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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.