Slide14.JPG

Slide15.JPG
Slide16.JPG Slide17.JPG Slide18.JPG Slide19.JPG Slide20.JPG Slide21.JPG

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

Transcript

Ex 7.7, 14 (Supplementary NCERT) (๐‘ฅ+3) โˆš(3โˆ’4๐‘ฅใ€–โˆ’๐‘ฅใ€—^2 ) (๐‘ฅ+3) โˆš(3โˆ’4๐‘ฅใ€–โˆ’๐‘ฅใ€—^2 ) We can write it as:- x + 3 = A [๐‘‘/๐‘‘๐‘ฅ (3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 )]+ B x + 3 = A [0โˆ’4โˆ’2๐‘ฅ]+ B x + 3 = A [โˆ’4โˆ’2๐‘ฅ]+ B x + 3 = โˆ’4"A"โˆ’2"A" ๐‘ฅ+ B x = (โˆ’2A) x ๐‘ฅ/๐‘ฅ = โˆ’2A 1 = โˆ’2A A = (โˆ’1)/2 3 = โˆ’4A + B 3 = โˆ’4((โˆ’1)/2) + B 3 = 2 + B B = 3 โˆ’ 2 B = 1 Thus, we can write x + 3 = A [โˆ’4โˆ’2๐‘ฅ] + B x + 3 = (โˆ’1)/2 [โˆ’4โˆ’2๐‘ฅ] + 1 Integrating the function w.r.t.x โˆซ1โ–’ใ€–(๐‘ฅ+3) โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ใ€— ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–[โˆ’1/2 [โˆ’4 โˆ’2๐‘ฅ]+1] ใ€— โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–[โˆ’1/2 [โˆ’4 โˆ’2๐‘ฅ] โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 )+1โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 )] ใ€— ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–โˆ’1/2 [โˆ’4 โˆ’2๐‘ฅ] โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ+ใ€— โˆซ1โ–’โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ = โˆ’1/2 โˆซ1โ–’ใ€–[โˆ’4 โˆ’2๐‘ฅ] โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ+ใ€— โˆซ1โ–’โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ Solving ๐‘ฐ_๐Ÿ I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[โˆ’4โˆ’2๐‘ฅ] โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ Let 3 โˆ’ 4๐‘ฅ โˆ’ ๐‘ฅ^2 = t Diff. both sides w.r.t.x 0 โˆ’ 4 โˆ’2x = ๐‘‘๐‘ก/๐‘‘๐‘ฅ โˆ’ 4 โˆ’ 2x = ๐‘‘๐‘ก/๐‘‘๐‘ฅ dx = ๐‘‘๐‘ก/(โˆ’4 โˆ’ 2๐‘ฅ) Thus, our equation becomes I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[โˆ’4โˆ’2๐‘ฅ] โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ Putting the value if (3โˆ’4๐‘ฅโˆ’๐‘ฅ^2) and dx, we get I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[โˆ’4โˆ’2๐‘ฅ] โˆš๐‘กใ€—. ๐‘‘๐‘ฅ I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[โˆ’4โˆ’2๐‘ฅ] โˆš๐‘กใ€—. ๐‘‘๐‘ก/[โˆ’4โˆ’2๐‘ฅ] Integrating the function w.r.t.x โˆซ1โ–’ใ€–(๐‘ฅ+3) โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ใ€— ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–[โˆ’1/2 [โˆ’4 โˆ’2๐‘ฅ]+1] ใ€— โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–[โˆ’1/2 [โˆ’4 โˆ’2๐‘ฅ] โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 )+1โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 )] ใ€— ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–โˆ’1/2 [โˆ’4 โˆ’2๐‘ฅ] โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ+ใ€— โˆซ1โ–’โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ = โˆ’1/2 โˆซ1โ–’ใ€–[โˆ’4 โˆ’2๐‘ฅ] โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ+ใ€— โˆซ1โ–’โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ Solving ๐‘ฐ_๐Ÿ I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[โˆ’4โˆ’2๐‘ฅ] โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ Let 3 โˆ’ 4๐‘ฅ โˆ’ ๐‘ฅ^2 = t Diff. both sides w.r.t.x 0 โˆ’ 4 โˆ’2x = ๐‘‘๐‘ก/๐‘‘๐‘ฅ โˆ’ 4 โˆ’ 2x = ๐‘‘๐‘ก/๐‘‘๐‘ฅ dx = ๐‘‘๐‘ก/(โˆ’4 โˆ’ 2๐‘ฅ) Thus, our equation becomes I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[โˆ’4โˆ’2๐‘ฅ] โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ Putting the value if (3โˆ’4๐‘ฅโˆ’๐‘ฅ^2) and dx, we get I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[โˆ’4โˆ’2๐‘ฅ] โˆš๐‘กใ€—. ๐‘‘๐‘ฅ I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[โˆ’4โˆ’2๐‘ฅ] โˆš๐‘กใ€—. ๐‘‘๐‘ก/[โˆ’4โˆ’2๐‘ฅ] ๐ผ_1 = (โˆ’1)/2 โˆซ1โ–’โˆš๐‘ก. ๐‘‘๐‘ก I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–โˆš๐‘ก ใ€—^(1/2) ๐‘‘๐‘ก I_1 = (โˆ’1)/2 ใ€–๐‘ก ใ€—^(1/2 + 1)/((1/2 + 1) )+ C_1 I_1 = (โˆ’1)/2 ใ€–๐‘ก ใ€—^((1 + 2)/2 )/(((1 + 2)/2) )+ C_1 I_1 = (โˆ’1)/2 ใ€–๐‘ก ใ€—^(3/2 )/((3/2) )+ C_1 I_1 = (โˆ’1)/3 ใ€–๐‘ก ใ€—^(3/2 )+ C_1 I_1 = (โˆ’1)/3 ใ€–(3โˆ’4๐‘ฅ โˆ’๐‘ฅ^2 ) ใ€—^(3/2 )+ C_1 Solving ๐‘ฐ_๐Ÿ I_2 = โˆซ1โ–’โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ I_2 = โˆซ1โ–’โˆš(3โˆ’(4๐‘ฅ+๐‘ฅ^2)) ๐‘‘๐‘ฅ I_2 = โˆซ1โ–’โˆš(3โˆ’(๐‘ฅ^2+4๐‘ฅ)) ๐‘‘๐‘ฅ I_2 = โˆซ1โ–’ใ€–โˆš(3โˆ’[๐‘ฅ^2+2(2) (๐‘ฅ))] ใ€— ๐‘‘๐‘ฅ I_2 = โˆซ1โ–’ใ€–โˆš(3โˆ’[๐‘ฅ^2+2(2)+(2)^2 โˆ’ใ€–(2)ใ€—^2 )] ใ€— ๐‘‘๐‘ฅ I_2 = โˆซ1โ–’ใ€–โˆš(3โˆ’[(ใ€–๐‘ฅ+2)ใ€—^2โˆ’ ใ€–(2)ใ€—^2 )] ใ€— ๐‘‘๐‘ฅ I_2 = โˆซ1โ–’ใ€–โˆš(3โˆ’(๐‘ฅ+2)^2+ ใ€–(2)ใ€—^2 ) ใ€— ๐‘‘๐‘ฅ I_2 = โˆซ1โ–’ใ€–โˆš(3โˆ’(๐‘ฅ+2)^2+4) ใ€— ๐‘‘๐‘ฅ I_2 = โˆซ1โ–’ใ€–โˆš(7โˆ’(๐‘ฅ+2)^2 ) ใ€— ๐‘‘๐‘ฅ I_2 = โˆซ1โ–’ใ€–โˆš(ใ€–(โˆš7)ใ€—^2โˆ’(๐‘ฅ+2)^2 ) ใ€— ๐‘‘๐‘ฅ I_2 = (๐‘ฅ + 2)/2 โˆš((โˆš7)^2โˆ’(๐‘ฅ+2)^2 )+ใ€–(7)ใ€—^2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((๐‘ฅ + 2)/โˆš7)+ C_2 I_2 = (๐‘ฅ + 2)/2 โˆš(7โˆ’ใ€–(๐‘ฅใ€—^2 + 4๐‘ฅ +4)) +7/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((๐‘ฅ + 2)/โˆš7)+ C_2 I_2 = (๐‘ฅ + 2)/2 โˆš(7โˆ’๐‘ฅ^2 + 4๐‘ฅ +4) +7/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((๐‘ฅ + 2)/โˆš7)+ C_2 I_2 = (๐‘ฅ + 2)/2 โˆš(3โˆ’ 4๐‘ฅโˆ’๐‘ฅ^2 ) +7/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((๐‘ฅ + 2)/โˆš7)+ C_2 It is of form โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ=1/2 ๐‘ฅโˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )+๐‘Ž^2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) (๐‘ฅ/๐‘Ž)+ C_2 Replacing x by (x + 2) a by โˆš7 , we get Putting the value of I_1 and I_2 in eq. (1) , we get โˆซ1โ–’(๐‘ฅ+3) โˆš(3โˆ’ 4๐‘ฅโˆ’๐‘ฅ^2 ) d๐‘ฅ = (โˆ’1)/2 โˆซ1โ–’ใ€–[โˆ’4โˆ’2๐‘ฅ] โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ+โˆซ1โ–’โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ = (โˆ’1)/3 ใ€–(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2)ใ€—^(3/2) + C_1+ ((๐‘ฅ +2) โˆš(3โˆ’4๐‘ฅโˆ’๐‘ฅ^2 ))/2+7/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((๐‘ฅ + 2)/โˆš7)+ C_3 = (โˆ’๐Ÿ)/๐Ÿ‘ ใ€–(๐Ÿ‘โˆ’๐Ÿ’๐’™โˆ’๐’™^๐Ÿ)ใ€—^(๐Ÿ‘/๐Ÿ) +๐Ÿ•/๐Ÿ ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ) ((๐’™ + ๐Ÿ)/โˆš๐Ÿ•)+ ((๐’™ +๐Ÿ) โˆš(๐Ÿ‘โˆ’๐Ÿ’๐’™โˆ’๐’™^๐Ÿ ))/๐Ÿ+ ๐‚

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.