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Ex 7.2, 12 - Integrate (x3 - 1)1/3 .x5 - Chapter 7 CBSE - Integration by substitution - x^n

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.2, 12 Integrate the function: (๐‘ฅ3 โ€“ 1)^(1/3) . ๐‘ฅ5 (๐‘ฅ3 โ€“ 1)^(1/3) . ๐‘ฅ5 Step 1: Let ๐‘ฅ3= ๐‘ก Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ 3๐‘ฅ^2= ๐‘‘๐‘ก/๐‘‘๐‘ฅ 3๐‘ฅ^2. d๐‘ฅ=๐‘‘๐‘ก ๐‘‘๐‘ฅ = ๐‘‘๐‘ก/(3๐‘ฅ^2 ) Step 2: Integrating the function โˆซ1โ–’ใ€–" " (๐‘ฅ3 โ€“ 1)^(1/3) . ๐‘ฅ5" " ใ€— . ๐‘‘๐‘ฅ Putting the value of ๐‘ฅ^3 & ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(3๐‘ฅ^2 ) = โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) . ๐‘ฅ5ใ€— . ๐‘‘๐‘ก/(3๐‘ฅ^2 ) = โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) . ๐‘ฅ^2. ๐‘ฅ^3 ใ€—. ๐‘‘๐‘ก/(3๐‘ฅ^2 ) = โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) ใ€— . ๐‘ฅ^3/3 . ๐‘‘๐‘ก = โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) ใ€— . ๐‘ก/3 . ๐‘‘๐‘ก = 1/3 โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) ใ€— . ๐‘ก . ๐‘‘๐‘ก = 1/3 โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) ใ€— . (๐‘กโˆ’1+1) ๐‘‘๐‘ก = 1/3 โˆซ1โ–’ใ€–" " (๐‘ก โ€“ 1)^(1/3) ใ€— . ((๐‘กโˆ’1)+1) ๐‘‘๐‘ก = 1/3 โˆซ1โ–’((๐‘ก โ€“ 1)^(1/3) (๐‘กโˆ’1)+(๐‘กโˆ’1)^(1/3) ) ๐‘‘๐‘ก = 1/3 โˆซ1โ–’((๐‘ก โ€“ 1)^(1/3 +1)+(๐‘กโˆ’1)^(1/3) ) ๐‘‘๐‘ก = 1/3 โˆซ1โ–’((๐‘ก โ€“ 1)^(4/3 )+(๐‘กโˆ’1)^(1/3) ) ๐‘‘๐‘ก = 1/3 โˆซ1โ–’ใ€– (๐‘ก โ€“ 1)^(4/3 ). ๐‘‘๐‘กใ€— + 1/3 โˆซ1โ–’ใ€– (๐‘ก โ€“ 1)^(1/3 ). ๐‘‘๐‘กใ€— = 1/3 โˆซ1โ–’ใ€– (๐‘ก โ€“1)^(4/3 ). ๐‘‘๐‘กใ€— + 1/3 โˆซ1โ–’ใ€– (๐‘ก โ€“1)^(1/3 ). ๐‘‘๐‘กใ€— = 1/3 (๐‘ก โ€“1)^(4/3 + 1)/(4/3 + 1) + 1/3 (๐‘ก โ€“1)^(1/3 + 1)/(1/3 + 1) + ๐ถ = 1/3 (๐‘ก โˆ’ 1)^(7/3)/(7/3) + 1/3 (๐‘ก โˆ’ 1)^(4/3)/(4/3) + ๐ถ

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