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Ex 7.10, 8 - Evaluate integral (1/x - 1/ 2x2) e2x dx - Ex 7.10

Ex 7.10, 8 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.10, 8 - Chapter 7 Class 12 Integrals - Part 3


Transcript

Ex 7.10, 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’

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.