How do you evaluate the integral #int e^(-2x)dx# from #0# to #oo#?
1 Answer
Oct 21, 2016
Explanation:
First without the bounds:
#I=inte^(-2x)dx#
With
#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
#J=0+1/2=1/2#