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

Question

Scarlett Allison · Accepted Answer

#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#

Charles Arocha · Answer

undefined To evaluate the integral ∫₀^∞ 17e^(-10s)ds, we use the formula for the integral of an exponential function. The integral converges, so we proceed with the evaluation:

∫₀^∞ 17e^(-10s)ds = lim┬(b→∞)⁡〖∫_0^b 17e^(-10s)ds〗

Using the integral of e^(-10s), which is -1/10e^(-10s), we get:

= lim┬(b→∞)⁡[17(-1/10)e^(-10s)]₀^b

= lim┬(b→∞)⁡[(-17/10)e^(-10b) - (-17/10)e^(0)]

As b approaches infinity, e^(-10b) approaches 0, so the term (-17/10)e^(-10b) approaches 0. Thus, the integral simplifies to:

= (-17/10)e^(0) - 0

= -17/10 × 1

= -17/10

Therefore, ∫₀^∞ 17e^(-10s)ds = -17/10.