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State whetever the given integral converges or diverges, and jusify your claim....
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(Level 1) Associate Educator, Expert, Newton
First we split the integral into two pieces:
`int_0^oo = int_0^1 + int_1^oo`
Consider the first integral, `int_0^1`
If `0<=x<=1` , then `1 <= e^x <= e`
So `1 >= 1/(e^x) >= 1/e`
`int_0^1 1/x 1/(e^x) dx >= int_0^1 1/x 1/e dx = 1/e int_0^1 1/x dx`
`int 1/x dx = "ln" |x|`
`lim_(a->o) int_a^1 1/x dx =`
`lim_(a->0) ["ln" |x| ] |_a^1`
`=lim_(a->0) (0-"ln" a)`
`int_0^1 1/x dx` diverges, so `1/e int_0^1 1/x dx` diverges
Thus `int_0^1 1/(xe^x) dx >=1/e int_0^1 1/x dx ` diverges
And we don't even have to figure out the second integral, `int_1^oo`
Posted by mlehuzzah on December 13, 2012 at 12:14 AM (Answer #1)
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