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Question 2 Find the area of the region bounded by y2 = 9๐‘ฅ, ๐‘ฅ = 2, ๐‘ฅ = 4 and the ๐‘ฅ-axis in the first quadrant. Given curve ๐‘ฆ^2=9๐‘ฅ We have to find area between x = 2 and x = 4 โˆด We have to find area of BCFE Area of BCFE = โˆซ_2^4โ–’ใ€–๐‘ฆ .ใ€— ๐‘‘๐‘ฅ We know that ๐‘ฆ^2=9๐‘ฅ Taking square root on both sides ๐‘ฆ=ยฑโˆš9๐‘ฅ ๐‘ฆ=ยฑ3โˆš๐‘ฅ Since BCFE is in 1st Quadrant We take positive value of y โˆด ๐‘ฆ=3โˆš๐‘ฅ Area of BCFE = โˆซ_2^4โ–’ใ€–๐‘ฆ .ใ€— ๐‘‘๐‘ฅ = 3โˆซ_2^4โ–’โˆš๐‘ฅ ๐‘‘๐‘ฅ = 3โˆซ_2^4โ–’ใ€–(๐‘ฅ)^(1/2) ๐‘‘๐‘ฅใ€— = 3 [๐‘ฅ^(1/2 + 1)/(1/2 + 1 )]_2^4 = 3 [๐‘ฅ^(3/2 )/(3/2)]_2^4 = 3 ร— 2/3 [๐‘ฅ^(3/2) ]_2^4 = 2 [(4)^(3/2 )โˆ’(2)^(3/2) ] = 2 [((4)^(1/2) )^3โˆ’((2)^(1/2) )^3 ] = 2 [(2)^3โˆ’(โˆš2)^3 ] = 2 [8 โˆ’2โˆš2] = 16 โ€“ 4โˆš2 Thus, Area = 16 โ€“ 4โˆš๐Ÿ square units

  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise

About the Author

Davneet Singh

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