What is the antiderivative of $e^{- 3 x}$ $e^(-3x)$ ?

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What is the antiderivative of $e^{- 3 x}$ $e^(-3x)$ ?

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Answer 1 · 2016-03-26T13:51:26

$- \frac{1}{3} e^{- 3 x} + C$ $-1/3e^(-3x)+C$

Explanation:

The question you are asking is:

$\int e^{- 3 x} d x$ $inte^(-3x)dx$

We will want to put this in the form

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

So, let $u = - 3 x$ $u=-3x$ , then $d u = - 3 d x$ $du=-3dx$ .

So that there can be a $- 3 d x$ $-3dx$ inside the integral, multiply by $- 3$ $-3$ . Compensate for this by multiplying the outside of the integral by $- 1 / 3$ $-1//3$ .

$= - \frac{1}{3} \int e^{- 3 x} (- 3) d x$ $=-1/3inte^(-3x)(-3)dx$

Substitute using the $u$ $u$ and $d u$ $du$ values we defined earlier.

$= - \frac{1}{3} \int e^{u} d u$ $=-1/3inte^udu$

This becomes

$= - \frac{1}{3} e^{- 3 x} + C$ $=-1/3e^(-3x)+C$

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What is the antiderivative of $e^{- 3 x}$ $e^(-3x)$ ?

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