I'm having a problem with solving this double integral...can someone help me?

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I'm having a problem with solving this double integral...can someone help me?

Mathematics
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Hard to tell if we can't see it ;)
ok there's no way i can put it up like im seeing it because there's integral sign but i'll try my best...
integral 0-2 (integral 0-x/2 5(e)^(x)^2 dy dx)...help me please

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Other answers:

first we integrate inside ignore the outside for right now
since we are integrating with repsect to y we treat everything thats not y ish as a constant
i under stand that. There's no y so I dont understand if i just put a y in there?
what? so you are saying you don't see the inside being 5ye^x^2?
no, there's not a y in the whole problem
whats the antiderivative of 5 with repect to x
0
no
5x?
its 5x since the derivative of 5x is 5
so now you understand how we got 5ye^x^2 for inner integral (we haven't used upper and lower limit yet for inner integral)
ok so you added a y because we were doing respect to y?
yes and the derivative of 5ye^x^2 with respect to y will be 5e^x^2
so the inside integral becomes (5(x/2)e^x^2-0)
now we have the outside integral that we integrating with respect to x
use a substitution for x^2
du/2=x
du/2=x dx yes
dont forget to put the limits in u terms
so the upper limit will be 4 and the lower limit will be 0
so we have int((5/4) *e^u, u=0..4)=(5/4)e^u, u=0..4= (5/4)e^4-(5/4)e^0=(5/4)e^4
-1
i forgot about the minus 1 sorry
e^0=1
thank you very much
np don't be afraid to ask any other questions or this one over if you want
is this a 24 hour website?
on this problem, and when doing substitution is it always required to put the limits in u terms?
i don't know. people come and go as they please. the tutors work whenever they want because we don't work for the website we work because we think its fun lol
if you don't put the limits in terms of u, you have to change it back to x and use the limits for x
is there any way you can explain that?
its the same thing
lets do an example...
this seems like it's going to be a really hard problem
no its not hard i'm just busy doing something else at sametime
pretend we have int(5x,x=0..1)
if we integrate this normal way we get 5x^2/2 x=0..1 = 5/2
we could also do u substitution if we just wanted to for fun so let u=5x
du=5dx => du/5=dx
ok
so we get int(1/5 (u), u=blah blah) =1/5 *u^2/2 let's change this back to x so we will have 1/5*((5x))^2/2 x=0..1 but we could have already done the 5x if we had used u=5x
ok
is it ok to ask another question or should i just retype it on the "ask a question"? you seem to really know your stuff thats why im asking
that didn't answer it?
i mean i have another question.
oh lol yeah I have to do something. I'm fixing to have to move a mattress
ok, thanks for all your help. hope you make lots of money with your math skill one day

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