Solve all your doubts with Teachoo Black (new monthly pack available now!)
Miscellaneous
Misc 1 (ii) Important
Misc 2 Deleted for CBSE Board 2023 Exams
Misc 3
Misc 4 Important
Misc 5 Important
Misc 6 Important
Misc 7
Misc 8
Misc 9 Important
Misc 10
Misc 11 Important You are here
Misc 12 Important Deleted for CBSE Board 2023 Exams
Misc 13
Misc 14 Important
Misc 15 Deleted for CBSE Board 2023 Exams
Misc 16 (MCQ)
Misc 17 (MCQ) Important
Misc 18 (MCQ) Deleted for CBSE Board 2023 Exams
Misc 19 (MCQ) Important Deleted for CBSE Board 2023 Exams
Miscellaneous
Last updated at Dec. 12, 2019 by Teachoo
Misc 11 Using the method of integration find the area bounded by the curve |𝑥|+|𝑦|=1 [Hint: The required region is bounded by lines 𝑥+𝑦= 1, 𝑥 –𝑦=1, –𝑥+𝑦 =1 and −𝑥 −𝑦=1 ] We know that "│" 𝑥"│"={█(𝑥, 𝑥≥0@&−𝑥, 𝑥<0)┤ & "│" 𝑦"│"={█(𝑦, 𝑦≥0@&−𝑦, 𝑦<0)┤ So, we can write │𝑥"│+│" 𝑦"│"=1 as {█(█(█( 𝑥+𝑦=1 𝑓𝑜𝑟 𝑥>0 , 𝑦>0@−𝑥+𝑦=1 𝑓𝑜𝑟 𝑥<0 𝑦>0)@█( 𝑥−𝑦 =1 𝑓𝑜𝑟 𝑥>0 , 𝑦<0@−𝑥−𝑦=1 𝑓𝑜𝑟 𝑥<0 𝑦<0)))┤ For 𝒙+𝒚=𝟏 For −𝒙+𝒚=𝟏 For −𝒙−𝒚=𝟏 For 𝒙−𝒚=𝟏 Joining them, we get our diagram Since the Curve symmetrical about 𝑥 & 𝑦−𝑎𝑥𝑖𝑠 Required Area = 4 × Area AOB Area AOB Area AOB = ∫_0^1▒〖𝑦 𝑑𝑥〗 where 𝑥+𝑦=1 𝑦=1−𝑥 Therefore, Area AOB = ∫_0^1▒〖(1−𝑥) 𝑑𝑥〗 = [𝑥−𝑥^2/2]_0^1 =1−〖 1〗^2/2−(0−0^2/2)^2 =1−1/2 =1/2 Hence, Required Area = 4 × Area AOB = 4 × 1/2 = 2 square units