Integration Full Chapter Explained - Integration Class 12 - Everything you need

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

Transcript

Ex 7.8, 1 โˆซ1_๐‘Ž^๐‘โ–’ใ€–๐‘ฅ ๐‘‘๐‘ฅใ€— โˆซ1_๐‘Ž^๐‘โ–’ใ€–๐‘ฅ ๐‘‘๐‘ฅใ€— Putting ๐‘Ž =๐‘Ž ๐‘ =๐‘ โ„Ž=(๐‘ โˆ’ ๐‘Ž)/๐‘› ๐‘“(๐‘ฅ)=๐‘ฅ Ex 7.8, 1 โˆซ1_๐‘Ž^๐‘โ–’ใ€–๐‘ฅ ๐‘‘๐‘ฅใ€— โˆซ1_๐‘Ž^๐‘โ–’ใ€–๐‘ฅ ๐‘‘๐‘ฅใ€— Putting ๐‘Ž =๐‘Ž ๐‘ =๐‘ โ„Ž=(๐‘ โˆ’ ๐‘Ž)/๐‘› ๐‘“(๐‘ฅ)=๐‘ฅ 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_๐‘Ž^๐‘โ–’ใ€–๐‘ฅ ๐‘‘๐‘ฅใ€— =(๐‘โˆ’๐‘Ž) limโ”ฌ(nโ†’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) Here, ๐‘“(๐‘ฅ)=๐‘ฅ ๐‘“(๐‘Ž)=๐‘Ž ๐‘“(๐‘Ž+โ„Ž)=๐‘Ž+โ„Ž ๐‘“ (๐‘Ž+2โ„Ž)=๐‘Ž+2โ„Ž โ€ฆ ๐‘“(๐‘Ž+(๐‘›โˆ’1)โ„Ž)=๐‘Ž+(๐‘›โˆ’1)โ„Ž Hence, our equation becomes โˆด โˆซ_0^๐‘Žโ–’๐‘ฅ ๐‘‘๐‘ฅ = (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)) = (๐‘โˆ’๐‘Ž) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› (๐‘Ž+(๐‘Ž+โ„Ž)+(๐‘Ž+2โ„Ž)+ โ€ฆ+(๐‘Ž+(๐‘›โˆ’1)โ„Ž)) = (๐‘โˆ’๐‘Ž) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› ( ๐‘Ž+๐‘Ž+ โ€ฆ+๐‘Ž +โ„Ž+2โ„Ž+ โ€ฆโ€ฆ+(๐‘›โˆ’1)โ„Ž) = (๐‘โˆ’๐‘Ž) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› ( ๐‘›๐‘Ž+โ„Ž (1+2+ โ€ฆโ€ฆโ€ฆ+(๐‘›โˆ’1))) 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‘โˆ’๐‘Ž) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) 1/๐‘› ( ๐‘›๐‘Ž+(โ„Ž . ๐‘›(๐‘› โˆ’ 1))/2) = (๐‘โˆ’๐‘Ž) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) ( ๐‘›๐‘Ž/๐‘›+๐‘›(๐‘› โˆ’ 1)โ„Ž/2๐‘›) = (๐‘โˆ’๐‘Ž) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) ( ๐‘Ž+(๐‘› โˆ’ 1)โ„Ž/2) = (๐‘โˆ’๐‘Ž) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) ( ๐‘Ž+(๐‘› โˆ’ 1)(๐‘ โˆ’๐‘Ž)/(2 . ๐‘›)) = (๐‘โˆ’๐‘Ž) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) ( ๐‘Ž+(๐‘›/๐‘› โˆ’ 1/๐‘›) ((๐‘ โˆ’ ๐‘Ž) )/2) [๐‘ˆ๐‘ ๐‘–๐‘›๐‘” โ„Ž=(๐‘ โˆ’ ๐‘Ž)/๐‘›] = (๐‘โˆ’๐‘Ž) (๐‘™๐‘–๐‘š)โ”ฌ(๐‘›โ†’โˆž) ( ๐‘Ž+(1โˆ’ 1/๐‘›) ((๐‘ โˆ’ ๐‘Ž) )/2) = (๐‘โˆ’๐‘Ž)( ๐‘Ž+(1โˆ’ 1/โˆž) ((๐‘ โˆ’ ๐‘Ž) )/2) = (๐‘โˆ’๐‘Ž)( ๐‘Ž+(1โˆ’0) ((๐‘ โˆ’ ๐‘Ž) )/2) = (๐‘โˆ’๐‘Ž)( ๐‘Ž+ (๐‘ โˆ’ ๐‘Ž )/2) = (๐‘โˆ’๐‘Ž)((2๐‘Ž + ๐‘ โˆ’ ๐‘Ž )/2) = (๐‘ โˆ’ ๐‘Ž)(๐‘ + ๐‘Ž)/2 = (๐’ƒ^๐Ÿ โˆ’ ๐’‚^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.