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Examples
Example 1 (ii)
Example 1 (iii)
Example 2 (i)
Example 2 (ii)
Example 2 (iii) Important
Example 3 (i)
Example 3 (ii) Important
Example 3 (iii)
Example 4
Example 5 (i)
Example 5 (ii)
Example 5 (iii) Important
Example 5 (iv) Important
Example 6 (i)
Example 6 (ii) Important
Example 6 (iii) Important
Example 7 (i)
Example 7 (ii) Important
Example 7 (iii)
Example 8 (i)
Example 8 (ii) Important
Example 9 (i)
Example 9 (ii) Important
Example 9 (iii) Important
Example 10 (i)
Example 10 (ii) Important
Example 11
Example 12
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Example 14
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Example 16 Important
Example 17
Example 18 Important
Example 19
Example 20 Important
Example 21 Important
Example 22 Important
Example 23
Example 24
Example 25 (i)
Example 25 (ii) Important
Example 25 (iii)
Example 25 (iv) Important
Example 26
Example 27
Example 28 Important
Example 29
Example 30
Example 31
Example 32 Important
Example 33 Important
Example 34 Important
Example 35
Example 36 Important
Example 37 Important
Example 38 Important
Example 39 Important
Example 40 Important
Example 41 Important
Example 42 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams You are here
Question 3 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams
Last updated at May 29, 2023 by Teachoo
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