Subscribe to our Youtube Channel -

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


Misc 16 Area bounded by the curve ๐‘ฆ=๐‘ฅ3, the ๐‘ฅ-axis and the ordinates ๐‘ฅ = โ€“ 2 and ๐‘ฅ = 1 is (A) โ€“ 9 (B) (โˆ’15)/4 (C) 15/4 (D) 17/4 Area Required = Area ABO + Area DCO Area ABO Area ABO =โˆซ_(โˆ’2)^0โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— Here, ๐‘ฆ=๐‘ฅ^3 Therefore, Area ABO =โˆซ_(โˆ’2)^0โ–’ใ€–๐‘ฅ^3 ๐‘‘๐‘ฅใ€— ใ€–=[๐‘ฅ^4/4]ใ€—_(โˆ’2)^0 =1/4 [0โˆ’(โˆ’2)^4 ] =1/4 ร— (โˆ’16) =โˆ’4 Since Area is always positive, Area ABO = 4 Area DCO Area DCO = โˆซ_0^1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— =โˆซ_0^1โ–’ใ€–๐‘ฅ^3 ๐‘‘๐‘ฅใ€— =[๐‘ฅ^4/4]_0^1 =1/4 [1^3โˆ’0^3 ] =1/4 โˆด Area Required = Area ABO + Area DCO =4+1/4 =17/4 โˆด D is the Correct Option

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.