Examples

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

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