Triple integral logistics? halp please ^.^

so i integrated -z+4dz = -z^2/2 + 4z |limits
can i take out 1/2 from the first term and 4 from the second term before i plug my limits in? and then combine them? so i'd be putting a 2 to the outside of the next integral..

or do i have plug the limits in first?

example inside ^.^

Question

Triple integral logistics? halp please ^.^

so i integrated -z+4dz = -z^2/2 + 4z |limits
can i take out 1/2 from the first term and 4 from the second term before i plug my limits in? and then combine them? so i'd be putting a 2 to the outside of the next integral.. 

or do i have plug the limits in first?

example inside ^.^

anonymous · Answer

$$\int\limits_{0}^{2x}-z+4dz = -z^2/2 + 4z $$
can i take out 2 here, before plugging in the limits? so i'm plugging 2x into -z^2+z instead?

anonymous · Answer

and i'm actually trying to take out 4 and 1/2 individually, and it ends up being 2..
$$(1/2) (4) \int\limits_{0}^{1}dx$$

anonymous · Answer

i would do it as a common factor....

anonymous · Answer

so take the 1/2 out and turn 4z to 2z

anonymous · Answer

and then plug in the limits?

anonymous · Answer

yeah that would be your next integral

anonymous · Answer

but waht if they don't have a common factor, like what if it was 2/3z and 4/7z ?

anonymous · Answer

wait 4z to 8z my bad

anonymous · Answer

then you would probably have to find the gcf for hat

anonymous · Answer

your probably is making me nervous haha as if you're not 100% sure

anonymous · Answer

yeah...this is more of a calc 1 issue but it folows all of the normal distributive rules when moving a factor outside of the integral

anonymous · Answer

so if you were distributing something like  (4x + 6x) you wouldnt do 4(6)(x), you would do 2(2x+ 3x)...same logic

anonymous · Answer

but i don't think i'm factoring anything out... =\ am i?

anonymous · Answer

taking the coefficicients to the front is basically doing that

anonymous · Answer

but.. you can like take out 2/3 out 2/3x+y, right? and then that's not factoring, cuz there isn't a common factor... 

i've seriously confused myself over this lol

anonymous · Answer

if that was in an integral? no you wouldnt be able to do that

anonymous · Answer

no? really? shoot...

anonymous · Answer

yeah. if you can take the integral of the original equation, it would probably be less confusing for you...

anonymous · Answer

what do you mean?

anonymous · Answer

its possible to take the original integral you gave

anonymous · Answer

but then doesn't it get super messy?

anonymous · Answer

my teacher is always taking stuff out, and i just realized, i don't understand her method lol