How do you integrate $e^{- x} d x$ $e^(-x)dx$ ?

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How do you integrate $e^{- x} d x$ $e^(-x)dx$ ?

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Answer 1 · 2016-04-26T16:35:15

$- e^{- x} + C$ $-e^-x+C$

Explanation:

We will use the following integral rule:

$\int e^{u} d u = e^{u} + C$ $inte^udu=e^u+C$

Thus, to integrate

$\int e^{- x} d x$ $inte^-xdx$

We set $u = - x$ $u=-x$ . Thus, we have:

$u = - x \Rightarrow \frac{d u}{d x} = - 1 \Rightarrow d u = - d x$ $color(red)(u=-x)" "=>" "(du)/dx=-1" "=>" "color(blue)(du=-dx)$

Since we have only $d x$ $dx$ in the integral and not $- d x$ $-dx$ , multiply the interior of the integral by $- 1$ $-1$ . Balance this by multiplying the exterior by $- 1$ $-1$ as well.

$\int e^{- x} d x = - \int e^{- x} (- 1) d x = - \int e^{u} d u$ $inte^-xdx=-inte^color(red)(-x)color(blue)((-1)dx)=-inte^color(red)ucolor(blue)(du)$

This is the rule we knew originally. Don't forget that the integral is multiplied by $- 1$ $-1$ :

$-inte^udu=-e^u+C=barul|color(white)(a/a)-e^-x+Ccolor(white)(a/a)|$

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How do you integrate $e^{- x} d x$ $e^(-x)dx$ ?

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