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Hard to tell if we can't see it ;)

integral 0-2 (integral 0-x/2 5(e)^(x)^2 dy dx)...help me please

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

no

5x?

its 5x since the derivative of 5x is 5

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?

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

ok

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