1. The problem statement, all variables and given/known data 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 3. 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)... im not sure why but thats what i got. if i sub in the lower bound 5 i get -e^-1-e^-1 which doesnt match anything :S stuckkkkkk.