Question 5 - Area as a sum - Chapter 7 Class 12 Integrals

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