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

Transcript

Ex 8.1, 3 Find the area of the region bounded by ๐‘ฅ2= 4๐‘ฆ , ๐‘ฆ = 2 , ๐‘ฆ=4 and the ๐‘ฆ-axis in the first quadrant. Here, The curve is ๐‘ฅ^2=4๐‘ฆ We have to find area between y = 2 and y = 4 โˆด We have to find area of BCFE Area of BCFE = โˆซ_2^4โ–’๐‘ฅ ๐‘‘๐‘ฆ We know that ๐‘ฅ^2=4๐‘ฆ Taking square root on both sides ๐‘ฅ="ยฑ" โˆš4๐‘ฆโ€– ๐‘ฅ="ยฑ" 2โˆš๐‘ฆ Since, BCEF is in 1st Quadrant We take positive value of x โˆด ๐‘ฅ=2โˆš๐‘ฆ Area of BCFE = โˆซ_2^4โ–’๐‘ฅ ๐‘‘๐‘ฆ = โˆซ_2^4โ–’ใ€–2โˆš๐‘ฆใ€— ๐‘‘๐‘ฆ = 2โˆซ_2^4โ–’ใ€–(๐‘ฆ)^(1/2) ๐‘‘๐‘ฆใ€— = 2 [๐‘ฆ^(3/2)/(3/2)]_2^4 = 2 ร— 2/3 [๐‘ฆ^(3/2) ]_2^4 = 4/3 [(4)^(3/2 )โˆ’(2)^(3/2) ] = 4/3 [((4)^(1/2) )^3โˆ’((2)^(1/2) )^3 ] = 4/3 [(2)^3โˆ’(โˆš2)^3 ] = 4/3 [8 โˆ’2โˆš2] = (32 โˆ’ 8โˆš2)/3 Thus, Area = (๐Ÿ‘๐Ÿ โˆ’ ๐Ÿ–โˆš๐Ÿ)/๐Ÿ‘ square units

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.