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

Transcript

Misc 17 The area bounded by the curve ๐‘ฆ = ๐‘ฅ |๐‘ฅ| , ๐‘ฅโˆ’๐‘Ž๐‘ฅ๐‘–๐‘  and the ordinates ๐‘ฅ = โ€“ 1 and ๐‘ฅ=1 is given by (A) 0 (B) 1/3 (C) 2/3 (D) 4/3 [Hint : ๐‘ฆ=๐‘ฅ2 if ๐‘ฅ > 0 ๐‘Ž๐‘›๐‘‘ ๐‘ฆ =โˆ’๐‘ฅ2 if ๐‘ฅ < 0] We know that |๐‘ฅ|={โ–ˆ(๐‘ฅ, ๐‘ฅโ‰ฅ0@&โˆ’๐‘ฅ, ๐‘ฅ<0)โ”ค Therefore, y = x|๐‘ฅ|={โ–ˆ(๐‘ฅ๐‘ฅ, ๐‘ฅโ‰ฅ0@&๐‘ฅ(โˆ’๐‘ฅ), ๐‘ฅ<0)โ”ค y ={โ–ˆ(๐‘ฅ^2, ๐‘ฅโ‰ฅ0@&โˆ’๐‘ฅ^2, ๐‘ฅ<0)โ”ค Area Required = Area ABO + Area DCO Area ABO Area ABO =โˆซ_(โˆ’1)^0โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— Here, ๐‘ฆ=ใ€–โˆ’๐‘ฅใ€—^2 Therefore, Area ABO =โˆซ_(โˆ’1)^0โ–’ใ€–ใ€–โˆ’๐‘ฅใ€—^2 ๐‘‘๐‘ฅใ€— ใ€–=โˆ’[๐‘ฅ^3/3]ใ€—_(โˆ’1)^0 =โˆ’[0^3/3โˆ’(โˆ’1)^3/3] =(โˆ’1)/3 Since Area is always positive, Area ABO = 1/3 Area DCO Area DCO =โˆซ_0^1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— Here, ๐‘ฆ=๐‘ฅ^2 Therefore, Area DCO =โˆซ_0^1โ–’ใ€–๐‘ฅ^2 ๐‘‘๐‘ฅใ€— ใ€–=[๐‘ฅ^3/3]ใ€—_0^1 =1/3 [1^3โˆ’0^3 ] =1/3 [1โˆ’0] =1/3 โˆด 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.