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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.7, 13 (Supplementary NCERT) ๐‘ฅ+1 โˆš(2๐‘ฅ^2+3) ๐‘ฅ+1 โˆš( 2๐‘ฅ^2+3) Integrating the function w.r.t.x โˆซ1โ–’ใ€–(๐‘ฅ+1) โˆš(ใ€–2๐‘ฅใ€—^2+3)ใ€— ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–[๐‘ฅโˆš(ใ€–2๐‘ฅใ€—^2+3)+1โˆš(ใ€–2๐‘ฅใ€—^2+3)] ๐‘‘๐‘ฅใ€— = โˆซ1โ–’ใ€–๐‘ฅ โˆš(ใ€–2๐‘ฅใ€—^2+3) ๐‘‘๐‘ฅ+โˆซ1โ–’ใ€–โˆš(ใ€–2๐‘ฅใ€—^2+3 ) ๐‘‘๐‘ฅใ€—ใ€— Solving ๐ˆ_๐Ÿ ๐ˆ_๐Ÿ=โˆซ1โ–’ใ€–๐‘ฅ โˆš(ใ€–2๐‘ฅใ€—^2+3)ใ€— ๐‘‘๐‘ฅ Let ใ€–2๐‘ฅใ€—^2+3=๐‘ก Diff. both sides w.r.t.x 4x + 0 = ๐‘‘๐‘ก/๐‘‘๐‘ฅ dx = ๐‘‘๐‘ก/4๐‘ฅ Now, equation becomes I_1 = โˆซ1โ–’ใ€–๐‘ฅ โˆš(ใ€–2๐‘ฅใ€—^2+3)ใ€— ๐‘‘๐‘ฅ Putting the value of (ใ€–2๐‘ฅใ€—^2+3) and dx I_1 = โˆซ1โ–’ใ€–๐‘ฅ โˆš๐‘กใ€— . ๐‘‘๐‘ฅ I_1 = โˆซ1โ–’ใ€–๐‘ฅ โˆš๐‘กใ€— ๐‘‘๐‘ก/4๐‘ฅ I_1 = 1/4 โˆซ1โ–’ใ€– โˆš๐‘ก ใ€—. ๐‘‘๐‘ก I_1 = 1/4 โˆซ1โ–’ใ€– ๐‘ก^(1/2) ใ€— . ๐‘‘๐‘ก I_1 = 1/4 [ใ€–๐‘ก ใ€—^(1/2 + 1)/((1/2 + 1) )]+C_1 I_1 = 1/4 [ใ€–๐‘ก ใ€—^(3/2 )/(3/2)]+C_1 I_1 = ใ€–๐‘ก ใ€—^(3/2 )/6+C_1 I_1 = (ใ€–2๐‘ฅใ€—^2+3)^(3/2)/6+C_1 Solving ๐‘ฐ_๐Ÿ I_2 = โˆซ1โ–’โˆš(ใ€–2๐‘ฅใ€—^2+3) ๐‘‘๐‘ฅ I_2 = โˆซ1โ–’โˆš(2 (ใ€– ๐‘ฅใ€—^2+3/2) ) ๐‘‘๐‘ฅ I_2 = โˆซ1โ–’ใ€–โˆš2 โˆš( ใ€– ๐‘ฅใ€—^2+3/2)ใ€— ๐‘‘๐‘ฅ I_2 =โˆš2 โˆซ1โ–’โˆš(ใ€– ๐‘ฅใ€—^2+3/2) ๐‘‘๐‘ฅ I_2 =โˆš2 โˆซ1โ–’โˆš(ใ€– ๐‘ฅใ€—^2+(โˆš(3/2))^2 ) ๐‘‘๐‘ฅ It is of form โˆซ1โ–’ใ€–โˆš(๐‘ฅ^2+๐‘Ž^2 ) ใ€— ๐‘‘๐‘ฅ=1/2 ๐‘ฅ โˆš(๐‘ฅ^2+๐‘Ž^2 )+๐‘Ž^2/2 logโก|๐‘ฅ+โˆš(๐‘ฅ^2+๐‘Ž^2 )|+C_2 Replacing a by โˆš(3/2), we get I_2 =โˆš2 [๐‘ฅ/2 โˆš(ใ€– ๐‘ฅใ€—^2+(โˆš(3/2))^2 )]+โˆš2 ใ€–((โˆš(3/2 )))/2ใ€—^2 logโก|๐‘ฅ+โˆš(ใ€– ๐‘ฅใ€—^2+(โˆš(3/2))^2 )|+C_2 I_2 =โˆš2 [๐‘ฅ/2 โˆš(ใ€– ๐‘ฅใ€—^2+3/2)]+โˆš2 (3/2)/2 logโก|๐‘ฅ+โˆš(ใ€– ๐‘ฅใ€—^2+3/2)|+C_2 I_2 =(โˆš2 ๐‘ฅ)/2.โˆš(ใ€– 2๐‘ฅใ€—^2+3)/โˆš2+(3โˆš2)/4 logโก|๐‘ฅ+โˆš(ใ€– ๐‘ฅใ€—^2+3/2)|+ใ€–โˆš2 Cใ€—_2 I_2 =๐‘ฅ/2.โˆš(ใ€– 2๐‘ฅใ€—^2+3)+(3โˆš2)/4 logโก|๐‘ฅ+โˆš(ใ€– ๐‘ฅใ€—^2+3/2)|+ใ€– Cใ€—_3 Now, put the value of I_1 and I_2 in eq. (1), we get โˆซ1โ–’ใ€–(๐‘ฅ+1) โˆš(ใ€–2๐‘ฅใ€—^2+3)ใ€— ๐‘‘๐‘ฅ=โˆซ1โ–’๐‘ฅ โˆš(ใ€–2๐‘ฅใ€—^2+3) ๐‘‘๐‘ฅ+โˆซ1โ–’โˆš(ใ€–2๐‘ฅใ€—^2+3) ๐‘‘๐‘ฅ = (ใ€–2๐‘ฅใ€—^2+3)^(3/2)/6+ใ€– Cใ€—_1+๐‘ฅ/2 โˆš(ใ€–2๐‘ฅใ€—^2+3)+(3โˆš2)/4 logโก|๐‘ฅ+โˆš(ใ€– ๐‘ฅใ€—^2+3/2)|+ใ€– Cใ€—_3 = (ใ€–๐Ÿ๐’™ใ€—^๐Ÿ+๐Ÿ‘)^(๐Ÿ‘/๐Ÿ)/๐Ÿ”+๐’™/๐Ÿ โˆš(ใ€–๐Ÿ๐’™ใ€—^๐Ÿ+๐Ÿ‘)+(๐Ÿ‘โˆš๐Ÿ)/๐Ÿ’ ๐’๐’๐’ˆโก|๐’™+โˆš(ใ€– ๐’™ใ€—^๐Ÿ+๐Ÿ‘/๐Ÿ)|+๐‚

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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.