Integration Full Chapter Explained - https://you.tube/Integration-Class-12

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

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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.