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

Misc 40 Evaluate ∫_0^1▒𝑒^(2 βˆ’3π‘₯)⁑𝑑π‘₯ as a limit of a sum . I=∫_0^1▒𝑒^(2 βˆ’3π‘₯)⁑𝑑π‘₯ I=∫_0^1▒〖𝑒^2 . 𝑒^(βˆ’3π‘₯)〗⁑𝑑π‘₯ I=𝑒^2 ∫_0^1▒𝑒^(βˆ’3π‘₯)⁑𝑑π‘₯ Solving I1 separately ∫_0^1▒𝑒^(βˆ’3π‘₯) 𝑑π‘₯ Putting π‘Ž = 0 𝑏 =1 β„Ž = (𝑏 βˆ’ π‘Ž)/𝑛 = (1 βˆ’ 0)/𝑛 = 1/𝑛 𝑓(π‘₯)=𝑒^(βˆ’3π‘₯) 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 ∫_0^1▒𝑒^(βˆ’3π‘₯) 𝑑π‘₯ =(1βˆ’0) lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(0+β„Ž)+𝑓(0+2β„Ž)+… +𝑓(0+(π‘›βˆ’1)β„Ž) =lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(β„Ž)+𝑓(2β„Ž)……+𝑓((π‘›βˆ’1)β„Ž) Here, 𝑓(π‘₯)=𝑒^(βˆ’3π‘₯) 𝑓(0)=𝑒^(βˆ’3(0))=1 𝑓(β„Ž)=𝑒^(βˆ’3β„Ž) 𝑓(2β„Ž)=𝑒^(βˆ’3(2β„Ž))=𝑒^(βˆ’6β„Ž) 𝑓((π‘›βˆ’1)β„Ž)=𝑒^(βˆ’3(π‘›βˆ’1)β„Ž) Hence, our equation becomes ∫_0^1▒𝑒^(βˆ’3π‘₯) 𝑑π‘₯ =lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(β„Ž)+𝑓(2β„Ž)……+𝑓(π‘›βˆ’1)β„Ž) = lim┬(nβ†’βˆž) 1/𝑛 (1+𝑒^(βˆ’3β„Ž)+𝑒^(βˆ’6β„Ž)+ ……+𝑒^(βˆ’3(𝑛 βˆ’ 1) β„Ž) ) Let S = 1+𝑒^(βˆ’3β„Ž)+𝑒^(βˆ’6β„Ž)+ ……+𝑒^(βˆ’3(𝑛 βˆ’ 1) β„Ž) It is a G.P. with common ratio (r) r = 𝑒^(βˆ’3β„Ž)/1 = 𝑒^(βˆ’3β„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^(βˆ’3β„Ž) , we get S = 1(((𝑒^(βˆ’3β„Ž) )^𝑛 βˆ’ 1)/(𝑒^(βˆ’3β„Ž) βˆ’ 1))= (𝑒^(βˆ’3π‘›β„Ž) βˆ’ 1)/(𝑒^(βˆ’3β„Ž) βˆ’ 1) Thus ∴ ∫_0^1▒𝑒^(βˆ’3π‘₯) 𝑑π‘₯ =lim┬(nβ†’βˆž) 1/𝑛 (1+𝑒^(βˆ’3β„Ž)+𝑒^(βˆ’6β„Ž)+ …+𝑒^(βˆ’3(𝑛 βˆ’ 1) β„Ž) ) Putting the value of S, we get =lim┬(nβ†’βˆž) 1/𝑛 ((𝑒^(βˆ’3π‘›β„Ž) βˆ’ 1)/(𝑒^(βˆ’3β„Ž) βˆ’ 1)) = (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 ((𝑒^(βˆ’3π‘›β„Ž) βˆ’ 1)/(βˆ’3β„Ž . (𝑒^(βˆ’3β„Ž) βˆ’ 1)/(βˆ’3β„Ž))) = (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^(βˆ’3π‘›β„Ž) βˆ’ 1)/(βˆ’3π‘›β„Ž) . 1/( (𝑒^(βˆ’3β„Ž) βˆ’ 1)/(βˆ’3β„Ž)) = (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^(βˆ’3π‘›β„Ž) βˆ’ 1)/(βˆ’3π‘›β„Ž) . (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/( (𝑒^(βˆ’3β„Ž) βˆ’ 1)/(βˆ’3β„Ž)) Solving (π₯𝐒𝐦)┬(π§β†’βˆž) ( 𝟏)/(( 𝒆^(βˆ’πŸ‘π’‰) βˆ’ 𝟏)/(βˆ’πŸ‘π’‰)) As nβ†’βˆž β‡’ 1/β„Ž β†’βˆž β‡’ β„Ž β†’0 ∴ lim┬(nβ†’βˆž) ( 1)/(( 𝑒^(βˆ’3β„Ž) βˆ’ 1)/(βˆ’3β„Ž)) = lim┬(hβ†’0) ( 1)/(( 𝑒^(βˆ’3β„Ž) βˆ’ 1)/(βˆ’3β„Ž)) = 1/1 = 1 Thus, our equation becomes ∫1_0^1▒〖𝑒^(βˆ’3π‘₯) 𝑑π‘₯γ€— =(π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^(βˆ’3π‘›β„Ž) βˆ’ 1)/(βˆ’3π‘›β„Ž).(π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/( (𝑒^(βˆ’3β„Ž) βˆ’ 1)/(βˆ’3β„Ž)) = (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^(βˆ’3π‘›β„Ž) βˆ’ 1)/(βˆ’3π‘›β„Ž). 1 = (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^(βˆ’3𝑛 . 1/𝑛) βˆ’ 1)/(βˆ’3𝑛 (1/𝑛) ) = (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^(βˆ’3) βˆ’ 1)/(βˆ’3) = 1/1 = 1 Thus, our equation becomes ∫1_0^1▒〖𝑒^(βˆ’3π‘₯) 𝑑π‘₯γ€— =(π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^(βˆ’3π‘›β„Ž) βˆ’ 1)/(βˆ’3π‘›β„Ž).(π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/( (𝑒^(βˆ’3β„Ž) βˆ’ 1)/(βˆ’3β„Ž)) = (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^(βˆ’3π‘›β„Ž) βˆ’ 1)/(βˆ’3π‘›β„Ž). 1 = (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^(βˆ’3𝑛 . 1/𝑛) βˆ’ 1)/(βˆ’3𝑛 (1/𝑛) ) = (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^(βˆ’3) βˆ’ 1)/(βˆ’3) = (𝑒^(βˆ’3) βˆ’ 1)/(βˆ’3) = (1 βˆ’ 𝑒^(βˆ’3))/3 = (1 βˆ’ 1/𝑒^3 )/3 = (𝑒^3 βˆ’ 1)/(3𝑒^3 ) Putting the values of I1 in (1) I=𝑒^2Γ—1/3 [(𝑒^3 βˆ’ 1)/𝑒^3 ] I1=1/3 [(𝑒^3 βˆ’ 1)/𝑒] 𝐈=𝟏/πŸ‘ [𝒆^πŸβˆ’ 𝟏/𝒆]

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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.