How do you find the antiderivative of $e^{- 5} x$ $e^-5x$ using substitution?

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How do you find the antiderivative of $e^{- 5} x$ $e^-5x$ using substitution?

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Answer 1 · 2015-02-03T14:04:36

I suspect that you want the antiderivative of $e^{- 5 x}$ $e^(-5x)$ ...
In fact the anti derivative of $e^{- 5} x$ $e^-5x$ is simply: $e^{- 5} \frac{x^{2}}{2}$ $e^-5x^2/2$ (you don't need substitution... $e^{- 5}$ $e^-5$ is a constant).
However, if you want the antiderivative of $e^{- 5 x}$ $e^(-5x)$ you have (if you really want to use substitution):
$\int e^{- 5 x} d x =$ $inte^(-5x)dx=$
set $- 5 x = t$ $-5x=t$
$x = - \frac{t}{5}$ $x=-t/5$
$d x = - \frac{d t}{5}$ $dx=-dt/5$
So you get:
$\int e^{t} \cdot - \frac{d t}{5} = - \frac{1}{5} e^{t} + c$ $inte^t*-dt/5=-1/5e^t+c$
and substituting back:
$= - \frac{1}{5} e^{- 5 x} + c$ $=-1/5e^(-5x)+c$

Hope it helps

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How do you find the antiderivative of $e^{- 5} x$ $e^-5x$ using substitution?

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