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


Misc 11 Using the method of integration find the area bounded by the curve |๐‘ฅ|+|๐‘ฆ|=1 [Hint: The required region is bounded by lines ๐‘ฅ+๐‘ฆ= 1, ๐‘ฅ โ€“๐‘ฆ=1, โ€“๐‘ฅ+๐‘ฆ =1 and โˆ’๐‘ฅ โˆ’๐‘ฆ=1 ] We know that "โ”‚" ๐‘ฅ"โ”‚"={โ–ˆ(๐‘ฅ, ๐‘ฅโ‰ฅ0@&โˆ’๐‘ฅ, ๐‘ฅ<0)โ”ค & "โ”‚" ๐‘ฆ"โ”‚"={โ–ˆ(๐‘ฆ, ๐‘ฆโ‰ฅ0@&โˆ’๐‘ฆ, ๐‘ฆ<0)โ”ค So, we can write โ”‚๐‘ฅ"โ”‚+โ”‚" ๐‘ฆ"โ”‚"=1 as {โ–ˆ(โ–ˆ(โ–ˆ( ๐‘ฅ+๐‘ฆ=1 ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ>0 , ๐‘ฆ>0@โˆ’๐‘ฅ+๐‘ฆ=1 ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ<0 ๐‘ฆ>0)@โ–ˆ( ๐‘ฅโˆ’๐‘ฆ =1 ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ>0 , ๐‘ฆ<0@โˆ’๐‘ฅโˆ’๐‘ฆ=1 ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ<0 ๐‘ฆ<0)))โ”ค For ๐’™+๐’š=๐Ÿ For โˆ’๐’™+๐’š=๐Ÿ For โˆ’๐’™โˆ’๐’š=๐Ÿ For ๐’™โˆ’๐’š=๐Ÿ Joining them, we get our diagram Since the Curve symmetrical about ๐‘ฅ & ๐‘ฆโˆ’๐‘Ž๐‘ฅ๐‘–๐‘  Required Area = 4 ร— Area AOB Area AOB Area AOB = โˆซ_0^1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— where ๐‘ฅ+๐‘ฆ=1 ๐‘ฆ=1โˆ’๐‘ฅ Therefore, Area AOB = โˆซ_0^1โ–’ใ€–(1โˆ’๐‘ฅ) ๐‘‘๐‘ฅใ€— = [๐‘ฅโˆ’๐‘ฅ^2/2]_0^1 =1โˆ’ใ€– 1ใ€—^2/2โˆ’(0โˆ’0^2/2)^2 =1โˆ’1/2 =1/2 Hence, Required Area = 4 ร— Area AOB = 4 ร— 1/2 = 2 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 10 years. He provides courses for Maths and Science at Teachoo.