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

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  1. Chapter 7 Class 12 Integrals
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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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