# Misc 12 - Chapter 8 Class 12 Application of Integrals

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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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.