Question 5 - Area as a sum - Chapter 7 Class 12 Integrals

Last updated at April 16, 2024 by

Ex 7.8, 5 - Evaluate ex dx from -1 to 1 by limit as sum - Ex 7.8

Ex 7.8, 5 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.8, 5 - Chapter 7 Class 12 Integrals - Part 3
Ex 7.8, 5 - Chapter 7 Class 12 Integrals - Part 4
Ex 7.8, 5 - Chapter 7 Class 12 Integrals - Part 5 Ex 7.8, 5 - Chapter 7 Class 12 Integrals - Part 6 Ex 7.8, 5 - Chapter 7 Class 12 Integrals - Part 7

Transcript

Question 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.