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Evaluate ∫2−1 (e3x+7x−5) dx as a limit of sums - Mathematics

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Sum

Evaluate `int_(-1)^2(e^3x+7x-5)dx` as a limit of sums

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Solution

`int_(-1)^2(e^3x+7x-5)dx`

Here ` f(x)=e^(3x)+7x-5`

a=-1, b=2, h=(b-a)/n =3/n

By defination `int_(-1)^2(e^3x+7x-5)dx=lim_(n->oo)sum_(r=a)1^nh.f(a+rh)`

`lim_(n->oo)sum_(r=a)1^nh.f(-1+rh)=lim_(n->oo)sum_(r=a)1^nh.(e^3(-1+rh)+7(-1+rh)-5)`

`=lim_(n->oo)[h.e^(-3).e^(3h)(1+e^(3h)+3^(6h)+.....+e^(3nh))+7h^2(1+2+3+....+n)-12nh]`

`=lim_(n->oo)[(he^(3h))/(n.e^3)xx(e^(3nh)-1)/(e^(3h)-1)+7h^2(n(n+1))/2-12nh]`

`=lim_(n->oo)[((3e^(3xx3/n))/(n.e^3)xx(e^(3nxx3/n)-1)xx((3h)/(e^(3h)-1))xxn/(3xx3))+63/n^2xx(n(n+1))/2-12xx3]`

Now applying the limit we get

`=(e^9-1)/(3e^3)+63/2-36`

`=(e^9-1)/(3e^3)  - 9/2`

Concept: Definite Integral as the Limit of a Sum
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