# Example 6 - Chapter 8 Class 12 Application of Integrals

Last updated at Dec. 12, 2019 by Teachoo

Last updated at Dec. 12, 2019 by Teachoo

Transcript

Example 6 Find the area of the region bounded by the two parabolas ๐ฆ=๐ฅ2 and ๐ฆ2 = ๐ฅ Drawing figure Here, we have parabolas ๐ฆ^2=๐ฅ ๐ฅ^2=๐ฆ Area required = Area OABC Finding Point of intersection B Solving ๐ฆ2 = ๐ฅ ๐ฅ2 =๐ฆ Put (2) in (1) ๐ฆ2 = ๐ฅ (๐ฅ^2 )^2=๐ฅ ๐ฅ^4โ๐ฅ=0 ๐ฅ(๐ฅ^3โ1)=0 Finding y โ coordinate For ๐=๐ ๐ฆ=๐ฅ^2=0^2= 0 So, coordinates are (0 , 0) For ๐=๐ ๐ฆ=๐ฅ^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 =โซ_0^1โใ๐ฆ ๐๐ฅใ Here, ๐ฆ^2=๐ฅ ๐ฆ=ยฑโ๐ฅ As OABD is in 1st quadrant, value of y is positive โด ๐ฆ=โ๐ฅ Area OBQP =โซ_0^1โใโ๐ฅ ๐๐ฅใ =โซ_0^1โใโ๐ฅ ๐๐ฅใ =โซ_0^1โใ๐ฅ^(1/2) ๐๐ฅใ = [๐ฅ^(1/2 + 1)/(1/2 + 1)]_0^1 = [๐ฅ^(3/2)/(3/2)]_0^1 = 2/3 [๐ฅ^(3/2) ]_0^1 =2/3 [(1)^(3/2)โ(0)^(3/2) ] =2/3 [1โ0] =2/3 Area OCBD Area OCBD =โซ_0^1โใ๐ฆ ๐๐ฅใ Here, ๐ฅ^2=๐ฆ ๐ฆ=๐ฅ^2 Area OAQP =โซ_0^1โใ๐ฅ^2 ๐๐ฅใ =[๐ฅ^(2 + 1)/(2 + 1)]_0^1 =1/3 [๐ฅ^3 ]_0^1 =1/3 [1^3โ0^3 ] =๐/๐ Therefore, Area OABC = Area OABD โ Area OCBD = 2/3โ1/3 = ๐/๐ 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.