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


Transcript

Ex 7.8, 19 โˆซ_0^2โ–’(6๐‘ฅ + 3)/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅ Let F(๐‘ฅ)=โˆซ1โ–’ใ€–(6๐‘ฅ + 3)/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–6๐‘ฅ/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅ+โˆซ1โ–’ใ€–3/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅ ใ€—ใ€— Solving ๐‘ฐ๐Ÿ ๐ผ1=โˆซ1โ–’ใ€–6๐‘ฅ/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅใ€— Put ๐‘ฅ^2 + 4=๐‘ก Differentiating w.r.t.๐‘ฅ ๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ^2+4)=๐‘‘๐‘ก/๐‘‘๐‘ฅ 2๐‘ฅ+0=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/2๐‘ฅ Therefore, โˆซ1โ–’ใ€–6๐‘ฅ/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅ=โˆซ1โ–’ใ€–6๐‘ฅ/๐‘ก ๐‘‘๐‘ก/2๐‘ฅใ€—ใ€— =โˆซ1โ–’ใ€–3/๐‘ก ๐‘‘๐‘กใ€— =3 ๐‘™๐‘œ๐‘”|๐‘ก| =3 ๐‘™๐‘œ๐‘”|๐‘ฅ^2+4| Solving ๐‘ฐ๐Ÿ ๐ผ2=โˆซ1โ–’ใ€–3/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅใ€— =3โˆซ1โ–’1/(๐‘ฅ^2+4) ๐‘‘๐‘ฅ =3โˆซ1โ–’1/(๐‘ฅ^2 + 2^2 ) ๐‘‘๐‘ฅ =3 ร— 1/2 tan^(โˆ’1)โกใ€–๐‘ฅ/2ใ€— =3/2 tan^(โˆ’1)โกใ€–๐‘ฅ/2ใ€— Therefore F(๐‘ฅ)= ๐ผ1+๐ผ2 F(๐‘ฅ)=3๐‘™๐‘œ๐‘”|๐‘ฅ^2+4|+3/2 tan^(โˆ’1)โกใ€–๐‘ฅ/2ใ€— Now, โˆซ_0^2โ–’ใ€–(6๐‘ฅ + 3)/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅ=๐น(2)โˆ’๐น(0) ใ€— =3๐‘™๐‘œ๐‘”|2^2+4|+3/2 tan^(โˆ’1)โกใ€–2/2โˆ’3๐‘™๐‘œ๐‘”|0+4|โˆ’3/2 tan^(โˆ’1)โก(0/2) ใ€— =3๐‘™๐‘œ๐‘”|4+4|+3/2 tan^(โˆ’1)โกใ€–1โˆ’3๐‘™๐‘œ๐‘”|4|โˆ’3/2 ร— 0ใ€— =3๐‘™๐‘œ๐‘”|8|โˆ’3๐‘™๐‘œ๐‘”|4|+3/2 ๐œ‹/4 =3(๐‘™๐‘œ๐‘”|8|โˆ’๐‘™๐‘œ๐‘”|4|)+3๐œ‹/8 =3๐‘™๐‘œ๐‘”|8/4|+3๐œ‹/8 =๐Ÿ‘ ๐ฅ๐จ๐ โกใ€–๐Ÿ+๐Ÿ‘๐…/๐Ÿ–ใ€—

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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.