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

Transcript

Ex 7.8, 5 ∫1_(βˆ’1)^1▒〖𝑒π‘₯ 𝑑π‘₯γ€— ∫1_(βˆ’1)^1▒〖𝑒π‘₯ 𝑑π‘₯γ€— Putting π‘Ž =βˆ’1 𝑏 =1 β„Ž=(𝑏 βˆ’ π‘Ž)/𝑛 =(1 βˆ’ (βˆ’1))/𝑛 =(1 + 1)/𝑛=2/𝑛 𝑓(π‘₯)=𝑒^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_(βˆ’1)^1▒〖𝑒π‘₯ 𝑑π‘₯γ€— =(1 βˆ’(βˆ’1)) (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(βˆ’1)+𝑓(βˆ’1+β„Ž)+𝑓(βˆ’1+2β„Ž)+ …+𝑓(βˆ’1+(π‘›βˆ’1)β„Ž)) =2 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(βˆ’1)+𝑓(βˆ’1+β„Ž)+𝑓(βˆ’1+2β„Ž)+ …+𝑓(βˆ’1+(π‘›βˆ’1)β„Ž)) Here, 𝑓(π‘₯)=𝑒^π‘₯ 𝑓(βˆ’1)=𝑒^(βˆ’1) 𝑓(βˆ’1+β„Ž)=𝑒^(βˆ’1 + β„Ž) =𝑒^(βˆ’1). 𝑒^β„Ž 𝑓 (βˆ’1+2β„Ž)=𝑒^(βˆ’1 + 2β„Ž)=𝑒^(βˆ’1). 𝑒^2β„Ž 𝑓(βˆ’1+(π‘›βˆ’1)β„Ž)=𝑒^(βˆ’1 + (𝑛 βˆ’ 1)β„Ž)=𝑒^(βˆ’1).𝑒^(𝑛 βˆ’ 1)β„Ž Hence, our equation becomes ∫1_(βˆ’1)^1▒〖𝑒π‘₯ 𝑑π‘₯γ€— =2 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(βˆ’1)+𝑓(βˆ’1+β„Ž)+𝑓(βˆ’1+2β„Ž)+ …+𝑓(βˆ’1+(π‘›βˆ’1)β„Ž)) =2 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑒^(βˆ’1)+𝑒^(βˆ’1). 𝑒^β„Ž+𝑒^(βˆ’1). 𝑒^2β„Ž+ …+𝑒^(βˆ’1).𝑒^(π‘›βˆ’1)β„Ž ) =2 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑒^(βˆ’1) (1+β„Ž+𝑒^2β„Ž+ …+𝑒^(π‘›βˆ’1)β„Ž )) =2𝑒^(βˆ’1) (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (1+β„Ž+𝑒^2β„Ž+ …+𝑒^(π‘›βˆ’1)β„Ž ) Let S = 1+β„Ž+𝑒^2β„Ž+ …+𝑒^(π‘›βˆ’1)β„Ž It is G.P with common ratio (r) π‘Ÿ = 𝑒^β„Ž/1 = 𝑒^β„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^𝑛 , we get S = 1(((𝑒^β„Ž )^𝑛 βˆ’ 1)/(𝑒^𝑛 βˆ’ 1))= (𝑒^π‘›β„Ž βˆ’ 1)/(𝑒^𝑛 βˆ’ 1) Thus, ∫1_(βˆ’1)^1▒〖𝑒π‘₯ 𝑑π‘₯γ€— =2 . 𝑒^(βˆ’1) (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (1+𝑒^𝑛+𝑒^2β„Ž+ …+𝑒^(π‘›βˆ’1)β„Ž ) = 2/𝑒 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 ((𝑒^π‘›β„Ž βˆ’ 1)/(𝑒^𝑛 βˆ’ 1)) = 2/𝑒 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 ((𝑒^π‘›β„Ž βˆ’ 1)/(β„Ž . (𝑒^β„Ž βˆ’ 1)/β„Ž)) = 2/𝑒 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^π‘›β„Ž βˆ’ 1)/π‘›β„Ž . 1/( (𝑒^β„Ž βˆ’ 1)/β„Ž) = 2/𝑒 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^π‘›β„Ž βˆ’ 1)/π‘›β„Ž . (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/( (𝑒^β„Ž βˆ’ 1)/β„Ž) Solving (π₯𝐒𝐦)┬(π§β†’βˆž) ( 𝟏)/(( 𝒆^𝒉 βˆ’ 𝟏)/𝒉) As nβ†’βˆž β‡’ 2/β„Ž β†’βˆž β‡’ β„Ž β†’0 ∴ lim┬(nβ†’βˆž) ( 1)/(( 𝑒^β„Ž βˆ’ 1)/β„Ž) = lim┬(hβ†’0) ( 1)/(( 𝑒^β„Ž βˆ’ 1)/β„Ž) = 1/1 = 1 Thus, our equation becomes ∫1_(βˆ’1)^1▒〖𝑒π‘₯ 𝑑π‘₯γ€— = 2/𝑒 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^π‘›β„Ž βˆ’ 1)/π‘›β„Ž . (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/( (𝑒^β„Ž βˆ’ 1)/β„Ž) = 2/𝑒 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^π‘›β„Ž βˆ’ 1)/π‘›β„Ž . 1 = 2/𝑒 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^(𝑛 . 2/𝑛) βˆ’ 1)/(𝑛 (2/𝑛) ) = 2/𝑒 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (𝑒^2 βˆ’ 1)/2 (π‘ˆπ‘ π‘–π‘›π‘” (π‘™π‘–π‘š)┬(𝑑→0) (𝑒^𝑑 βˆ’ 1)/𝑑 =1) (π‘ˆπ‘ π‘–π‘›π‘” β„Ž=2/𝑛) = 2/𝑒 . (𝑒^2 βˆ’ 1)/2 = (𝑒^2 βˆ’ 1)/𝑒 = 𝑒^2/𝑒 βˆ’ 1/𝑒 =𝒆 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.