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Misc 17 - Area bounded by y = x|x|, x-axis and x = -1, x = 1 is given

Misc 17 - Chapter 8 Class 12 Application of Integrals - Part 2
Misc 17 - Chapter 8 Class 12 Application of Integrals - Part 3
Misc 17 - Chapter 8 Class 12 Application of Integrals - Part 4

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Misc 17 The area bounded by the curve 𝑦 = π‘₯ |π‘₯| , π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 and the ordinates π‘₯ = – 1 and π‘₯=1 is given by (A) 0 (B) 1/3 (C) 2/3 (D) 4/3 [Hint : 𝑦=π‘₯2 if π‘₯ > 0 π‘Žπ‘›π‘‘ 𝑦 =βˆ’π‘₯2 if π‘₯ < 0] We know that |π‘₯|={β–ˆ(π‘₯, π‘₯β‰₯0@&βˆ’π‘₯, π‘₯<0)─ Therefore, y = x|π‘₯|={β–ˆ(π‘₯π‘₯, π‘₯β‰₯0@&π‘₯(βˆ’π‘₯), π‘₯<0)─ y ={β–ˆ(π‘₯^2, π‘₯β‰₯0@&βˆ’π‘₯^2, π‘₯<0)─ Area Required = Area ABO + Area DCO Area ABO Area ABO =∫_(βˆ’1)^0▒〖𝑦 𝑑π‘₯γ€— Here, 𝑦=γ€–βˆ’π‘₯γ€—^2 Therefore, Area ABO =∫_(βˆ’1)^0β–’γ€–γ€–βˆ’π‘₯γ€—^2 𝑑π‘₯γ€— γ€–=βˆ’[π‘₯^3/3]γ€—_(βˆ’1)^0 =βˆ’[0^3/3βˆ’(βˆ’1)^3/3] =(βˆ’1)/3 Since Area is always positive, Area ABO = 1/3 Area DCO Area DCO =∫_0^1▒〖𝑦 𝑑π‘₯γ€— Here, 𝑦=π‘₯^2 Therefore, Area DCO =∫_0^1β–’γ€–π‘₯^2 𝑑π‘₯γ€— γ€–=[π‘₯^3/3]γ€—_0^1 =1/3 [1^3βˆ’0^3 ] =1/3 [1βˆ’0] =1/3 ∴ Required Area = Area ABO + Area DCO =1/3+1/3 =2/3 So, Option C is Correct

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Davneet Singh

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.