How do you find the integral of $e^{- x} d x$ $e^(-x)dx$ ?

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How do you find the integral of $e^{- x} d x$ $e^(-x)dx$ ?

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Answer 1 · 2018-05-15T07:29:04

it's $- e^{- x} + c$ $-e^(-x) + c$

Explanation:

let $u = - x$ $u = -x$ then $\frac{d u}{d x} = - 1$ $(du)/dx = -1$ which means $d x = - d u$ $dx = -du$
then
$\int (e^{- x}) d x = \int (- e^{u}) d u = - \int (e^{u}) d u = - e^{u} + c = - e^{- x} + c$ $int (e^(-x))dx = int (-e^u)du = -int (e^u)du = -e^u + c= -e^(-x) + c$

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How do you find the integral of $e^{- x} d x$ $e^(-x)dx$ ?

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