Chapter 7 Class 12 Integrals
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Misc 40 - Evaluate 0->1 e2 - 3x dx as a limit of a sum - Miscellaneous

Misc 40 - Chapter 7 Class 12 Integrals - Part 2
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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Question 3 Evaluate ∫_0^1▒𝑒^(2 −3𝑥)⁡𝑑𝑥 as a limit of a sum . I=∫_0^1▒𝑒^(2 −3𝑥)⁡𝑑𝑥 I=∫_0^1▒〖𝑒^2 . 𝑒^(−3𝑥)〗⁡𝑑𝑥 I=𝑒^2 ∫_0^1▒𝑒^(−3𝑥)⁡𝑑𝑥 Solving I1 separately ∫_0^1▒𝑒^(−3𝑥) 𝑑𝑥 Putting 𝑎 = 0 𝑏 =1 ℎ = (𝑏 − 𝑎)/𝑛 = (1 − 0)/𝑛 = 1/𝑛 𝑓(𝑥)=𝑒^(−3𝑥) We know that ∫1_𝑎^𝑏▒〖𝑥 𝑑𝑥〗 =(𝑏−𝑎) (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(𝑎)+𝑓(𝑎+ℎ)+𝑓(𝑎+2ℎ)…+𝑓(𝑎+(𝑛−1)ℎ)) Hence we can write ∫_0^1▒𝑒^(−3𝑥) 𝑑𝑥 =(1−0) lim┬(n→∞) 1/𝑛 (𝑓(0)+𝑓(0+ℎ)+𝑓(0+2ℎ)+… +𝑓(0+(𝑛−1)ℎ) =lim┬(n→∞) 1/𝑛 (𝑓(0)+𝑓(ℎ)+𝑓(2ℎ)……+𝑓((𝑛−1)ℎ) Here, 𝑓(𝑥)=𝑒^(−3𝑥) 𝑓(0)=𝑒^(−3(0))=1 𝑓(ℎ)=𝑒^(−3ℎ) 𝑓(2ℎ)=𝑒^(−3(2ℎ))=𝑒^(−6ℎ) 𝑓((𝑛−1)ℎ)=𝑒^(−3(𝑛−1)ℎ) Hence, our equation becomes ∫_0^1▒𝑒^(−3𝑥) 𝑑𝑥 =lim┬(n→∞) 1/𝑛 (𝑓(0)+𝑓(ℎ)+𝑓(2ℎ)……+𝑓(𝑛−1)ℎ) = lim┬(n→∞) 1/𝑛 (1+𝑒^(−3ℎ)+𝑒^(−6ℎ)+ ……+𝑒^(−3(𝑛 − 1) ℎ) ) Let S = 1+𝑒^(−3ℎ)+𝑒^(−6ℎ)+ ……+𝑒^(−3(𝑛 − 1) ℎ) It is a G.P. with common ratio (r) r = 𝑒^(−3ℎ)/1 = 𝑒^(−3ℎ) We know Sum of G.P = a((𝑟^𝑛 − 1)/(𝑟 − 1)) Replacing a by 1 and r by 𝑒^(−3ℎ) , we get S = 1(((𝑒^(−3ℎ) )^𝑛 − 1)/(𝑒^(−3ℎ) − 1))= (𝑒^(−3𝑛ℎ) − 1)/(𝑒^(−3ℎ) − 1) Thus ∴ ∫_0^1▒𝑒^(−3𝑥) 𝑑𝑥 =lim┬(n→∞) 1/𝑛 (1+𝑒^(−3ℎ)+𝑒^(−6ℎ)+ …+𝑒^(−3(𝑛 − 1) ℎ) ) Putting the value of S, we get =lim┬(n→∞) 1/𝑛 ((𝑒^(−3𝑛ℎ) − 1)/(𝑒^(−3ℎ) − 1)) = (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 ((𝑒^(−3𝑛ℎ) − 1)/(−3ℎ . (𝑒^(−3ℎ) − 1)/(−3ℎ))) = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛ℎ) − 1)/(−3𝑛ℎ) . 1/( (𝑒^(−3ℎ) − 1)/(−3ℎ)) = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛ℎ) − 1)/(−3𝑛ℎ) . (𝑙𝑖𝑚)┬(𝑛→∞) 1/( (𝑒^(−3ℎ) − 1)/(−3ℎ)) Solving (𝐥𝐢𝐦)┬(𝐧→∞) ( 𝟏)/(( 𝒆^(−𝟑𝒉) − 𝟏)/(−𝟑𝒉)) As n→∞ ⇒ 1/ℎ →∞ ⇒ ℎ →0 ∴ lim┬(n→∞) ( 1)/(( 𝑒^(−3ℎ) − 1)/(−3ℎ)) = lim┬(h→0) ( 1)/(( 𝑒^(−3ℎ) − 1)/(−3ℎ)) = 1/1 = 1 Thus, our equation becomes ∫1_0^1▒〖𝑒^(−3𝑥) 𝑑𝑥〗 =(𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛ℎ) − 1)/(−3𝑛ℎ).(𝑙𝑖𝑚)┬(𝑛→∞) 1/( (𝑒^(−3ℎ) − 1)/(−3ℎ)) = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛ℎ) − 1)/(−3𝑛ℎ). 1 = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛 . 1/𝑛) − 1)/(−3𝑛 (1/𝑛) ) = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3) − 1)/(−3) = 1/1 = 1 Thus, our equation becomes ∫1_0^1▒〖𝑒^(−3𝑥) 𝑑𝑥〗 =(𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛ℎ) − 1)/(−3𝑛ℎ).(𝑙𝑖𝑚)┬(𝑛→∞) 1/( (𝑒^(−3ℎ) − 1)/(−3ℎ)) = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛ℎ) − 1)/(−3𝑛ℎ). 1 = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛 . 1/𝑛) − 1)/(−3𝑛 (1/𝑛) ) = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3) − 1)/(−3) = (𝑒^(−3) − 1)/(−3) = (1 − 𝑒^(−3))/3 = (1 − 1/𝑒^3 )/3 = (𝑒^3 − 1)/(3𝑒^3 ) Putting the values of I1 in (1) I=𝑒^2×1/3 [(𝑒^3 − 1)/𝑒^3 ] I1=1/3 [(𝑒^3 − 1)/𝑒] 𝐈=𝟏/𝟑 [𝒆^𝟐− 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.