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 $0$ :

$J=0+1/2=1/2$

How do you evaluate the integral int e^(-2x)dxint e^(-2x)dx from 00 to oooo?