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Ex 7.8, 6 -  Integrate (x + e2x) dx from 0 to 4 by limit as a sum

Ex 7.8, 6 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.8, 6 - Chapter 7 Class 12 Integrals - Part 3
Ex 7.8, 6 - Chapter 7 Class 12 Integrals - Part 4
Ex 7.8, 6 - Chapter 7 Class 12 Integrals - Part 5
Ex 7.8, 6 - Chapter 7 Class 12 Integrals - Part 6
Ex 7.8, 6 - Chapter 7 Class 12 Integrals - Part 7
Ex 7.8, 6 - Chapter 7 Class 12 Integrals - Part 8
Ex 7.8, 6 - Chapter 7 Class 12 Integrals - Part 9
Ex 7.8, 6 - Chapter 7 Class 12 Integrals - Part 10
Ex 7.8, 6 - Chapter 7 Class 12 Integrals - Part 11
Ex 7.8, 6 - Chapter 7 Class 12 Integrals - Part 12
Ex 7.8, 6 - Chapter 7 Class 12 Integrals - Part 13

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Transcript

Ex 7.8, 6 ∫1_0^4β–’(π‘₯+𝑒2π‘₯)𝑑π‘₯ Let I = ∫1_0^4β–’(π‘₯+𝑒2π‘₯)𝑑π‘₯ I = ∫1_0^4β–’γ€–π‘₯ . 𝑑π‘₯γ€—+∫1_0^4β–’γ€– 𝑒2π‘₯ . 𝑑π‘₯γ€— Solving I1 and I2 separately Solving I1 ∫1_0^4β–’γ€–π‘₯ 𝑑π‘₯γ€— Putting π‘Ž =0 𝑏 =4 β„Ž=(𝑏 βˆ’ π‘Ž)/𝑛 =(4 βˆ’ 0)/𝑛 =4/𝑛 𝑓(π‘₯)=π‘₯ We know that ∫1_π‘Ž^𝑏▒〖π‘₯ 𝑑π‘₯γ€— =(π‘βˆ’π‘Ž) (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(π‘Ž)+𝑓(π‘Ž+β„Ž)+𝑓(π‘Ž+2β„Ž)…+𝑓(π‘Ž+(π‘›βˆ’1)β„Ž)) Hence we can write ∫1_0^4β–’γ€–π‘₯ 𝑑π‘₯γ€— =(4βˆ’0) lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(0+β„Ž)+𝑓(0+2β„Ž)+… +𝑓(0+(π‘›βˆ’1)β„Ž) =4 lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(β„Ž)+𝑓(2β„Ž)+… +𝑓((π‘›βˆ’1)β„Ž) Here, 𝑓(π‘₯)=π‘₯ 𝑓(0)=0 𝑓(β„Ž)=β„Ž 𝑓 (2β„Ž)=2β„Ž 𝑓((π‘›βˆ’1)β„Ž)=(π‘›βˆ’1)β„Ž Hence, our equation becomes ∴ ∫_0^4β–’π‘₯ 𝑑π‘₯ =4 lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(β„Ž)+𝑓(2β„Ž)+… +𝑓((π‘›βˆ’1)β„Ž) = 4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (0+β„Ž+2β„Ž+ ……+(π‘›βˆ’1)β„Ž) = 4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 ( β„Ž (1+2+ ………+(π‘›βˆ’1))) We know that 1+2+3+ ……+𝑛= (𝑛 (𝑛 + 1))/2 1+2+3+ ……+π‘›βˆ’1= ((𝑛 βˆ’ 1) (𝑛 βˆ’ 1 + 1))/2 = (𝑛 (𝑛 βˆ’ 1) )/2 = 4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 ( (β„Ž . 𝑛(𝑛 βˆ’ 1))/2) = 4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) ( 𝑛(𝑛 βˆ’ 1)β„Ž/2𝑛) = 4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) ( (𝑛 βˆ’ 1)β„Ž/2) = 4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) ( (𝑛 βˆ’ 1)4/(2 . 𝑛)) = 4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) ( 2(𝑛/𝑛 βˆ’ 1/𝑛)) = 4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) ( 2(1βˆ’ 1/𝑛)) = 4( 2(1βˆ’ 1/∞)) [π‘ˆπ‘ π‘–π‘›π‘” β„Ž=4/𝑛] = 4( 2(1βˆ’0)) = 4Γ—2 = πŸ– Solving I2 ∫_0^4▒𝑒^2π‘₯ 𝑑π‘₯ Putting π‘Ž = 0 𝑏 =4 β„Ž = (𝑏 βˆ’ π‘Ž)/𝑛 = (4 βˆ’ 0)/𝑛 = 4/𝑛 𝑓(π‘₯)=𝑒^2π‘₯ We know that ∫1_π‘Ž^𝑏▒〖π‘₯ 𝑑π‘₯γ€— =(π‘βˆ’π‘Ž) (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(π‘Ž)+𝑓(π‘Ž+β„Ž)+𝑓(π‘Ž+2β„Ž)…+𝑓(π‘Ž+(π‘›βˆ’1)β„Ž)) Hence