Misc 12 - Chapter 8 Class 12 Application of Integrals (Term 2)
Last updated at Dec. 12, 2019 by Teachoo
Last updated at Dec. 12, 2019 by Teachoo
Transcript
Misc 12 Find the area bounded by curves {(๐ฅ, ๐ฆ) :๐ฆโฅ ๐ฅ2 and ๐ฆ=|๐ฅ|} Here, ๐ฅ^2=๐ฆ is a parabola And y = |๐ฅ| ={โ(๐ฅ, ๐ฅโฅ0@&โ๐ฅ, ๐ฅ<0)โค So, we draw a parabola and two lines Point A is the intersection of parabola and line y = โx Point B is the intersection of parabola and line y = x Finding points A & B Point A Point A is intersection of y = x2 & y = โx Solving x2 = โx x2 + x = 0 x(x + 1) = 0 So, x = โ1 & x = 0 For x = โ1 y = โx = โ(โ1) = 1 So, point A (โ1, 1) Point B Point B is intersection of y = x2 & y = x Solving x2 = x x2 โ x = 0 x(x โ 1) = 0 So, x = 1 & x = 0 For x = 1 y = x = 1 So, point B (1, 1) Since Required area is symmetrical about y-axis Required Area = 2 ร Area ODBC Area ODBC Area ODBC = Area ODBE โ Area OCBE Area ODBE Area ODBE = โซ_0^1โใ๐ฆ ๐๐ฅใ y โ Equation of line y = x Area ODBE =โซ_0^1โใ๐ฅ ๐๐ฅใ =[๐ฅ^2/2]_0^1 =1^2/( 2)โ0^2/2 =1/2 Area OCBE Area OCBE = โซ_0^1โใ๐ฆ ๐๐ฅใ y โ Equation of parabola y = x2 Therefore, Area OCBE =โซ_0^1โใ๐ฅ^2 ๐๐ฅใ =[๐ฅ^3/3]_0^1 =1^3/3โ0^3/3 =1/3 Hence, Area ODBC = Area ODBE โ Area OCBE = 1/2โ1/3 = 1/6 Also, Required Area = 2 ร Area ODBC = 2 ร 1/6 = ๐/๐ square units
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