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Ex 7.9, 22 - Direct Integrate dx / 4 + 9x2 from 0 to 2/3 - Definate Integration - By Formulae

Ex 7.9, 22 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.9, 22 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.9, 22 - Chapter 7 Class 12 Integrals - Part 4


Transcript

Ex7.9, 22 Choose the correct answer โˆซ_0^(2/3)โ–’๐‘‘๐‘ฅ/(4 +9๐‘ฅ^2 ) equals (A) ๐œ‹/6 (B) ๐œ‹/12 (C) ๐œ‹/24 (D) ๐œ‹/4 Step 1 :- Let F(๐‘ฅ)=โˆซ1โ–’๐‘‘๐‘ฅ/(4 + 9๐‘ฅ^2 ) Divide and multiply the integrate by 4 =โˆซ1โ–’โ–ˆ(๐‘‘๐‘ฅ@4)/((4 + 9๐‘ฅ^2)/4) =1/4 โˆซ1โ–’๐‘‘๐‘ฅ/(1 + 9/4 ๐‘ฅ^2 ) =1/4 โˆซ1โ–’๐‘‘๐‘ฅ/(1 + (3/2 ๐‘ฅ)^2 ) Put 3/2 ๐‘ฅ=๐‘ก Differentiating w.r.t.๐‘ฅ ๐‘‘(3/2 ๐‘ฅ)/๐‘‘๐‘ฅ=๐‘‘๐‘ก/๐‘‘๐‘ฅ 3/2=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(3/2) ๐‘‘๐‘ฅ=2/3 ๐‘‘๐‘ก Hence 1/4 โˆซ1โ–’ใ€–๐‘‘๐‘ฅ/(1+(3/2 ๐‘ฅ)^2 )=1/4 โˆซ1โ–’ใ€–1/(1+๐‘ก^2 ) 2/3ใ€— ๐‘‘๐‘กใ€— =1/4 ร— 2/3 โˆซ1โ–’๐‘‘๐‘ก/(1+๐‘ก^2 ) =1/6 tan^(โˆ’1)โก๐‘ก Putting ๐‘ก=3/2 ๐‘ฅ =1/6 tan^(โˆ’1)โกใ€–3/2 ๐‘ฅใ€— Hence F(๐‘ฅ)=1/6 tan^(โˆ’1)โกใ€–3/2 ๐‘ฅใ€— Step 2 :- โˆซ_0^(2/3)โ–’ใ€–๐‘‘๐‘ฅ/(4+9๐‘ฅ^2 )=๐น(2/3)โˆ’๐น(0) ใ€— =1/6 tan^(โˆ’1)โกใ€–(3/2 ร— 2/3)โˆ’1/6 tan^(โˆ’1)โก(0) ใ€— =1/6 tan^(โˆ’1)โกใ€–(1)โˆ’1/6 tan^(โˆ’1)โก(0) ใ€— =1/6 ร— ๐œ‹/4โˆ’0 =๐œ‹/24 โˆต Option (C) is correct.

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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.