What is the antiderivative of $e^{- x}$ $e^-x$ ?

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What is the antiderivative of $e^{- x}$ $e^-x$ ?

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Answer 1 · 2016-01-27T15:37:31

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

Explanation:

The problem, written out, is:

$\int e^{- x} d x$ $inte^(-x)dx$

The following rule will be used:

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

Using substitution:
Set $u = - x$ $u=-x$ , so we know $\frac{d u}{d x} = - 1$ $(du)/dx=-1$ and $- d u = d x$ $-du=dx$ .

This gives us a simplified integral of

$= \int e^{u} (- d u)$ $=inte^u(-du)$

We can bring the negative sign out:

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

Using intuition:

We know that the derivative of $e^{- x}$ $e^(-x)$ is $- e^{- x}$ $-e^(-x)$ , but we want a derivative of the positive version $e^{- x}$ $e^(-x)$ . So, we make the antiderivative negative, causing the derivative of be positive:

$\frac{d}{d x} - e^{- x} = e^{- x}$ $d/dx-e^(-x)=e^(-x)$

Then, we simply add $C$ $C$ , the constant of integration.

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What is the antiderivative of $e^{- x}$ $e^-x$ ?

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