Question

Does anyone know the integral of -e^{-2y}

Answer 1

i think I can do that....

Answer 2

(1/2)e^(-2y)

Answer 3

[S] -e^-2y dy

(1/2) [S] -2 e^-2y dy

e^-2y
----- is what I get if I did it right :)
2

Answer 4

We are having a dispute as to whether or not the 2y stays negative

Answer 5

+C ....dont forget the +C lol

Answer 6

arrrgh

Answer 7

yeah thats right....i got it....go me.... its not my birthday but it was at somepoint....uhhunh

Answer 8

i h8 this someday i'll start writing alculus and someone'll go "+C!!!"

Answer 9

LOL

Answer 10

So the integral of e^{-x) = -e^{-x}+c doesnt e^{-2y} end up -(1/2)e^{-2y}+c? Or am I wrong about the -e?

Answer 11

thats exponential shouldnt change so thats correct, but it is divided by constant -2 so it works out to be positive