Example 25 (Supplementary NCERT) - Integrate x root(1 + x - x2) dx

Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 2
Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 3 Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 4 Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 5 Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 6 Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 7 Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 8 Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 9

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 3 (Supplementary NCERT) โˆซ1โ–’๐‘ฅ โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ โˆซ1โ–’๐‘ฅ โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ We can write it as:- x = A [๐‘‘/๐‘‘๐‘ฅ (1+๐‘ฅโˆ’๐‘ฅ^2 )]+ B x = A [0+1โˆ’2๐‘ฅ]+ B x = A [1โˆ’2๐‘ฅ]+ B x = "A"โˆ’2"A" ๐‘ฅ+ B x = โˆ’2"A" ๐‘ฅ+(๐ด+๐ต) Comparing x and constant term Thus, we can write x = A [1โˆ’2๐‘ฅ] + B x = ((โˆ’1)/2)[1โˆ’2๐‘ฅ] + 1 x = (โˆ’2A) x ๐‘ฅ/๐‘ฅ = โˆ’2A 1 = โˆ’2A A = (โˆ’1)/2 0 = A + B B = โˆ’A B = โˆ’((โˆ’1)/2) B = 1/2 Integrating โˆซ1โ–’ใ€–๐‘ฅโˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ใ€— ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–[(โˆ’1/2)[1โˆ’2๐‘ฅ]+1/2] ใ€— โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–[(โˆ’1/2)[1 โˆ’2๐‘ฅ] โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 )+1/2 โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 )] ใ€— ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–(โˆ’1/2)[1โˆ’2๐‘ฅ] โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ+ใ€— โˆซ1โ–’ใ€–1/2 โˆš(1โˆ’๐‘ฅโˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ = โˆ’1/2 โˆซ1โ–’ใ€–[1โˆ’2๐‘ฅ] โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ+ใ€— 1/2 โˆซ1โ–’โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ Solving ๐‘ฐ_๐Ÿ I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[1โˆ’2๐‘ฅ] โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ Let 1 + ๐‘ฅ โˆ’ ๐‘ฅ^2 = t Diff. both sides w.r.t.x 0 + 1 โˆ’2x = ๐‘‘๐‘ก/๐‘‘๐‘ฅ (1 โˆ’ 2x) dx = dt dx = ๐‘‘๐‘ก/(1 โˆ’ 2๐‘ฅ) Thus, our equation becomes I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[1โˆ’2๐‘ฅ] โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ Putting the value if (1+๐‘ฅโˆ’๐‘ฅ^2) and dx, we get I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[1โˆ’2๐‘ฅ] โˆš๐‘กใ€—. ๐‘‘๐‘ฅ I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[1โˆ’2๐‘ฅ] โˆš๐‘กใ€—. ๐‘‘๐‘ก/[1 โˆ’ 2๐‘ฅ] ๐ผ_1 = (โˆ’1)/2 โˆซ1โ–’โˆš๐‘ก. ๐‘‘๐‘ก I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–(๐‘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€—^(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€–(1โˆ’๐‘ฅ โˆ’๐‘ฅ^2 ) ใ€—^(3/2 )+ C_1 ("Using t = " 1โˆ’๐‘ฅ โˆ’๐‘ฅ^2 ) Solving ๐‘ฐ_๐Ÿ I_2 = 1/2 โˆซ1โ–’โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’โˆš(โˆ’(๐‘ฅ^2โˆ’๐‘ฅโˆ’1)) ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’โˆš(โˆ’[๐‘ฅ^2โˆ’2(๐‘ฅ)(1/2)โˆ’1] ) ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(โˆ’[๐‘ฅ^2โˆ’2(๐‘ฅ)(1/2)+(1/2)^2โˆ’(1/2)^2โˆ’1] ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(โˆ’[(๐‘ฅโˆ’1/2)^2โˆ’(1/2)^2โˆ’1] ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(โˆ’[ใ€– (๐‘ฅโˆ’1/2)ใ€—^2+(โˆ’1 โˆ’4)/4] ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(โˆ’[ใ€– (๐‘ฅโˆ’1/2)ใ€—^2 โˆ’5/4] ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(5/4 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš((โˆš5/4)^2 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 ((๐‘ฅ โˆ’ 1/2)/2 โˆš((โˆš5/4)^2 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 )+(โˆš5/2)^2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((๐‘ฅ + 1/2)/(โˆš5/2))+ C_2 ) I_2 = 1/2 (((2๐‘ฅ โˆ’1)/2)/2 โˆš(5/4โˆ’[๐‘ฅ^2 +1/4โˆ’2๐‘ฅ(1/2)] ) +(5/4)/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) (((2๐‘ฅ โˆ’ 1)/2)/(โˆš5/2))+ 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 โ€“ 1/2) and a by โˆš5/2 , we get I_2 = 1/2 ((2๐‘ฅ โˆ’1)/4 โˆš(5/4โˆ’[๐‘ฅ^2 +1/4 โˆ’๐‘ฅ] ) +5/8 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((2๐‘ฅ โˆ’ 1)/โˆš5)+ C_2 ) I_2 " = " (2๐‘ฅ โˆ’1)/8 โˆš(1+๐‘ฅ+๐‘ฅ^2 ) +5/16 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((2๐‘ฅ โˆ’ 1)/โˆš5)+C_3 Putting the value of I_1 and I_2 in (1) โˆซ1โ–’๐‘ฅ โˆš(1+ ๐‘ฅโˆ’๐‘ฅ^2 ) d๐‘ฅ = (โˆ’1)/2 โˆซ1โ–’ใ€–[1โˆ’2๐‘ฅ] โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ+โˆซ1โ–’โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ = (โˆ’1)/3 ใ€–(1+๐‘ฅโˆ’๐‘ฅ^2)ใ€—^(3/2) + C_1 + ((2๐‘ฅ โˆ’1))/8 โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 )+5/16 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((2๐‘ฅ โˆ’ 1)/โˆš5)+ C_3 = (โˆ’๐Ÿ)/๐Ÿ‘ ใ€–(๐Ÿ+๐’™โˆ’๐’™^๐Ÿ)ใ€—^(๐Ÿ‘/๐Ÿ) +๐Ÿ/๐Ÿ– (๐Ÿ๐’™ โˆ’๐Ÿ) โˆš(๐Ÿ+๐’™โˆ’๐’™^๐Ÿ )+ ๐Ÿ“/๐Ÿ๐Ÿ” ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ) ((๐Ÿ๐’™ โˆ’ ๐Ÿ)/โˆš๐Ÿ“)+ ๐‘ช

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Davneet Singh

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.