1. ## quick integration question

i am doing a problem by parts, and my dv= e^-2t dt

i was just wondering how one knows that v, then, =-1/2 e^-2t

2. Originally Posted by buttonbear
i am doing a problem by parts, and my dv= e^-2t dt

i was just wondering how one knows that v, then, =-1/2 e^-2t
v=-1/2 e^-2t

because the derivative of v is dv.

v=-1/2 e^-2t
then $\displaystyle dv=e^{-2t} \cdot -2 \cdot \frac{-1}{2}=dv= e^{-2t}$

That's why.

3. i know that the derivative of v is dv

i'm sorry if this is a stupid question
but if i'm starting with dv, i just don't see how you integrate dv to get v..

4. Originally Posted by buttonbear
i am doing a problem by parts, and my dv= e^-2t dt

i was just wondering how one knows that v, then, =-1/2 e^-2t
b/c

$\displaystyle \int e^{-2t}dt$

$\displaystyle =e^{-2t} \cdot \frac{1}{-2}+C$

and double check you have the right integral by differentiating it.

there are many steps involving $\displaystyle \int e^{f(x)}dx$

5. Originally Posted by buttonbear
i know that the derivative of v is dv

i'm sorry if this is a stupid question
but if i'm starting with dv, i just don't see how you integrate dv to get v..
$\displaystyle \int e^{kx}~dx = \frac 1k e^{kx} + C$, for $\displaystyle k \ne 0$ a constant.

we obtain this by doing a substitution of $\displaystyle u = kx$