How do you evaluate the integral $\int e^{- x} d x$ $inte^(-x) dx$ ?

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Answer 1 · 2014-08-20T02:43:25

The answer is $I = - e^{- x} + C$ $I=-e^(-x)+C$ .

This integral can be solved by a substitution:

$u = - x$ $u=-x$
$d u = - d x$ $du=-dx$
$- d u = d x$ $-du=dx$

So, now we can substitute:

$\int e^{- x} d x = \int e^{u} (- d u)$ $int e^(-x)dx = int e^u (-du)$
$= - \int e^{u} d u$ $=-int e^u du$
$= - e^{u} + C$ $=-e^u + C$

and substitute back:
$= - e^{- x} + C$ $=-e^(-x) + C$

For simple looking integrands, you should try a quick check to see if substitution works before trying harder integration methods.

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