Question

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

Answer 1

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

Answer 2

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...

Answer 3

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

Answer 4

first we integrate inside ignore the outside for right now

Answer 5

since we are integrating with repsect to y we treat everything thats not y ish as a constant

Answer 6

i under stand that. There's no y so I dont understand if i just put a y in there?

Answer 7

what? so you are saying you don't see the inside being 5ye^x^2?

Answer 8

no, there's not a y in the whole problem

Answer 9

whats the antiderivative of 5 with repect to x

Answer 10

0

Answer 11

no

Answer 12

5x?

Answer 13

its 5x since the derivative of 5x is 5

Answer 14

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)

Answer 15

ok so you added a y because we were doing respect to y?

Answer 16

yes and the derivative of 5ye^x^2 with respect to y will be 5e^x^2

Answer 17

so the inside integral becomes (5(x/2)e^x^2-0)

Answer 18

now we have the outside integral that we integrating with respect to x

Answer 19

use a substitution for x^2

Answer 20

du/2=x

Answer 21

du/2=x dx yes

Answer 22

dont forget to put the limits in u terms

Answer 23

so the upper limit will be 4 and the lower limit will be 0

Answer 24

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

Answer 25

-1

Answer 26

i forgot about the minus 1 sorry

Answer 27

e^0=1

Answer 28

thank you very much

Answer 29

np don't be afraid to ask any other questions or this one over if you want

Answer 30

is this a 24 hour website?

Answer 31

on this problem, and when doing substitution is it always required to put the limits in u terms?

Answer 32

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

Answer 33

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

Answer 34

is there any way you can explain that?

Answer 35

its the same thing

Answer 36

lets do an example...

Answer 37

this seems like it's going to be a really hard problem

Answer 38

no its not hard i'm just busy doing something else at sametime

Answer 39

pretend we have int(5x,x=0..1)

Answer 40

if we integrate this normal way we get 5x^2/2 x=0..1 = 5/2

Answer 41

we could also do u substitution if we just wanted to for fun so let u=5x

Answer 42

du=5dx => du/5=dx

Answer 43

ok

Answer 44

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

Answer 45

ok

Answer 46

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

Answer 47

that didn't answer it?

Answer 48

i mean i have another question.

Answer 49

oh lol yeah I have to do something. I'm fixing to have to move a mattress

Answer 50

ok, thanks for all your help. hope you make lots of money with your math skill one day