Example 26 - Evaluate integral ex dx as the limit of a sum

Example 26 - Chapter 7 Class 12 Integrals - Part 2
Example 26 - Chapter 7 Class 12 Integrals - Part 3 Example 26 - Chapter 7 Class 12 Integrals - Part 4 Example 26 - Chapter 7 Class 12 Integrals - Part 5 Example 26 - Chapter 7 Class 12 Integrals - Part 6 Example 26 - Chapter 7 Class 12 Integrals - Part 7 Example 26 - Chapter 7 Class 12 Integrals - Part 8

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


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Question 2 Evaluate ∫_0^2▒𝑒^𝑥 𝑑𝑥 as the limit of a sum . ∫_0^2▒𝑒^𝑥 𝑑𝑥 Putting 𝑎 = 0 𝑏 = 2 ℎ = (𝑏 − 𝑎)/𝑛 = (2 − 0)/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 ∫_0^2▒𝑒^𝑥 𝑑𝑥 =(2−0) lim┬(n→∞) 1/𝑛 (𝑓(0)+𝑓(0+ℎ)+𝑓(0+2ℎ)+… +𝑓(0+(𝑛−1)ℎ) =2 lim┬(n→∞) 1/𝑛 (𝑓(0)+𝑓(ℎ)+𝑓(2ℎ)……+𝑓((𝑛−1)ℎ) Here, 𝑓(𝑥)=𝑒^𝑥 𝑓(0)=𝑒^0=1 𝑓(0+ℎ)=𝑒^(0 + ℎ)=𝑒^ℎ 𝑓(0+2ℎ)=𝑒^(0 + 2ℎ)=𝑒^2ℎ 𝑓(0+(𝑛−1)ℎ)=𝑒^(0 + (𝑛−1)ℎ)=𝑒^(𝑛−1)ℎ Hence, our equation becomes ∴ ∫_0^2▒𝑒^𝑥 𝑑𝑥 =2 lim┬(n→∞) 1/𝑛 (𝑓(0)+𝑓(ℎ)+𝑓(2ℎ)……+𝑓(𝑛−1)ℎ) = 2 .lim┬(n→∞) 1/𝑛 (1+𝑒^ℎ+𝑒^2ℎ+ ……+𝑒^((𝑛 − 1) ℎ) ) Let S = 𝑒^ℎ+𝑒^2ℎ+ ……+𝑒^((𝑛 − 1) ℎ) It is a G.P. with common ratio (r) r = 𝑒^2ℎ/𝑒^ℎ = 𝑒^(2ℎ − ℎ) = 𝑒^ℎ Sum of G.P. S = (𝑎 (1 − 𝑟^𝑛 ))/(1 − 𝑟) = (𝑒^ℎ (1 − (𝑒^ℎ )^𝑛 ))/(1 − 𝑒^ℎ ) = (𝑒^ℎ (1 − 𝑒^𝑛ℎ ))/(1 − 𝑒^ℎ ) = (2−0) lim┬(n→∞) 1/𝑛 (𝑓(0)+𝑓(0+ℎ)+𝑓(0+2ℎ)+……+𝑓(0+(𝑛−1)ℎ) = 2 .lim┬(n→∞) 1/𝑛 (1+𝑒^ℎ+𝑒^2ℎ+ ……+𝑒^((𝑛 − 1) ℎ) ) Let S = 𝑒^ℎ+𝑒^2ℎ+ ……+𝑒^((𝑛 − 1) ℎ) It is a G.P. with common ratio (r) r = 𝑒^2ℎ/𝑒^ℎ = 𝑒^(2ℎ − ℎ) = 𝑒^ℎ Sum of G.P. S = (𝑎 (1 − 𝑟^𝑛 ))/(1 − 𝑟) = (𝑒^ℎ (1 − (𝑒^ℎ )^𝑛 ))/(1 − 𝑒^ℎ ) = (𝑒^ℎ (1 − 𝑒^𝑛ℎ ))/(1 − 𝑒^ℎ ) Thus ∴ ∫_0^2▒𝑒^𝑥 𝑑𝑥 = 2 .lim┬(n→∞) 1/𝑛 (1+𝑒^ℎ+𝑒^2ℎ+ ……+𝑒^((𝑛 − 1) ℎ) ) Putting the value of S, we get = 2 .lim┬(n→∞) 1/𝑛 (1+ (𝑒^ℎ (1 − 𝑒^𝑛ℎ ))/(1 − 𝑒^ℎ )) = 2 (lim┬(n→∞) 1/𝑛 +lim┬(n→∞) 1/𝑛 (𝑒^ℎ ((1 − 𝑒^𝑛ℎ)/(1 − 𝑒^ℎ )))) "Taking −1 common from" "numerator and denominator " = 2 (1/∞ +lim┬(n→∞) (𝑒^ℎ/𝑛 . ( 𝑒^𝑛ℎ − 1)/(𝑒^ℎ − 1))) "Multiplying and dividing denominator by h" = 2 (0+lim┬(n→∞) (𝑒^ℎ/𝑛 . ( 𝑒^𝑛ℎ − 1)/(ℎ . ( 𝑒^ℎ − 1)/ℎ))) = 2 . lim┬(n→∞) 𝑒^ℎ/𝑛 (( 𝑒^𝑛ℎ − 1)/ℎ)(( 1)/(( 𝑒^ℎ − 1)/ℎ)) = 2 . lim┬(n→∞) 𝑒^ℎ/𝑛 (( 𝑒^𝑛ℎ − 1)/ℎ) . 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 (𝑈𝑠𝑖𝑛𝑔 lim┬(t → 0) ( 𝑒^𝑡 − 1)/𝑡=1) Thus, our equation becomes ∴ ∫_0^2▒𝑒^𝑥 𝑑𝑥 = 2 . lim┬(n→∞) 𝑒^ℎ/𝑛 (( 𝑒^𝑛ℎ − 1)/ℎ) . lim┬(n→∞) (( 1)/(( 𝑒^ℎ − 1)/ℎ)) = 2 . lim┬(n→∞) 𝑒^ℎ/𝑛 (( 𝑒^𝑛ℎ − 1)/ℎ) . 1 = 2 . lim┬(n→∞) 𝑒^( 2/𝑛)/𝑛 (( 𝑒^(𝑛 . 2/𝑛) − 1)/(2/𝑛)) = 2 . lim┬(n→∞) 〖 𝑒〗^( 2/𝑛) (( 𝑒^2 − 1)/2) = 2 (𝑒^( 2/∞) ((𝑒^2 − 1))/2) (𝑈𝑠𝑖𝑛𝑔 ℎ= 2/𝑛) = 2 . 𝑒^0 ((𝑒^2 − 1))/2 = 2/2 . 1 (𝑒^2−1) = 1 . 1 (𝑒^2−1) = 𝒆^𝟐−𝟏

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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.