Slide23.JPG

Slide24.JPG
Slide25.JPG


Transcript

Ex 7.9, 8 Evaluate the integrals using substitution โˆซ_1^(2 )โ–’ใ€– (1/๐‘ฅ โˆ’1/(2๐‘ฅ^2 )) ใ€— ๐‘’^2๐‘ฅ ๐‘‘๐‘ฅ Let ๐‘ก=2๐‘ฅ ๐‘‘๐‘ก/๐‘‘๐‘ฅ=2 ๐‘‘๐‘ก/2=๐‘‘๐‘ฅ Thus, when x varies from 1 to 2, t varies from 2 to 4 Substituting, โˆซ_1^(2 )โ–’ใ€– (1/๐‘ฅ โˆ’1/(2๐‘ฅ^2 )) ใ€— ๐‘’^2๐‘ฅ ๐‘‘๐‘ฅ = โˆซ_2^4โ–’ใ€–๐‘’^๐‘ก (1/(๐‘ก/2)โˆ’1/(2ใ€– (๐‘ก/2)ใ€—^2 )) ใ€— ๐‘‘๐‘ก/2 =โˆซ_2^4โ–’ใ€–๐‘’^๐‘ก (2/๐‘กโˆ’4/(2๐‘ก^2 )) ใ€— ๐‘‘๐‘ก/2 =โˆซ_2^4โ–’ใ€–๐‘’^๐‘ก (1/๐‘กโˆ’2/๐‘ก^2 ) ใ€— ๐‘‘๐‘ก It is of the form โˆซ1โ–’ใ€–๐‘’^๐‘ฅ [๐‘“(๐‘ฅ)+๐‘“^โ€ฒ (๐‘ฅ)] ใ€— ๐‘‘๐‘ฅ=๐‘’^๐‘ฅ ๐‘“(๐‘ฅ)+๐ถ Where ๐‘“(๐‘ฅ)=1/๐‘ก ๐‘“^โ€ฒ (๐‘ฅ)= (โˆ’1)/๐‘ก^2 Hence, our equation becomes โˆซ_2^4โ–’ใ€–๐‘’^๐‘ก (1/๐‘กโˆ’2/๐‘ก^2 ) ใ€— ๐‘‘๐‘ก = [๐‘’^๐‘กร—1/๐‘ก]_2^4 = (๐‘’^4/4โˆ’๐‘’^2/2) = (๐’†^๐Ÿ (๐’†^๐Ÿ โˆ’ ๐Ÿ))/๐Ÿ’

Go Ad-free
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.