we can write ∫_0^4▒𝑒^2π‘₯ 𝑑π‘₯ =(4βˆ’0) lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(0+β„Ž)+𝑓(0+2β„Ž)+… +𝑓(0+(π‘›βˆ’1)β„Ž) =4 lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(β„Ž)+𝑓(2β„Ž)……+𝑓((π‘›βˆ’1)β„Ž) Here, 𝑓(π‘₯)=𝑒^2π‘₯ 𝑓(0)=𝑒^(2(0))=1 𝑓(β„Ž)=𝑒^2β„Ž 𝑓(2β„Ž)=𝑒^(2(2β„Ž))=𝑒^4β„Ž 𝑓((π‘›βˆ’1)β„Ž)=𝑒^2(π‘›βˆ’1)β„Ž Hence we can write ∫_0^4▒𝑒^2π‘₯ 𝑑π‘₯ =(4βˆ’0) lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(0+β„Ž)+𝑓(0+2β„Ž)+… +𝑓(0+(π‘›βˆ’1)β„Ž) =4 lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(β„Ž)+𝑓(2β„Ž)……+𝑓((π‘›βˆ’1)β„Ž) Here, 𝑓(π‘₯)=𝑒^2π‘₯ 𝑓(0)=𝑒^(2(0))=1 𝑓(β„Ž)=𝑒^2β„Ž 𝑓(2β„Ž)=𝑒^(2(2β„Ž))=𝑒^4β„Ž 𝑓((π‘›βˆ’1)β„Ž)=𝑒^2(π‘›βˆ’1)β„Ž Hence, our equation becomes ∴ ∫_0^4▒𝑒^2π‘₯ 𝑑π‘₯ =4 lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(β„Ž)+𝑓(2β„Ž)……+𝑓(π‘›βˆ’1)β„Ž) = 4 .lim┬(nβ†’βˆž) 1/𝑛 (1+𝑒^2β„Ž+𝑒^4β„Ž+ ……+𝑒^(2(𝑛 βˆ’ 1) β„Ž) ) Let S = 1+𝑒^2β„Ž+𝑒^4β„Ž+ ……+𝑒^(2(𝑛 βˆ’ 1) β„Ž) It is a G.P. with common ratio (r) r = 𝑒^2β„Ž/1 = 𝑒^2β„Ž We know Sum of G.P = a((π‘Ÿ^𝑛 βˆ’ 1)/(π‘Ÿ βˆ’ 1)) Replacing a by 1 and r by 𝑒^2β„Ž , we get S = 1(((𝑒^2β„Ž )^𝑛 βˆ’ 1)/(𝑒^2β„Ž βˆ’ 1))= (𝑒^2π‘›β„Ž βˆ’ 1)/(𝑒^2β„Ž βˆ’ 1) Thus ∴ ∫_0^4▒𝑒^π‘₯ 𝑑π‘₯ = 4 lim┬(nβ†’βˆž) 1/𝑛 (1+𝑒^2β„Ž+𝑒^4β„Ž+ ……+𝑒^(2(𝑛 βˆ’ 1) β„Ž) ) Putting the value of S, we get = 4 .lim┬(nβ†’βˆž) 1/𝑛 ((𝑒^2π‘›β„Ž βˆ’ 1)/(𝑒^2β„Ž βˆ’ 1)) = 4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 ((𝑒^2π‘›β„Ž βˆ’ 1)/(2β„Ž . (𝑒^2β„Ž βˆ’ 1)/2β„Ž)) = 4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^2π‘›β„Ž βˆ’ 1)/2π‘›β„Ž . 1/( (𝑒^2β„Ž βˆ’ 1)/2β„Ž) = 4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^2π‘›β„Ž βˆ’ 1)/2π‘›β„Ž . (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/( (𝑒^2β„Ž βˆ’ 1)/2β„Ž) Solving (π₯𝐒𝐦)┬(π§β†’βˆž) ( 𝟏)/(( 𝒆^πŸπ’‰ βˆ’ 𝟏)/πŸπ’‰) As nβ†’βˆž β‡’ 2/β„Ž β†’βˆž β‡’ β„Ž β†’0 ∴ lim┬(nβ†’βˆž) ( 1)/(( 𝑒^2β„Ž βˆ’ 1)/2β„Ž) = lim┬(hβ†’0) ( 1)/(( 𝑒^2β„Ž βˆ’ 1)/2β„Ž) = 1/1 = 1 Thus, our equation becomes ∫1_0^4▒〖𝑒π‘₯ 𝑑π‘₯γ€— ="4" (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^2π‘›β„Ž βˆ’ 1)/2π‘›β„Ž . (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/( (𝑒^2β„Ž βˆ’ 1)/2β„Ž) " " = "4" (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^2π‘›β„Ž βˆ’ 1)/2π‘›β„Ž . 1 = 4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^(2𝑛 . 4/𝑛) βˆ’ 1)/(2𝑛 (4/𝑛) ) = 4 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^8 βˆ’ 1)/8 (π‘ˆπ‘ π‘–π‘›π‘” (π‘™π‘–π‘š)┬(𝑑→0) (𝑒^𝑑 βˆ’ 1)/𝑑 =1) (π‘ˆπ‘ π‘–π‘›π‘” β„Ž=4/𝑛) = 4 (𝑒^8 βˆ’ 1)/8 = (𝒆^πŸ– βˆ’ 𝟏)/𝟐 Putting the values of I1 and I2 in I ∴ I = ∫1_0^4β–’γ€–π‘₯ . 𝑑π‘₯γ€—+∫1_0^4β–’γ€– 𝑒2π‘₯ . 𝑑π‘₯γ€— = 8 + (𝑒^8 βˆ’ 1)/2 = (16 + 𝑒^8 βˆ’ 1)/2 = (πŸπŸ“ + 𝒆^πŸ–)/𝟐

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.