Misc 40 - Chapter 7 Class 12 Integrals

Transcript

Misc 40 Evaluate ∫_0^1▒𝑒^(2 −3𝑥)⁡𝑑𝑥 as a limit of a sum . I=∫_0^1▒𝑒^(2 −3𝑥)⁡𝑑𝑥 I=∫_0^1▒〖𝑒^2 . 𝑒^(−3𝑥)〗⁡𝑑𝑥 I=𝑒^2 ∫_0^1▒𝑒^(−3𝑥)⁡𝑑𝑥 Solving I1 separately ∫_0^1▒𝑒^(−3𝑥) 𝑑𝑥 Putting 𝑎 = 0 𝑏 =1 ℎ = (𝑏 − 𝑎)/𝑛 = (1 − 0)/𝑛 = 1/𝑛 𝑓(𝑥)=𝑒^(−3𝑥) 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 ∫_0^1▒𝑒^(−3𝑥) 𝑑𝑥 =(1−0) lim┬(n→∞) 1/𝑛 (𝑓(0)+𝑓(0+ℎ)+𝑓(0+2ℎ)+… +𝑓(0+(𝑛−1)ℎ) =lim┬(n→∞) 1/𝑛 (𝑓(0)+𝑓(ℎ)+𝑓(2ℎ)……+𝑓((𝑛−1)ℎ) Here, 𝑓(𝑥)=𝑒^(−3𝑥) 𝑓(0)=𝑒^(−3(0))=1 𝑓(ℎ)=𝑒^(−3ℎ) 𝑓(2ℎ)=𝑒^(−3(2ℎ))=𝑒^(−6ℎ) 𝑓((𝑛−1)ℎ)=𝑒^(−3(𝑛−1)ℎ) Hence, our equation becomes ∫_0^1▒𝑒^(−3𝑥) 𝑑𝑥 =lim┬(n→∞) 1/𝑛 (𝑓(0)+𝑓(ℎ)+𝑓(2ℎ)……+𝑓(𝑛−1)ℎ) = lim┬(n→∞) 1/𝑛 (1+𝑒^(−3ℎ)+𝑒^(−6ℎ)+ ……+𝑒^(−3(𝑛 − 1) ℎ) ) Let S = 1+𝑒^(−3ℎ)+𝑒^(−6ℎ)+ ……+𝑒^(−3(𝑛 − 1) ℎ) It is a G.P. with common ratio (r) r = 𝑒^(−3ℎ)/1 = 𝑒^(−3ℎ) We know Sum of G.P = a((𝑟^𝑛 − 1)/(𝑟 − 1)) Replacing a by 1 and r by 𝑒^(−3ℎ) , we get S = 1(((𝑒^(−3ℎ) )^𝑛 − 1)/(𝑒^(−3ℎ) − 1))= (𝑒^(−3𝑛ℎ) − 1)/(𝑒^(−3ℎ) − 1) Thus ∴ ∫_0^1▒𝑒^(−3𝑥) 𝑑𝑥 =lim┬(n→∞) 1/𝑛 (1+𝑒^(−3ℎ)+𝑒^(−6ℎ)+ …+𝑒^(−3(𝑛 − 1) ℎ) ) Putting the value of S, we get =lim┬(n→∞) 1/𝑛 ((𝑒^(−3𝑛ℎ) − 1)/(𝑒^(−3ℎ) − 1)) = (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 ((𝑒^(−3𝑛ℎ) − 1)/(−3ℎ . (𝑒^(−3ℎ) − 1)/(−3ℎ))) = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛ℎ) − 1)/(−3𝑛ℎ) . 1/( (𝑒^(−3ℎ) − 1)/(−3ℎ)) = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛ℎ) − 1)/(−3𝑛ℎ) . (𝑙𝑖𝑚)┬(𝑛→∞) 1/( (𝑒^(−3ℎ) − 1)/(−3ℎ)) Solving (𝐥𝐢𝐦)┬(𝐧→∞) ( 𝟏)/(( 𝒆^(−𝟑𝒉) − 𝟏)/(−𝟑𝒉)) As n→∞ ⇒ 1/ℎ →∞ ⇒ ℎ →0 ∴ lim┬(n→∞) ( 1)/(( 𝑒^(−3ℎ) − 1)/(−3ℎ)) = lim┬(h→0) ( 1)/(( 𝑒^(−3ℎ) − 1)/(−3ℎ)) = 1/1 = 1 Thus, our equation becomes ∫1_0^1▒〖𝑒^(−3𝑥) 𝑑𝑥〗 =(𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛ℎ) − 1)/(−3𝑛ℎ).(𝑙𝑖𝑚)┬(𝑛→∞) 1/( (𝑒^(−3ℎ) − 1)/(−3ℎ)) = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛ℎ) − 1)/(−3𝑛ℎ). 1 = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛 . 1/𝑛) − 1)/(−3𝑛 (1/𝑛) ) = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3) − 1)/(−3) = 1/1 = 1 Thus, our equation becomes ∫1_0^1▒〖𝑒^(−3𝑥) 𝑑𝑥〗 =(𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛ℎ) − 1)/(−3𝑛ℎ).(𝑙𝑖𝑚)┬(𝑛→∞) 1/( (𝑒^(−3ℎ) − 1)/(−3ℎ)) = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛ℎ) − 1)/(−3𝑛ℎ). 1 = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3𝑛 . 1/𝑛) − 1)/(−3𝑛 (1/𝑛) ) = (𝑙𝑖𝑚)┬(𝑛→∞) (𝑒^(−3) − 1)/(−3) = (𝑒^(−3) − 1)/(−3) = (1 − 𝑒^(−3))/3 = (1 − 1/𝑒^3 )/3 = (𝑒^3 − 1)/(3𝑒^3 ) Putting the values of I1 in (1) I=𝑒^2×1/3 [(𝑒^3 − 1)/𝑒^3 ] I1=1/3 [(𝑒^3 − 1)/𝑒] 𝐈=𝟏/𝟑 [𝒆^𝟐− 𝟏/𝒆]