Chapter 7 Class 12 Integrals
Concept wise

  Slide27.JPG

Slide28.JPG
Slide29.JPG

Go Ad-free

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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.