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


Ex 7.9, 4 Evaluate the integrals using substitution โˆซ_0^2โ–’ใ€–๐‘ฅโˆš(๐‘ฅ+2)ใ€—โกใ€– (๐‘๐‘ข๐‘ก ๐‘ฅ+2=๐‘ก^2 )ใ€— โˆซ_0^2โ–’ใ€–๐‘ฅโˆš(๐‘ฅ+2)ใ€—โกใ€– ๐‘‘๐‘ฅใ€— Put ๐‘ฅ+2=๐‘ก^2 Differentiating w.r.t. ๐‘ฅ ๐‘‘(๐‘ฅ + 2)/๐‘‘๐‘ฅ=๐‘‘(๐‘ก^2 )/๐‘‘๐‘ก ร—๐‘‘๐‘ก/๐‘‘๐‘ฅ 1=2๐‘ก ร— ๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=2๐‘ก ๐‘‘๐‘ก Hence, when ๐‘ฅ varies from 0 to 2, then t varies from โˆš2 to 2 Therefore we can write โˆซ_0^2โ–’ใ€–๐‘ฅโˆš(๐‘ฅ+2) ๐‘‘๐‘ฅ =โˆซ_(โˆš2)^2โ–’ใ€–(๐‘ก^2โˆ’2) โˆš(๐‘ก^2 ) 2๐‘ก ๐‘‘๐‘กใ€—ใ€— =โˆซ_(โˆš2)^2โ–’ใ€–(๐‘ก^2โˆ’2)๐‘ก ร—2๐‘ก ๐‘‘๐‘กใ€— =2โˆซ_(โˆš2)^2โ–’ใ€–(๐‘ก^2โˆ’2) ๐‘ก^2 ๐‘‘๐‘กใ€— =2โˆซ_(โˆš2)^2โ–’ใ€–(๐‘ก^4โˆ’2๐‘ก^2 ) ๐‘‘๐‘กใ€— =2[๐‘ก^(4+1)/(4+1)โˆ’2 ๐‘ก^(2+1)/(2+1)]_(โˆš2)^2 =2[๐‘ก^5/5โˆ’2 ๐‘ก^3/3]_(โˆš2)^2 =2ร— [2^5/5โˆ’2/3 2^3โˆ’((โˆš2)^5/5โˆ’2/3 (โˆš2)^3 )] =2ร— [๐‘ฅ^5/2โˆ’2/3 2^3โˆ’((โˆš2)^5/2โˆ’2/3 (โˆš2)^3 )] =2ร— [32/5โˆ’16/3โˆ’(4โˆš2)/5+4/3 โˆš2] =2ร— [(96 โˆ’ 80 โˆ’ 12โˆš2 + 20โˆš2)/15] =2ร— [(16 + 8โˆš2)/15] =(2 ร— 8 โˆš2 (โˆš2 + 1))/15 =(๐Ÿ๐Ÿ”โˆš(๐Ÿ ) (โˆš๐Ÿ + ๐Ÿ))/๐Ÿ๐Ÿ“

Ask a doubt
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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.