Misc 4 (MCQ) - Chapter 8 Class 12 Application of Integrals
Last updated at April 16, 2024 by Teachoo
Miscellaneous
Misc 1 (ii) Important
Misc 2 Important
Misc 3 Important
Misc 4 (MCQ) You are here
Misc 5 (MCQ) Important
Question 1 Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 7 Deleted for CBSE Board 2024 Exams
Question 8 Important Deleted for CBSE Board 2024 Exams
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Question 10 Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Question 13 (MCQ) Deleted for CBSE Board 2024 Exams
Question 14 (MCQ) Important Deleted for CBSE Board 2024 Exams
Miscellaneous
Last updated at April 16, 2024 by Teachoo
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)