#int_0^oo 17 e^(-10s)ds = # ?

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Answer 1 · 2017-12-03T01:28:46

#17/10^2#

Considering

#I = int_0^oo c s e^(-alpha s) ds# we have

#d/(ds)(se^(-alpha s)) = e^(-alpha s) -alpha se^(-alpha s)# or

#c d/(ds)(se^(-alpha s)) = c e^(-alpha s) -alpha cse^(-alpha s)# or

#(c se^(-alpha s))_0^oo = int_0^oo c e^(-alpha s) ds -alpha I# and then

#I = 1/alpha(int_0^oo c e^(-alpha s) ds -(c se^(-alpha s))_0^oo)# but

#int_0^oo c e^(-alpha s) ds = -1/alpha c (e^(-alpha s))_0^oo# and then

#I = 1/alpha(-1/alpha(ce^(-alpha s))_0^oo-(c se^(-alpha s))_0^oo)# or

#I = 1/alpha(-1/alphac(0-1)) -c(0-0)) = c/alpha^2#

then

#int_0^oo 17 e^(-10s)ds = 17/10^2#