Ex 7.8, 5 - Evaluate ex dx from -1 to 1 by limit as sum - Ex 7.8

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.8, 5 โˆซ1_(โˆ’1)^1โ–’ใ€–๐‘’๐‘ฅ ๐‘‘๐‘ฅใ€— Hence we can write it as โˆซ1_(โˆ’1)^1โ–’ใ€–๐‘’๐‘ฅ ๐‘‘๐‘ฅใ€— =(1 โˆ’(โˆ’1)) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (๐‘“(โˆ’1)+๐‘“(โˆ’1+โ„Ž)+๐‘“(โˆ’1+2โ„Ž)+ โ€ฆ+๐‘“(โˆ’1+(๐‘›โˆ’1)โ„Ž)) =2 (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (๐‘“(โˆ’1)+๐‘“(โˆ’1+โ„Ž)+๐‘“(โˆ’1+2โ„Ž)+ โ€ฆ+๐‘“(โˆ’1+(๐‘›โˆ’1)โ„Ž)) ๐‘“(๐‘ฅ)=๐‘’^๐‘ฅ ๐‘“(โˆ’1)=๐‘’^(โˆ’1) ๐‘“(โˆ’1+โ„Ž)=๐‘’^(โˆ’1 + โ„Ž) =๐‘’^(โˆ’1). ๐‘’^โ„Ž ๐‘“ (โˆ’1+2โ„Ž)=๐‘’^(โˆ’1 + 2โ„Ž)=๐‘’^(โˆ’1). ๐‘’^2โ„Ž โ€ฆ. ๐‘“(โˆ’1+(๐‘›โˆ’1)โ„Ž)=๐‘’^(โˆ’1+(๐‘›โˆ’1)โ„Ž)=๐‘’^(โˆ’1).๐‘’^(๐‘›+1)โ„Ž Thus, โˆซ1_(โˆ’1)^1โ–’ใ€–๐‘’๐‘ฅ ๐‘‘๐‘ฅใ€— =2 (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (๐‘“(โˆ’1)+๐‘“(โˆ’1+โ„Ž)+๐‘“(โˆ’1+2โ„Ž)+ โ€ฆ+๐‘“(โˆ’1+(๐‘›โˆ’1)โ„Ž)) =2 (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (๐‘’^(โˆ’1)+๐‘’^(โˆ’1). ๐‘’^โ„Ž+๐‘’^(โˆ’1). ๐‘’^2โ„Ž+ โ€ฆ+๐‘’^(โˆ’1).๐‘’^(๐‘›โˆ’1)โ„Ž ) =2 (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (๐‘’^(โˆ’1) (1+โ„Ž+๐‘’^2โ„Ž+ โ€ฆ+๐‘’^(๐‘›โˆ’1)โ„Ž )) =2๐‘’^(โˆ’1) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (1+โ„Ž+๐‘’^2โ„Ž+ โ€ฆ+๐‘’^(๐‘›โˆ’1)โ„Ž ) Let S = 1+โ„Ž+๐‘’^2โ„Ž+ โ€ฆ+๐‘’^(๐‘›โˆ’1)โ„Ž It is G.P with common ratio (r) ๐‘Ÿ = ๐‘’^โ„Ž/1 = ๐‘’^โ„Ž We know Sum of G.P = a((๐‘Ÿ^๐‘› โˆ’ 1)/(๐‘Ÿ โˆ’ 1)) Replacing a by 1 and r by ๐‘’^๐‘› , we get S = 1(((๐‘’^โ„Ž )^๐‘› โˆ’ 1)/(๐‘’^๐‘› โˆ’ 1))= (๐‘’^๐‘›โ„Ž โˆ’ 1)/(๐‘’^๐‘› โˆ’ 1) Thus, โˆซ1_(โˆ’1)^1โ–’ใ€–๐‘’๐‘ฅ ๐‘‘๐‘ฅใ€— =2 . ๐‘’^(โˆ’1) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (1+๐‘’^๐‘›+๐‘’^2โ„Ž+ โ€ฆ+๐‘’^(๐‘›โˆ’1)โ„Ž ) = 2/๐‘’ (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› ((๐‘’^๐‘›โ„Ž โˆ’ 1)/(๐‘’^๐‘› โˆ’ 1)) = 2/๐‘’ (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› ((๐‘’^๐‘›โ„Ž โˆ’ 1)/(โ„Ž . (๐‘’^โ„Ž โˆ’ 1)/โ„Ž)) = 2/๐‘’ (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) (๐‘’^๐‘›โ„Ž โˆ’ 1)/๐‘›โ„Ž . 1/( (๐‘’^โ„Ž โˆ’ 1)/โ„Ž) = 2/๐‘’ (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) (๐‘’^๐‘›โ„Ž โˆ’ 1)/๐‘›โ„Ž . (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/( (๐‘’^โ„Ž โˆ’ 1)/โ„Ž) Taking (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/( (๐‘’^โ„Ž โˆ’ 1)/โ„Ž) As ๐‘›โ†’โˆž โ‡’ 2/๐‘› โ†’0 โ‡’ โ„Ž โ†’0 โˆด (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/( (๐‘’^โ„Ž โˆ’ 1)/โ„Ž) = (๐‘™๐‘–๐‘š)โ”ฌ(โ„Žโ†’0) 1/( (๐‘’^โ„Ž โˆ’ 1)/โ„Ž) = 1/1 = 1 Now, โˆซ1_(โˆ’1)^1โ–’ใ€–๐‘’๐‘ฅ ๐‘‘๐‘ฅใ€— = 2/๐‘’ (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) (๐‘’^๐‘›โ„Ž โˆ’ 1)/๐‘›โ„Ž . (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/( (๐‘’^โ„Ž โˆ’ 1)/โ„Ž) = 2/๐‘’ (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) (๐‘’^๐‘›โ„Ž โˆ’ 1)/๐‘›โ„Ž . 1 = 2/๐‘’ (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) (๐‘’^(๐‘› . 2/๐‘›) โˆ’ 1)/(๐‘› (2/๐‘›) ) = 2/๐‘’ (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) (๐‘’^2 โˆ’ 1)/2 = 2/๐‘’ . (๐‘’^2 โˆ’ 1)/2 = (๐‘’^2 โˆ’ 1)/๐‘’ = ๐‘’^2/๐‘’ โˆ’ 1/๐‘’ =๐’† โˆ’ ๐Ÿ/๐’†

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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.