Question 2 - Miscellaneous - Chapter 8 Class 12 Application of Integrals
Last updated at Dec. 16, 2024 by Teachoo
Miscellaneous
Misc 1 (ii) Important
Misc 2 Important
Misc 3 Important
Misc 4 (MCQ)
Misc 5 (MCQ) Important
Question 1
Question 2 You are here
Question 3 Important
Question 4
Question 5
Question 6 Important
Question 7
Question 8 Important
Question 9 Important
Question 10
Question 11 Important
Question 12
Question 13 (MCQ)
Question 14 (MCQ) Important
Miscellaneous
Last updated at Dec. 16, 2024 by Teachoo
Question 2 Find the area of the region lying in the first quadrant and bounded by =4 2 , = 0, =1 and =4 Area required = Area ABCD = 1 4 y equation of parabola = 4 2 4x2 = y x2 = 4 x = 4 x = 2 Thus, Area required = 1 4 = 1 2 4 2 =4 3 3 1 2 = 1 3 2 3 1 3 = 1 3 8 1 = 7 3 Thus, Area required = 1 4 = 1 4 2 = 1 2 1 4 1 2 = 1 2 1 2 +1 1 2 +1 1 4 = 1 2 3 2 3 2 1 4 = 1 2 2 3 3 2 1 4 = 1 3 4 3 2 1 3 2 = 2 3 = 1 3 2 2 3 2 1 = 1 3 2 3 1 =