๐‘ฅใ€— =1/4 [๐‘ฅ^(2+1)/(2+1)]_0^4 =1/(4 ร—3) [๐‘ฅ^3 ]_0^4 =1/12 [4^3โˆ’0^3 ] Area OAQP Area OAQP =โˆซ_0^4โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— Here, ๐‘ฅ^2=4๐‘ฆ ๐‘ฆ=๐‘ฅ^2/4 Area OAQP =โˆซ_0^4โ–’ใ€–๐‘ฅ^2/4 ๐‘‘๐‘ฅใ€— =1/4 [๐‘ฅ^(2+1)/(2+1)]_0^4 =1/(4 ร—3) [๐‘ฅ^3 ]_0^4 =1/12 [4^3โˆ’0^3 ] =1/12 ร— [64โˆ’0] =16/3 โˆด Area OAQB = Area OBQD โ€“ Area OAQP = 32/3โˆ’16/3 = 16/3 So, Area OAQB = Area OAQP = Area OBRQ = ๐Ÿ๐Ÿ”/๐Ÿ‘ square units Hence Proved.

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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.