How do you evaluate #\int _ { 0} ^ { \infty } [ \frac { 2} { e ^ { x } ] ] d x#?

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Answer 1 · 2017-12-07T13:29:02

Please see below.

#int_0^oo 2/e^x dx = lim_(brarroo) int_0^b 2/e^x dx#

if the limit exists.

Use substitution to find the indefinite integral

#int 2/e^x dx = int 2e^-x dx = underbrace(-2 int e^-x (-dx))_(u=-x) = -2e^-x = -2/e^x#.

So we have

#int_0^oo 2/e^x dx = {:lim_(brarroo)-2/e^x]_0^b#

# = lim_(brarroo)-2/e^b - (-2/e^0)#

As #brarroo#, #e^brarroo#, so #2/e^brarr0# (And, #e^0 = 1#) So

#int_0^oo 2/e^x dx = 0-(-2) = 2#

Answer 2 · 2017-12-07T14:12:35

#int_0^oo 2/e^x*dx=2#

#int_0^oo 2/e^x*dx#

=#int_0^oo 2e^(-x)*dx#

=#[-2e^(-x)]_0^oo#

=#2#