Ex 7.8, 13 - Chapter 7 Class 12 Integrals

Last updated at April 16, 2024 by

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Transcript

Ex 7.8, 13 โˆซ_2^3โ–’๐‘ฅ๐‘‘๐‘ฅ/(๐‘ฅ2+1) Step 1 :- Let F(๐‘ฅ)=โˆซ1โ–’ใ€–๐‘ฅ/(๐‘ฅ^2 + 1) ๐‘‘๐‘ฅใ€— Let ๐‘ฅ^2 + 1=๐‘ก Differentiating w.r.t.๐‘ฅ ๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ^2+1)=๐‘‘๐‘ก/๐‘‘๐‘ฅ 2๐‘ฅ=๐‘‘๐‘ก/๐‘‘๐‘ฅ d๐‘ฅ=๐‘‘๐‘ก/2๐‘ฅ Putting Values of ๐‘ฅ^2+1=๐‘ก & ๐‘‘๐‘ฅ=๐‘‘๐‘ก/2๐‘ฅ โˆซ1โ–’ใ€–(๐‘ฅ ๐‘‘๐‘ฅ)/(๐‘ฅ^2 + 1)=โˆซ1โ–’ใ€–๐‘ฅ/๐‘ก ๐‘‘๐‘ก/2๐‘ฅใ€—ใ€— =1/2 โˆซ1โ–’๐‘‘๐‘ก/๐‘ก =1/2 ๐‘™๐‘œ๐‘”|๐‘ก| =1/2 ๐‘™๐‘œ๐‘”|๐‘ฅ^2+1| Hence F(๐‘ฅ)=1/2 ๐‘™๐‘œ๐‘”|๐‘ฅ^2+1| Step 2 :- โˆซ_2^3โ–’(๐‘ฅ ๐‘‘๐‘ฅ)/(๐‘ฅ^2+1)=๐น(3)โˆ’๐น(2) =1/2 ๐‘™๐‘œ๐‘”|3^2+1|โˆ’1/2 ๐‘™๐‘œ๐‘”|2^2+1| =1/2 logโกใ€–10โˆ’1/2 ๐‘™๐‘œ๐‘”(5)ใ€— =1/2 (logโกใ€–10โˆ’logโก5 ใ€— ) =1/2 ๐‘™๐‘œ๐‘” 10/5 =๐Ÿ/๐Ÿ ๐ฅ๐จ๐ โก๐Ÿ

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.