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Last updated at Dec. 12, 2019 by Teachoo

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Ex 8.1, 3 Find the area of the region bounded by ๐ฅ2= 4๐ฆ , ๐ฆ = 2 , ๐ฆ=4 and the ๐ฆ-axis in the first quadrant. Here, The curve is ๐ฅ^2=4๐ฆ We have to find area between y = 2 and y = 4 โด We have to find area of BCFE Area of BCFE = โซ_2^4โ๐ฅ ๐๐ฆ We know that ๐ฅ^2=4๐ฆ Taking square root on both sides ๐ฅ="ยฑ" โ4๐ฆโ ๐ฅ="ยฑ" 2โ๐ฆ Since, BCEF is in 1st Quadrant We take positive value of x โด ๐ฅ=2โ๐ฆ Area of BCFE = โซ_2^4โ๐ฅ ๐๐ฆ = โซ_2^4โใ2โ๐ฆใ ๐๐ฆ = 2โซ_2^4โใ(๐ฆ)^(1/2) ๐๐ฆใ = 2 [๐ฆ^(3/2)/(3/2)]_2^4 = 2 ร 2/3 [๐ฆ^(3/2) ]_2^4 = 4/3 [(4)^(3/2 )โ(2)^(3/2) ] = 4/3 [((4)^(1/2) )^3โ((2)^(1/2) )^3 ] = 4/3 [(2)^3โ(โ2)^3 ] = 4/3 [8 โ2โ2] = (32 โ 8โ2)/3 Thus, Area = (๐๐ โ ๐โ๐)/๐ 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.