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Question
Evaluate `int_1^4 ( 1+ x +e^(2x)) dx` as limit of sums.
Solution
`I = int_1^4 (1 + x +e^(2x)) dx = int_1^4 (1+x) dx + int _1^4 e^(2x) dx`
`= I_1 +I_2`
`h = (b-a)/n = (4-1)/n = 3/n `
`I_1 = int_1^4 (1 + x) dx = lim_(n->oo)[3/n[f(1) + f(1 +3/n) + ....... f (1 +((n-1))/n 3)]]`
` = 3"lim_(n-> oo) [1/n[(1 + 1) +(1+(1+3/n)+....)(1+((n-1))/n3)]]`
` = 3"lim_(n-> oo) [(2n)/n + 1/n [0.(3/n)+1(3/n)+....(n-1) 3/n]]`
` = 3"lim_(n-> oo) [2 + 1/n^2 [3 + 2(3) + .... (n -1 )3]]`
` = 3"lim_(n-> oo) [2 +3/n^2 [((n-1)n)/2]]`
` = 3["lim_(n-> oo) (2+3/2(1-1/n))]`
`=3(2+3/2)=6 +9/2=21/2`
`I_2 = int_1^4 e^(2x) dx `
`I_2 = "lim_(n-> oo) [3/n[f(1)+......+f(1+((n-1))/n3)]]`
` = "lim_(n-> oo) [3/n[e^2 + e^(2(1+3/n)) + ......e^(2(1+((n-1))/n 3)]]]`
`= 3e^2 "lim_(n -> oo )[1/n[1 + e^(2 (3/n))+....e^(2(3((n-1))/n))]]`
` = 3e^2 lim_(n -> oo ) 1/n [(e^(2(3/n)^n) -1)/(e^(2(3/n)^n)-1) ]`
` = 3e^2 lim_(n -> oo ) 1/n[(e^6 -1)/(e^(6/n) -1)]`
` = 3e^2(e^6 - 1) "lim_(n -> oo) (1/n)/(e^(6/n)-1)`
`=3e^2 (e^6 - 1) "lim_(n -> oo )((-1)/n^2)/(e^(6/n)((-1)/n^2)xx6)`
`=1/2e^2 (e^6-1)= (e^8 - e^2 )/2`
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