- #1

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.