Ex 7.2, 11 - Integrate x / root (x+4) - Class 12 NCERT - Ex 7.2

Slide33.JPG
Slide34.JPG Slide35.JPG Slide36.JPG Slide37.JPG

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Ex 7.2, 11 (Method 1) Integrate the function: ๐‘ฅ/โˆš(๐‘ฅ + 4) , ๐‘ฅ>0 ๐‘ฅ/โˆš(๐‘ฅ + 4) , ๐‘ฅ>0 Step 1: Simplify the given function ๐‘ฅ/โˆš(๐‘ฅ + 4) = (๐‘ฅ + 4 โˆ’ 4)/โˆš(๐‘ฅ + 4) = (๐‘ฅ + 4)/โˆš(๐‘ฅ + 4) โˆ’ 4/โˆš(๐‘ฅ + 4) = (๐‘ฅ + 4)^(1/2 โˆ’ 1) โˆ’ 4(๐‘ฅ + 4)^(1/2) = (๐‘ฅ + 4)^(1/2) โˆ’ 4(๐‘ฅ + 4)^(โˆ’ 1/2) Step 2: Integrating the function โˆซ1โ–’ใ€–" " ๐‘ฅ/โˆš(๐‘ฅ + 4)ใ€— . ๐‘‘๐‘ฅ = โˆซ1โ–’((๐‘ฅ + 4)^(1/2) " โˆ’ " ใ€–4 (๐‘ฅ + 4)ใ€—^(โˆ’ 1/2) ) . ๐‘‘๐‘ฅ = โˆซ1โ–’(๐‘ฅ + 4)^(1/2) . ๐‘‘๐‘ฅ โˆ’ 4โˆซ1โ–’(๐‘ฅ + 4)^(โˆ’ 1/2) . ๐‘‘๐‘ฅ = (๐‘ฅ + 4)^(1/2 + 1)/(1/2 + 1) โˆ’ (4 (๐‘ฅ + 4)^(โˆ’ 1/2 + 1))/(โˆ’ 1/2 + 1) + C = (๐‘ฅ + 4)^(3/2)/(3/2 ) โˆ’ (4 (๐‘ฅ + 4)^(1/2))/( 1/2) + C = 2/3 (๐‘ฅ+4)^(3/2) โˆ’ 4.2 (๐‘ฅ+4)^(1/2) + C = ใ€–2(๐‘ฅ+4)ใ€—^(1/2) ((๐‘ฅ + 4)/3 โˆ’4) + ๐ถ = ใ€–2(๐‘ฅ+4)ใ€—^(1/2) ((๐‘ฅ + 4 โˆ’12)/3) + ๐ถ = ใ€–2(๐‘ฅ+4)ใ€—^(1/2) ((๐‘ฅ โˆ’ 8))/3 + ๐ถ = ๐Ÿ/๐Ÿ‘ โˆš(๐’™ + ๐Ÿ’) (๐’™โˆ’๐Ÿ–) + ๐‘ช Ex 7.2, 11 (Method 2) Integrate the function: ๐‘ฅ/โˆš(๐‘ฅ + 4) , ๐‘ฅ>0 Step 1: Let ๐‘ฅ+4=๐‘ก Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ 1+0= ๐‘‘๐‘ก/๐‘‘๐‘ฅ 1= ๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก Step 2: Integrating the function โˆซ1โ–’ใ€–" " ๐‘ฅ/โˆš(๐‘ฅ + 4)ใ€— . ๐‘‘๐‘ฅ Putting ๐‘ก=๐‘ฅ+4 & ๐‘‘๐‘ฅ=๐‘‘๐‘ก = โˆซ1โ–’๐‘ฅ/โˆš๐‘ก . ๐‘‘๐‘ก = โˆซ1โ–’(๐‘ก โˆ’ 4)/โˆš๐‘ก . ๐‘‘๐‘ก = โˆซ1โ–’(๐‘ก โˆ’ 4)/โˆš๐‘ก . ๐‘‘๐‘ก = โˆซ1โ–’(๐‘ก/โˆš๐‘ก โˆ’4/โˆš๐‘ก) ๐‘‘๐‘ก = โˆซ1โ–’๐‘ก/โˆš๐‘ก . ๐‘‘๐‘ก โˆ’ โˆซ1โ–’4/โˆš๐‘ก . ๐‘‘๐‘ก = โˆซ1โ–’๐‘ก^(1 โˆ’ 1/2) . ๐‘‘๐‘ก โˆ’ โˆซ1โ–’ใ€–4 . ๐‘ก^(โˆ’ 1/2) ใ€—. ๐‘‘๐‘ก = โˆซ1โ–’๐‘ก^(1/2) . ๐‘‘๐‘ก โˆ’ โˆซ1โ–’ใ€–4 . ๐‘ก^(โˆ’ 1/2) ใ€—. ๐‘‘๐‘ก = ๐‘ก^(1/2 + 1)/(1/2 + 1) โˆ’ 4 ๐‘ก^(โˆ’ 1/2 + 1)/(โˆ’ 1/2 + 1) +๐ถ = (๐‘ก^(3/2) )/(3/2) โˆ’ 4 ๐‘ก^(1/2)/(1/2) +๐ถ = 2/3 ๐‘ก^(3/2) โˆ’ 4 . 2๐‘ก^(1/2) +๐ถ Taking 2/3 . ๐‘ก^(1/2) as common , we get = 2/3 . ๐‘ก^(1/2) (๐‘กโˆ’4 . 3)+๐ถ = 2/3 . ๐‘ก^(1/2) (๐‘กโˆ’12)+๐ถ Putting the value of ๐‘ก=๐‘ฅ+4 = 2/3 . (๐‘ฅ+4)^(1/2) (๐‘ฅ+4โˆ’12)+๐ถ = ๐Ÿ/๐Ÿ‘ โˆš(๐’™+๐Ÿ’) (๐’™โˆ’๐Ÿ–)+๐‘ช

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.