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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Ex 7.8, 3 โˆซ1_2^3โ–’ใ€–๐‘ฅ2 ๐‘‘๐‘ฅใ€— โˆซ1_2^3โ–’ใ€–๐‘ฅ2 ๐‘‘๐‘ฅใ€— Putting ๐‘Ž =2 ๐‘ =3 โ„Ž=(๐‘ โˆ’ ๐‘Ž)/๐‘› =(3 โˆ’ 2)/๐‘› =1/๐‘› ๐‘“(๐‘ฅ)=๐‘ฅ^2 We know that โˆซ1_๐‘Ž^๐‘โ–’ใ€–๐‘ฅ ๐‘‘๐‘ฅใ€— =(๐‘โˆ’๐‘Ž) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (๐‘“(๐‘Ž)+๐‘“(๐‘Ž+โ„Ž)+๐‘“(๐‘Ž+2โ„Ž)โ€ฆ+๐‘“(๐‘Ž+(๐‘›โˆ’1)โ„Ž)) Hence we can write โˆซ1_2^3โ–’ใ€–๐‘ฅ2 ๐‘‘๐‘ฅใ€— =(3โˆ’2) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (๐‘“(2)+๐‘“(2+โ„Ž)+๐‘“(2+2โ„Ž)+ โ€ฆ+๐‘“(2+(๐‘›โˆ’1)โ„Ž)) =(๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (๐‘“(2)+๐‘“(2+โ„Ž)+๐‘“(2+2โ„Ž)+ โ€ฆ+๐‘“(2+(๐‘›โˆ’1)โ„Ž)) Here, ๐‘“(๐‘ฅ)=๐‘ฅ^2 ๐‘“(2)=(2)^2=4 ๐‘“(2+โ„Ž)=(2+โ„Ž)^2 ๐‘“ (2+2โ„Ž)=(2+2โ„Ž)^2 โ€ฆ ๐‘“(2+(๐‘›โˆ’1)โ„Ž)=(2+(๐‘›โˆ’1)โ„Ž)^2 Hence, our equation becomes โˆซ1_2^3โ–’ใ€–๐‘ฅ2 ๐‘‘๐‘ฅใ€— " " =(๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (๐‘“(2)+๐‘“(2+โ„Ž)+๐‘“(2+2โ„Ž)+ โ€ฆ+๐‘“(2+(๐‘›โˆ’1)โ„Ž)) =1 (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› ((2)^2+(2+โ„Ž)^2+(2+2โ„Ž)^2+ โ€ฆ+(2+(๐‘›โˆ’1)โ„Ž)^2 ) =(๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (โ–ˆ(2^2+(2^2+โ„Ž^2+4โ„Ž)+ใ€–(2ใ€—^2+(2โ„Ž)^2+8โ„Ž)+ โ€ฆโ€ฆ @ โ€ฆ+(2^2+((๐‘›โˆ’1)โ„Ž)^2+4(๐‘›โˆ’1) โ„Ž) )) =(๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› [2^2+2^2+ โ€ฆ +2^2 ] + โ„Ž^2+(2โ„Ž)^2+ โ€ฆ +(๐‘›โˆ’1)โ„Ž^2 + [4โ„Ž+8โ„Ž+ โ€ฆ +4(๐‘›โˆ’1)โ„Ž] =(๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (ใ€–๐‘›(2)ใ€—^2+[โ„Ž^2+(2)^2 . โ„Ž^2+ โ€ฆ +(๐‘›โˆ’1)^2 โ„Ž^2 ] +[4โ„Ž+2ร—4โ„Ž+ โ€ฆ +(๐‘›โˆ’1)ร—4โ„Ž] ) =(๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘›(๐‘›(2)^2+๐’‰^2 [(1)^2+(2)^2+ โ€ฆ+(๐‘›โˆ’1)^2 ] + ๐Ÿ’๐’‰ [1+2+ โ€ฆ+(๐‘›โˆ’1)]) =(๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (๐‘›(2)^2+โ„Ž^2 [๐‘›(๐‘› โˆ’ 1)(2๐‘› โˆ’ 1)/6]+4โ„Ž[๐‘›(๐‘› โˆ’ 1)/2] ) We know that 1^2+2^2+ โ€ฆ+๐‘›^2= (๐‘› (๐‘› + 1)(2๐‘› + 1))/6 1^2+2^2+ โ€ฆโ€ฆ+(๐‘›โˆ’1)^2 = ((๐‘› โˆ’ 1) (๐‘› โˆ’1 + 1)(2(๐‘› โˆ’ 1) + 1))/6 = ((๐‘› โˆ’ 1) ๐‘› (2๐‘› โˆ’ 2 + 1) )/6 = (๐‘› (๐‘› โˆ’ 1) (2๐‘› โˆ’ 1) )/6 We know that 1+2+3+ โ€ฆโ€ฆ+๐‘›= (๐‘› (๐‘› + 1))/2 1+2+3+ โ€ฆโ€ฆ+(๐‘›โˆ’1) = ((๐‘› โˆ’ 1) (๐‘› โˆ’ 1 + 1))/2 = (๐‘› (๐‘› โˆ’ 1) )/2 =(๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (4๐‘›+โ„Ž^2 [๐‘›(๐‘› โˆ’ 1)(2๐‘› โˆ’ 1)]/6+2โ„Ž[๐‘›(๐‘› โˆ’ 1)] ) =(๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) (4๐‘›/๐‘›+โ„Ž^2 [๐‘›(๐‘› โˆ’ 1)(2๐‘› โˆ’ 1)/6๐‘›]+2โ„Ž[๐‘›(๐‘› โˆ’ 1)/๐‘›]) =(๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) (4+โ„Ž^2 [(๐‘› โˆ’ 1)(2๐‘› โˆ’ 1)/6]+2โ„Ž[(๐‘› โˆ’ 1)]) =(๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) (4+(1/๐‘›)^2 (๐‘› โˆ’ 1)(2๐‘› โˆ’ 1)/6+2(1/๐‘›)(๐‘› โˆ’ 1)) =(๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) (4+1/๐‘›^2 . (๐‘› โˆ’ 1)(2๐‘› โˆ’ 1)/6 +2(1 โˆ’ 1/๐‘›)) =(๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) (4+ (1 โˆ’ 1/๐‘›)(2 โˆ’ 1/๐‘›)/6 +2(1 โˆ’ 1/๐‘›)) =4+ (1 โˆ’ 1/โˆž)(2 โˆ’ 1/โˆž)/6 +2(1 โˆ’ 1/โˆž) =4+ (1 โˆ’ 0)(2 โˆ’ 0)/6 +2(1 โˆ’0) =4+ (1 . 2)/6 +2 .1 =4+ 1/3 +2 =6+ 1/3 = ๐Ÿ๐Ÿ—/๐Ÿ‘

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.