Chapter 7 Class 12 Integrals
Concept wise

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

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 5 ∫1_(−1)^1▒〖𝑒𝑥 𝑑𝑥〗 ∫1_(−1)^1▒〖𝑒𝑥 𝑑𝑥〗 Putting 𝑎 =−1 𝑏 =1 ℎ=(𝑏 − 𝑎)/𝑛 =(1 − (−1))/𝑛 =(1 + 1)/𝑛=2/𝑛 𝑓(𝑥)=𝑒^𝑥 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 ∫1_(−1)^1▒〖𝑒𝑥 𝑑𝑥〗 =(1 −(−1)) (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(−1)+𝑓(−1+ℎ)+𝑓(−1+2ℎ)+ …+𝑓(−1+(𝑛−1)ℎ)) =2 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(−1)+𝑓(−1+ℎ)+𝑓(−1+2ℎ)+ …+𝑓(−1+(𝑛−1)ℎ)) Here, 𝑓(𝑥)=𝑒^𝑥 𝑓(−1)=𝑒^(−1) 𝑓(−1+ℎ)=𝑒^(−1 + ℎ) =𝑒^(−1). 𝑒^ℎ 𝑓 (−1+2ℎ)=𝑒^(−1 + 2ℎ)=𝑒^(−1). 𝑒^2ℎ 𝑓(−1+(𝑛−1)ℎ)=𝑒^(−1 + (𝑛 − 1)ℎ)=𝑒^(−1).𝑒^(𝑛 − 1)ℎ Hence, our equation becomes ∫1_(−1)^1▒〖𝑒𝑥 𝑑𝑥〗 =2 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(−1)+𝑓(−1+ℎ)+𝑓(−1+2ℎ)+ …+𝑓(−1+(𝑛−1)ℎ)) =2 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑒^(−1)+𝑒^(−1). 𝑒^ℎ+𝑒^(−1). 𝑒^2ℎ+ …+𝑒^(−1).𝑒^(𝑛−1)ℎ ) =2 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑒^(−1) (1+ℎ+𝑒^2ℎ+ …+𝑒^(𝑛−1)ℎ )) =2𝑒^(−1) (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (1+ℎ+𝑒^2ℎ+ …+𝑒^(𝑛−1)ℎ ) Let S = 1+ℎ+𝑒^2ℎ+ …+𝑒^(𝑛−1)ℎ It is G.P with common ratio (r) 𝑟 = 𝑒^ℎ/1 = 𝑒^ℎ We know Sum of G.P = a((𝑟^𝑛 − 1)/(𝑟 − 1)) Replacing a by 1 and r by 𝑒^𝑛 , we get S = 1(((𝑒^ℎ )^𝑛 − 1)/(𝑒^𝑛 − 1))= (𝑒^𝑛ℎ − 1)/(𝑒^𝑛 − 1) Thus, ∫1_(−1)^1▒〖𝑒𝑥 𝑑𝑥〗 =2 . 𝑒^(−1) (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (1+𝑒^𝑛+𝑒^2ℎ+ …+𝑒^(𝑛−1)ℎ ) = 2/𝑒 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 ((𝑒^𝑛ℎ − 1)/(𝑒^𝑛 − 1)) = 2/𝑒 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 ((𝑒^𝑛ℎ − 1)/(ℎ . (𝑒^ℎ − 1)/ℎ)) = 2/𝑒 (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^𝑛ℎ − 1)/𝑛ℎ . 1/( (𝑒^ℎ − 1)/ℎ) = 2/𝑒 (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^𝑛ℎ − 1)/𝑛ℎ . (𝑙𝑖𝑚)┬(𝑛→∞) 1/( (𝑒^ℎ − 1)/ℎ) Solving (𝐥𝐢𝐦)┬(𝐧→∞) ( 𝟏)/(( 𝒆^𝒉 − 𝟏)/𝒉) As n→∞ ⇒ 2/ℎ →∞ ⇒ ℎ →0 ∴ lim┬(n→∞) ( 1)/(( 𝑒^ℎ − 1)/ℎ) = lim┬(h→0) ( 1)/(( 𝑒^ℎ − 1)/ℎ) = 1/1 = 1 Thus, our equation becomes ∫1_(−1)^1▒〖𝑒𝑥 𝑑𝑥〗 = 2/𝑒 (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^𝑛ℎ − 1)/𝑛ℎ . (𝑙𝑖𝑚)┬(𝑛→∞) 1/( (𝑒^ℎ − 1)/ℎ) = 2/𝑒 (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^𝑛ℎ − 1)/𝑛ℎ . 1 = 2/𝑒 (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(𝑛 . 2/𝑛) − 1)/(𝑛 (2/𝑛) ) = 2/𝑒 (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^2 − 1)/2 (𝑈𝑠𝑖𝑛𝑔 (𝑙𝑖𝑚)┬(𝑡→0) (𝑒^𝑡 − 1)/𝑡 =1) (𝑈𝑠𝑖𝑛𝑔 ℎ=2/𝑛) = 2/𝑒 . (𝑒^2 − 1)/2 = (𝑒^2 − 1)/𝑒 = 𝑒^2/𝑒 − 1/𝑒 =𝒆 − 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.