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Misc 11 - Using integration find area bounded by |x| + |y| = 1

Misc 11 - Chapter 8 Class 12 Application of Integrals - Part 2
Misc 11 - Chapter 8 Class 12 Application of Integrals - Part 3
Misc 11 - Chapter 8 Class 12 Application of Integrals - Part 4
Misc 11 - Chapter 8 Class 12 Application of Integrals - Part 5

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Misc 11 Using the method of integration find the area bounded by the curve |π‘₯|+|𝑦|=1 [Hint: The required region is bounded by lines π‘₯+𝑦= 1, π‘₯ –𝑦=1, –π‘₯+𝑦 =1 and βˆ’π‘₯ βˆ’π‘¦=1 ] We know that "β”‚" π‘₯"β”‚"={β–ˆ(π‘₯, π‘₯β‰₯0@&βˆ’π‘₯, π‘₯<0)─ & "β”‚" 𝑦"β”‚"={β–ˆ(𝑦, 𝑦β‰₯0@&βˆ’π‘¦, 𝑦<0)─ So, we can write β”‚π‘₯"β”‚+β”‚" 𝑦"β”‚"=1 as {β–ˆ(β–ˆ(β–ˆ( π‘₯+𝑦=1 π‘“π‘œπ‘Ÿ π‘₯>0 , 𝑦>0@βˆ’π‘₯+𝑦=1 π‘“π‘œπ‘Ÿ π‘₯<0 𝑦>0)@β–ˆ( π‘₯βˆ’π‘¦ =1 π‘“π‘œπ‘Ÿ π‘₯>0 , 𝑦<0@βˆ’π‘₯βˆ’π‘¦=1 π‘“π‘œπ‘Ÿ π‘₯<0 𝑦<0)))─ For 𝒙+π’š=𝟏 For βˆ’π’™+π’š=𝟏 For βˆ’π’™βˆ’π’š=𝟏 For π’™βˆ’π’š=𝟏 Joining them, we get our diagram Since the Curve symmetrical about π‘₯ & π‘¦βˆ’π‘Žπ‘₯𝑖𝑠 Required Area = 4 Γ— Area AOB Area AOB Area AOB = ∫_0^1▒〖𝑦 𝑑π‘₯γ€— where π‘₯+𝑦=1 𝑦=1βˆ’π‘₯ Therefore, Area AOB = ∫_0^1β–’γ€–(1βˆ’π‘₯) 𝑑π‘₯γ€— = [π‘₯βˆ’π‘₯^2/2]_0^1 =1βˆ’γ€– 1γ€—^2/2βˆ’(0βˆ’0^2/2)^2 =1βˆ’1/2 =1/2 Hence, Required Area = 4 Γ— Area AOB = 4 Γ— 1/2 = 2 square units

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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.