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

Transcript

Misc 12 Find the area bounded by curves {(๐‘ฅ, ๐‘ฆ) :๐‘ฆโ‰ฅ ๐‘ฅ2 and ๐‘ฆ=|๐‘ฅ|} Here, ๐‘ฅ^2=๐‘ฆ is a parabola And y = |๐‘ฅ| ={โ–ˆ(๐‘ฅ, ๐‘ฅโ‰ฅ0@&โˆ’๐‘ฅ, ๐‘ฅ<0)โ”ค So, we draw a parabola and two lines Point A is the intersection of parabola and line y = โ€“x 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 = โ€“x Solving x2 = โ€“x x2 + x = 0 x(x + 1) = 0 So, x = โ€“1 & x = 0 For x = โ€“1 y = โ€“x = โ€“(โ€“1) = 1 So, point A (โ€“1, 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 = โˆซ_0^1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— y โ†’ Equation of line y = x Area ODBE =โˆซ_0^1โ–’ใ€–๐‘ฅ ๐‘‘๐‘ฅใ€— =[๐‘ฅ^2/2]_0^1 =1^2/( 2)โˆ’0^2/2 =1/2 Area OCBE Area OCBE = โˆซ_0^1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— y โ†’ Equation of parabola y = x2 Therefore, Area OCBE =โˆซ_0^1โ–’ใ€–๐‘ฅ^2 ๐‘‘๐‘ฅใ€— =[๐‘ฅ^3/3]_0^1 =1^3/3โˆ’0^3/3 =1/3 Hence, Area ODBC = Area ODBE โ€“ Area OCBE = 1/2โˆ’1/3 = 1/6 Also, Required Area = 2 ร— Area ODBC = 2 ร— 1/6 = ๐Ÿ/๐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.