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Misc 12 - Find area: {(x, y): y > x2 and y = |x|} - Class 12

Misc 12 - Chapter 8 Class 12 Application of Integrals - Part 2
Misc 12 - Chapter 8 Class 12 Application of Integrals - Part 3
Misc 12 - Chapter 8 Class 12 Application of Integrals - Part 4
Misc 12 - Chapter 8 Class 12 Application of Integrals - Part 5
Misc 12 - Chapter 8 Class 12 Application of Integrals - Part 6

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Misc 12 Find the area bounded by curves {(𝑥, 𝑦) :𝑦≥ 𝑥2 and 𝑦=|𝑥|} Here, 𝑥^2=𝑦 is a parabola And y = |𝑥| ={█(𝑥, 𝑥≥0@&−𝑥, 𝑥<0)┤ So, we draw a parabola and two lines Point A is the intersection of parabola and line y = –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 = –x Solving x2 = –x x2 + x = 0 x(x + 1) = 0 So, x = –1 & x = 0 For x = –1 y = –x = –(–1) = 1 So, point A (–1, 1) Point B Point B is intersection of y = x2 & y = x Solving x2 = x x2 – x = 0 x(x – 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 × Area ODBC Area ODBC Area ODBC = Area ODBE – Area OCBE Area ODBE Area ODBE = ∫_0^1▒〖𝑦 𝑑𝑥〗 y → Equation of line y = x Area ODBE =∫_0^1▒〖𝑥 𝑑𝑥〗 =[𝑥^2/2]_0^1 =1^2/( 2)−0^2/2 =1/2 Area OCBE Area OCBE = ∫_0^1▒〖𝑦 𝑑𝑥〗 y → Equation of parabola y = x2 Therefore, Area OCBE =∫_0^1▒〖𝑥^2 𝑑𝑥〗 =[𝑥^3/3]_0^1 =1^3/3−0^3/3 =1/3 Hence, Area ODBC = Area ODBE – Area OCBE = 1/2−1/3 = 1/6 Also, Required Area = 2 × Area ODBC = 2 × 1/6 = 𝟏/𝟑 square units

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.