Question 3 - Miscellaneous - Chapter 7 Class 12 Integrals

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Question 3 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 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.