How do you evaluate the integral #int e^(-2x)dx# from #0# to #oo#?

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Answer 1 · 2016-10-21T16:24:02

#int_0^ooe^(-2x)dx=1/2#

First without the bounds:

#I=inte^(-2x)dx#

With #u=-2x# and #du=-2dx#:

#I=-1/2inte^(-2x)(-2dx)=-1/2inte^udu=-1/2e^u+C#

Thus:

#I=-1/2e^(-2x)+C=-1/(2e^(2x))+C#

Applying the bounds:

#J=int_0^ooe^(-2x)dx=[-1/(2e^(2x))]_0^oo#

Evaluating and taking the limit at infinity:

#J=[lim_(xrarroo)(-1/(2e^(2x)))]-(-1/(2e^0))#

The limit goes to #0#:

#J=0+1/2=1/2#