Area bounded by curve and horizontal or vertical line

Chapter 8 Class 12 Application of Integrals
Concept wise

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### Transcript

Question 2 Find the area of the region bounded by y2 = 9π₯, π₯ = 2, π₯ = 4 and the π₯-axis in the first quadrant. Given curve π¦^2=9π₯ We have to find area between x = 2 and x = 4 β΄ We have to find area of BCFE Area of BCFE = β«_2^4βγπ¦ .γ ππ₯ We know that π¦^2=9π₯ Taking square root on both sides π¦=Β±β9π₯ π¦=Β±3βπ₯ Since BCFE is in 1st Quadrant We take positive value of y β΄ π¦=3βπ₯ Area of BCFE = β«_2^4βγπ¦ .γ ππ₯ = 3β«_2^4ββπ₯ ππ₯ = 3β«_2^4βγ(π₯)^(1/2) ππ₯γ = 3 [π₯^(1/2 + 1)/(1/2 + 1 )]_2^4 = 3 [π₯^(3/2 )/(3/2)]_2^4 = 3 Γ 2/3 [π₯^(3/2) ]_2^4 = 2 [(4)^(3/2 )β(2)^(3/2) ] = 2 [((4)^(1/2) )^3β((2)^(1/2) )^3 ] = 2 [(2)^3β(β2)^3 ] = 2 [8 β2β2] = 16 β 4β2 Thus, Area = 16 β 4βπ square units

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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.