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Misc 4 Area bounded by the curve 𝑦=𝑥3, the 𝑥-axis and the ordinates 𝑥 = –2 and 𝑥 = 1 is (A) – 9 (B) (−15)/4 (C) 15/4 (D) 17/4 Area Required = Area ABO + Area DCO Area ABO Area ABO =∫_(−2)^0▒〖𝑦 𝑑𝑥〗 Here, 𝑦=𝑥^3 Therefore, Area ABO =∫_(−𝟐)^𝟎▒〖𝒙^𝟑 𝒅𝒙〗 〖=[𝑥^4/4]〗_(−2)^0 =1/4 [0−(−2)^4 ] =1/4 × (−16) =−4 Since Area is always positive, Area ABO = 4 Area DCO Area DCO = ∫_0^1▒〖𝑦 𝑑𝑥〗 =∫_𝟎^𝟏▒〖𝒙^𝟑 𝒅𝒙〗 =[𝑥^4/4]_0^1 =1/4 [1^3−0^3 ] =𝟏/𝟒 Now, Area Required = Area ABO + Area DCO =4+1/4 =𝟏𝟕/𝟒 square units So, the correct answer is (d) ∴ D is the Correct Option So, the correct answer is (d)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.