1. integrate e^something

why does e^2x = 1/2 e^2x

does e^-x = -e^-x or -1/e^-x ?

2. Originally Posted by jeph
why does e^2x = 1/2 e^2x

does e^-x = -e^-x or -1/e^-x ?
First,
$e^{2x}$ DOES NOT EQUAL $1/2e^{2x}$

We need to find,
$\int e^{2x} dx$
Use the substitution $t=2x$
$\frac{1}{2} \int e^t dt = \frac{1}{2}e^t+C=\frac{1}{2}e^{2x}+C$

Similarly with the other one.

3. i dont get how the 1/2 gets there...

4. Originally Posted by jeph
i dont get how the 1/2 gets there...
Because if $t=2x$ then $t'=2$ so we need to introduce a factor of two into that expression. Thus, I simply multiply it in and divide it out.

5. $\int e^{kx}\,dx=\frac1ke^{kx}+c$

where $k$ is an arbitrary constant.

6. Originally Posted by Krizalid
$\int e^{kx}\,dx=\frac1ke^{kx}+c$

where $k$ is an arbitrary constant.
And $k\not = 0$

7. Originally Posted by ThePerfectHacker
And $k\not = 0$
but of course

8. Yeah, that's what I said "arbitrary"

(common sense)

9. Originally Posted by Krizalid
Yeah, that's what I said "arbitrary"

(common sense)
i guess, but mathematicians don't take anything for granted, and therefore, they are in the habit of rigorously defining common sense things.

for instance, to define a rational number, a mathematician would say:

a rational number $r$ is one that can be expressed as $r = \frac {a}{b} \mbox { for } a,b \in \mathbb {Z}, b \neq 0$

even though the $b \neq 0$ is the "common sense" part

10. Ahahahaha, got it!!

You're right, we gotta be stricts.