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Last updated at Dec. 12, 2019 by Teachoo

Transcript

Example 14 Prove that curves ๐ฆ2=4๐ฅ and ๐ฅ2=4๐ฆ divide the area of the square bounded by ๐ฅ=0, ๐ฅ=4, ๐ฆ=4 and ๐ฆ=0 into three equal parts Drawing figure Here, we have parabolas ๐ฆ^2=4๐ฅ ๐ฅ^2=4๐ฆ 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 = โซ_0^4โใ๐ฆ ๐๐ฅใ Here, 4๐ฆ=๐ฅ^2 ๐ฆ=๐ฅ^2/4 So, Area OPQA =โซ_0^4โใ๐ฅ^2/4 ๐๐ฅใ = 1/4 [๐ฅ^3/3]_0^4 =1/12ร[4^3โ0^3 ] =1/12 ร [64โ0] =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 ๐ฆโ๐๐ฅ๐๐ , we use formula โซ1โใ๐ฅ ๐๐ฆใ Area OBQR = โซ_0^4โใ๐ฅ ๐๐ฆใ Here, ๐ฆ^2=4๐ฅ ๐ฅ=๐ฆ^2/4 So, Area OBQR =โซ_0^4โใ๐ฆ^2/4 ๐๐ฆใ = 1/4 [๐ฆ^3/3]_0^4=1/12ร[4^3โ0^3 ]=1/12 ร[64โ0]=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 =โซ_0^4โใ๐ฆ ๐๐ฅใ Here, ๐ฆ^2=4๐ฅ ๐ฆ=ยฑโ4๐ฅ As OBQP is in 1st quadrant, value of y is positive โด ๐ฆ=โ4๐ฅ Area OBQP =โซ_0^4โใโ4๐ฅ ๐๐ฅใ =2โซ_0^4โใโ๐ฅ ๐๐ฅใ =2โซ_0^4โใ๐ฅ^(1/2) ๐๐ฅใ =2 ร [๐ฅ^(1/2+1)/(1/2+1)]_0^4 =2 ร [๐ฅ^(3/2)/(3/2)]_0^4 =2 ร 2/3 [๐ฅ^(3/2) ]_0^4 =4/3 [(4)^(3/2)โ(0)^(3/2) ] =4/3 [8โ0] =32/3 Area OAQP Area OAQP =โซ_0^4โใ๐ฆ ๐๐ฅใ Here, ๐ฅ^2=4๐ฆ ๐ฆ=๐ฅ^2/4 Area OAQP =โซ_0^4โใ๐ฅ^2/4 ๐๐ฅใ =1/4 [๐ฅ^(2+1)/(2+1)]_0^4 =1/(4 ร3) [๐ฅ^3 ]_0^4 =1/12 [4^3โ0^3 ] Area OAQP Area OAQP =โซ_0^4โใ๐ฆ ๐๐ฅใ Here, ๐ฅ^2=4๐ฆ ๐ฆ=๐ฅ^2/4 Area OAQP =โซ_0^4โใ๐ฅ^2/4 ๐๐ฅใ =1/4 [๐ฅ^(2+1)/(2+1)]_0^4 =1/(4 ร3) [๐ฅ^3 ]_0^4 =1/12 [4^3โ0^3 ] =1/12 ร [64โ0] =16/3 โด Area OAQB = Area OBQD โ Area OAQP = 32/3โ16/3 = 16/3 So, Area OAQB = Area OAQP = Area OBRQ = ๐๐/๐ 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.