Ex 7.7, 14 (Supplementary NCERT) - Integrate (x + 3) root(3 - 4x - x2)

Ex 7.7, 14 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 2
Ex 7.7, 14 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 3 Ex 7.7, 14 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 4 Ex 7.7, 14 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 5 Ex 7.7, 14 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 6 Ex 7.7, 14 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 7 Ex 7.7, 14 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 8

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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 + 3 = โˆ’2"A" ๐‘ฅ+(โˆ’4"A"+๐ต) Comparing x and constant term Thus, we can write x + 3 = A [โˆ’4โˆ’2๐‘ฅ] + B x + 3 = (โˆ’1)/2 [โˆ’4โˆ’2๐‘ฅ] + 1 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 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] ("Using t = " 3โˆ’4๐‘ฅ โˆ’๐‘ฅ^2 ) 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)+ ๐ถ_2 Replacing x by (x + 2) a by โˆš7 , we get Putting the value of I_1 and I_2 in (1) 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(3 โˆ’ 4๐‘ฅ โˆ’ 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‚

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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.