# Misc 17 - Chapter 8 Class 12 Application of Integrals

Last updated at Dec. 12, 2019 by Teachoo

Last updated at Dec. 12, 2019 by Teachoo

Transcript

Misc 17 The area bounded by the curve ๐ฆ = ๐ฅ |๐ฅ| , ๐ฅโ๐๐ฅ๐๐ and the ordinates ๐ฅ = โ 1 and ๐ฅ=1 is given by (A) 0 (B) 1/3 (C) 2/3 (D) 4/3 [Hint : ๐ฆ=๐ฅ2 if ๐ฅ > 0 ๐๐๐ ๐ฆ =โ๐ฅ2 if ๐ฅ < 0] We know that |๐ฅ|={โ(๐ฅ, ๐ฅโฅ0@&โ๐ฅ, ๐ฅ<0)โค Therefore, y = x|๐ฅ|={โ(๐ฅ๐ฅ, ๐ฅโฅ0@&๐ฅ(โ๐ฅ), ๐ฅ<0)โค y ={โ(๐ฅ^2, ๐ฅโฅ0@&โ๐ฅ^2, ๐ฅ<0)โค Area Required = Area ABO + Area DCO Area ABO Area ABO =โซ_(โ1)^0โใ๐ฆ ๐๐ฅใ Here, ๐ฆ=ใโ๐ฅใ^2 Therefore, Area ABO =โซ_(โ1)^0โใใโ๐ฅใ^2 ๐๐ฅใ ใ=โ[๐ฅ^3/3]ใ_(โ1)^0 =โ[0^3/3โ(โ1)^3/3] =(โ1)/3 Since Area is always positive, Area ABO = 1/3 Area DCO Area DCO =โซ_0^1โใ๐ฆ ๐๐ฅใ Here, ๐ฆ=๐ฅ^2 Therefore, Area DCO =โซ_0^1โใ๐ฅ^2 ๐๐ฅใ ใ=[๐ฅ^3/3]ใ_0^1 =1/3 [1^3โ0^3 ] =1/3 [1โ0] =1/3 โด Required Area = Area ABO + Area DCO =1/3+1/3 =2/3 So, Option C is Correct

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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.