Question 2 - Area bounded by curve and horizontal or vertical line - Chapter 8 Class 12 Application of Integrals

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Misc 3 - Find area bounded by y = 4x2, x = 0, y = 1, y = 4 - Miscellaneous

Misc 3 - Chapter 8 Class 12 Application of Integrals - Part 2
Misc 3 - Chapter 8 Class 12 Application of Integrals - Part 3

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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 =

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo