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ProPatto16
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Homework Statement
determine the value of the improper integral when using the integral test to show that [tex]\sum[/tex] k / e^k/5 is convergent.
answers are given as
a) 50/e
b) -1 / 5e^1/5
c) 5
d) 5e
e)1/50e
The Attempt at a Solution
f(x) = xe^-x/5 is continuos and positive for all values x>0
to see where its decreasing, take derivitive and see where f'(x) < 0
f'(x) = -x/5*e^-x/5 + e^-1/5
this function is < 0 only when x>5... so f(x) is continuous, positive and decreasing when x>5... therefore lower limit is 5.
[tex]\int[/tex]f(x).dx = [tex]\int[/tex] xe^-x/5 .dx
= [-x/5*e^-x/5 + e^-1/5] between b and 5 where b replaces infinity.
* Note here that f'(x) = [tex]\int[/tex] f(x)... I am not sure why but that's what i got.
if i sub in the lower bound 5 i get -e^-1-e^-1
which doesn't match anything :S
stuckkkkkk.