Question

What is the antiderivative of e^-2x/5 ?

Answer 1

let u = -2x/5

can you work it?

Answer 2

Is it \[\LARGE e^{-\frac{2x}{5}}\]
or
\[\LARGE\frac{e^{-2x}}{5}\]

Answer 3

So that would be \[e ^{u} * \frac{ du }{ dx } = f'(x) ... e ^{u} * -\frac{ 2 }{ 5 } ... -\frac{ 2 }{ 5 }e ^{2x/5}\]

Answer 4

Maybe? lol

Answer 5

Wait! Crap! That's the derivative not the antiderivative. ><;

Answer 6

that actually is a good way to see it tho

Answer 7

e^u had to come from something akin to k e^u

Answer 8

\[\frac d{dx}[k~e^{-2x/5}]\to \frac{-2}{5}k~e^{-2x/5}\]
\[\frac{-2}{5}k~e^{-2x/5}=e^{-2x/5},~solve~for~k\]

Answer 9

I'm getting \[-\frac{ 5 }{ 2}e ^{-2x/5} = ke ^{-2x/5}\]