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Chapter 7 Class 12 Integrals
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Example 26 (Method 1) Evaluate โˆซ_(โˆ’1)^1โ–’ใ€–5๐‘ฅ^4 โˆš(๐‘ฅ^5+1)ใ€— ๐‘‘๐‘ฅ Step 1 :- Let F(๐‘ฅ)=โˆซ1โ–’ใ€–5๐‘ฅ^4 โˆš(๐‘ฅ^5+1)ใ€— ๐‘‘๐‘ฅ Putting ๐‘ก=๐‘ฅ^5+1 Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ก/๐‘‘๐‘ฅ=5๐‘ฅ^4 ๐‘‘๐‘ก/(5๐‘ฅ^4 )=๐‘‘๐‘ฅ Therefore we can write โˆซ1โ–’ใ€–5๐‘ฅ^4 โˆš(๐‘ฅ^5+1) ๐‘‘๐‘ฅ=โˆซ1โ–’ใ€–5๐‘ฅ^4 โˆš๐‘ก . ๐‘‘๐‘ก/(5๐‘ฅ^4 )ใ€—ใ€— =โˆซ1โ–’โˆš๐‘ก ๐‘‘๐‘ก =โˆซ1โ–’ใ€–๐‘ก^(1/2) ๐‘‘๐‘กใ€— =ใ€–๐‘ก ใ€—^(1/2 +1)/(1/2 +1) =2/3 ๐‘ก^(3/2) Putting back ๐‘ก=๐‘ฅ^5+1 =2/3 (๐‘ฅ^5+1)^(3/2) Hence , F(๐‘ฅ)=2/3 (๐‘ฅ^5+1)^(3/2) Step 2 :- โˆซ_(โˆ’1)^1โ–’ใ€–5๐‘ฅ^4 ใ€— โˆš(๐‘ฅ^5+1) ๐‘‘๐‘ฅ=๐น(1)โˆ’๐น(โˆ’1) =2/3 (1^5+1)^(3/2)โˆ’2/3 ((โˆ’1)^5+1)^(3/2) =2/3 (1+1)^(3/2)โˆ’2/3 (โˆ’1+1)^(3/2) =2/3 (2)^(3/2)โˆ’0 =2/3 2โˆš2 =(๐Ÿ’โˆš๐Ÿ)/๐Ÿ‘ Example 26 (Method 2) Evaluate โˆซ_(โˆ’1)^1โ–’ใ€–5๐‘ฅ^4 โˆš(๐‘ฅ^5+1)ใ€— ๐‘‘๐‘ฅ Put ๐‘ก=๐‘ฅ^5+1 Differentiating w.r.t. ๐‘ฅ ๐‘‘๐‘ก/๐‘‘๐‘ฅ=๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ^5+1) ๐‘‘๐‘ก/๐‘‘๐‘ฅ=5๐‘ฅ^4 ๐‘‘๐‘ก/(5๐‘ฅ^4 )=๐‘‘๐‘ฅ Hence when ๐‘ฅ varies from ๐‘ฅ=โˆ’1 to 1, ๐‘ก varies from 0 to 2 Therefore, โˆซ_(โˆ’1)^1โ–’ใ€–5๐‘ฅ^4 โˆš(1+๐‘ฅ^5 ) ๐‘‘๐‘ฅ=โˆซ_0^2โ–’ใ€–5๐‘ฅ^4 โˆš๐‘ก ๐‘‘๐‘ก/(5๐‘ฅ^4 )ใ€—ใ€— =โˆซ1_0^2โ–’ใ€–โˆš๐‘ก ๐‘‘๐‘กใ€— =[๐‘ก^(1/2 + 1)/(1/2 +1)]_0^2 =[๐‘ก^(3/2)/(3/2)]_0^2 =[2/3 ๐‘ก^(3/2) ]_0^2 =2/3 (2^(3/2)โˆ’0^(3/2) ) =2/3 2^(3/2) =2/3 ร—2โˆš2 =๐Ÿ’/๐Ÿ‘ โˆš๐Ÿ

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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.