Misc 11 - Using integration find area bounded by |x| + |y| = 1 - Area between curve and curve

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  1. Chapter 8 Class 12 Application of Integrals
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Misc 11 Using the method of integration find the area bounded by the curve ๐‘ฅ๏ทฏ+ ๐‘ฆ๏ทฏ=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‘ฅโ”‚= ๐‘ฅ, ๐‘ฅโ‰ฅ0๏ทฎ&โˆ’๐‘ฅ, ๐‘ฅ<0๏ทฏ๏ทฏ & โ”‚๐‘ฆโ”‚= ๐‘ฆ, ๐‘ฆโ‰ฅ0๏ทฎ&โˆ’๐‘ฆ, ๐‘ฆ<0๏ทฏ๏ทฏ So, we can write โ”‚๐‘ฅโ”‚+โ”‚๐‘ฆโ”‚=1 as ๐‘ฅ+๐‘ฆ=1 ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ>0 , ๐‘ฆ>0๏ทฎโˆ’๐‘ฅ+๐‘ฆ=1 ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ<0 ๐‘ฆ>0๏ทฏ๏ทฎ ๐‘ฅโˆ’๐‘ฆ =1 ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ>0 , ๐‘ฆ<0๏ทฎโˆ’๐‘ฅโˆ’๐‘ฆ=1 ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ<0 ๐‘ฆ<0๏ทฏ๏ทฏ๏ทฏ๏ทฏ For ๐’™+๐’š=๐Ÿ For โˆ’๐’™+๐’š=๐Ÿ Hence the figure is Since the Curve symmetrical about ๐‘ฅ & ๐‘ฆโˆ’๐‘Ž๐‘ฅ๐‘–๐‘  Required Area = 4 ร— Area AOB Area AOB Area ABO = 0๏ทฎ1๏ทฎ๐‘ฆ ๐‘‘๐‘ฅ๏ทฏ where ๐‘ฅ+๐‘ฆ=1 ๐‘ฆ=1โˆ’๐‘ฅ Therefore, Area ABO = 0๏ทฎ1๏ทฎ 1โˆ’๐‘ฅ๏ทฏ ๐‘‘๐‘ฅ๏ทฏ = ๐‘ฅโˆ’ ๐‘ฅ๏ทฎ2๏ทฏ๏ทฎ2๏ทฏ๏ทฏ๏ทฎ0๏ทฎ1๏ทฏ =1โˆ’ 1๏ทฎ2๏ทฏ๏ทฎ2๏ทฏโˆ’ 0โˆ’ 0๏ทฎ2๏ทฏ๏ทฎ2๏ทฏ๏ทฏ๏ทฎ2๏ทฏ =1โˆ’ 1๏ทฎ2๏ทฏ = 1๏ทฎ2๏ทฏ Hence, Required Area = 4 ร— Area AOB = 4 ร— 1๏ทฎ2๏ทฏ = 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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.