Misc 4 (MCQ) - Chapter 8 Class 12 Application of Integrals

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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)