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

Transcript

Ex 8.1, 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

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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.