Question 2 - Ex 8.1 - Chapter 8 Class 12 Application of Integrals
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Question 2 Find the area of the region bounded by y2 = 9๐ฅ, ๐ฅ = 2, ๐ฅ = 4 and the ๐ฅ-axis in the first quadrant. Given curve ๐ฆ^2=9๐ฅ We have to find area between x = 2 and x = 4 โด We have to find area of BCFE Area of BCFE = โซ_2^4โใ๐ฆ .ใ ๐๐ฅ We know that ๐ฆ^2=9๐ฅ Taking square root on both sides ๐ฆ=ยฑโ9๐ฅ ๐ฆ=ยฑ3โ๐ฅ Since BCFE is in 1st Quadrant We take positive value of y โด ๐ฆ=3โ๐ฅ Area of BCFE = โซ_2^4โใ๐ฆ .ใ ๐๐ฅ = 3โซ_2^4โโ๐ฅ ๐๐ฅ = 3โซ_2^4โใ(๐ฅ)^(1/2) ๐๐ฅใ = 3 [๐ฅ^(1/2 + 1)/(1/2 + 1 )]_2^4 = 3 [๐ฅ^(3/2 )/(3/2)]_2^4 = 3 ร 2/3 [๐ฅ^(3/2) ]_2^4 = 2 [(4)^(3/2 )โ(2)^(3/2) ] = 2 [((4)^(1/2) )^3โ((2)^(1/2) )^3 ] = 2 [(2)^3โ(โ2)^3 ] = 2 [8 โ2โ2] = 16 โ 4โ2 Thus, Area = 16 โ 4โ๐ square units
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