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either evaluate the given improper integral or show that it diverges int^(+00_1)...

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rachell65 | Student | eNoter

Posted August 19, 2012 at 5:23 PM via web

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either evaluate the given improper integral or show that it diverges

int^(+00_1) xe^(-2x) dx

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 19, 2012 at 7:22 PM (Answer #1)

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You need to evaluate the improper integral such that:

`int_1^oo xe^(-2x) dx = lim_(x->oo) int_1^x xe^(-2x) dx`

You should use integration by parts such that:

`int udv = uv - int vdu`

`u = x => du = dx`

`dv = e^(-2x) dx => v = -(e^(-2x))/2`

`int_1^x xe^(-2x) dx = -x*(e^(-2x))/2|_1^x + int_1^x e^(-2x) dx` 

`int_1^x xe^(-2x) dx = -x*(e^(-2x))/2|_1^x - (e^(-2x))/2|_1^x`

` int_1^x xe^(-2x) dx = -(e^(-2x))/2*(x + 1)|_1^x`

`int_1^x xe^(-2x) dx = -(e^(-2x))/2*(x + 1) + (e^(-2))/2*(1 + 1)`

You may evaluate the limit such that:

`lim_(x->oo) int_1^x xe^(-2x) dx = lim_(x->oo) -(e^(-2x))/2*(x + 1) + lim_(x->oo) 1/(4e^2)`

`lim_(x->oo) int_1^x xe^(-2x) dx = 1/(4e^2) - lim_(x->oo) (e^(-2x))/2*(x + 1)`

You need to evaluate `lim_(x->oo) (e^(-2x))/2*(x + 1)`  such that:

`lim_(x->oo) (e^(-2x))/2*(x + 1) = oo/oo`

You should use l'Hospital's theorem such that:

`lim_(x->oo) ((x+1)')/((2e^(2x))') = lim_(x->oo) 1/(4e^(2x)) = 1/oo = 0`

`lim_(x->oo) int_1^x xe^(-2x) dx = 1/(4e^2) - 0`

`lim_(x->oo) int_1^x xe^(-2x) dx = 1/(4e^2)`

Hence, evaluating the limit of definite integral yields `lim_(x->oo) int_1^x xe^(-2x) dx = 1/(4e^2)` , hence, the integral converges.

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