Question

Evaluate the double integral

Answer 1

\[\int\limits_{}^{}\int\limits_{D}^{}x \cos(y) dA\]

where D is bounded by y=0, y=x^2, x=1

Answer 2

Have you tried graphing the region D?

Answer 3

I got this:


When zoomed in:


So then I think the limits are:
\[0 \le y \le x^2\\0 \le x \le 1\]

And the integration would look like:
\[\int\limits_{0}^{1}\int\limits_{0}^{x^2}x \cos(y) dydx\]


How is that so far?

Answer 4

you are right, continue, just take int

Answer 5

Oh thanks @Hoa

Then I ended with (I hope I didn't make any mistakes along the way...)
\[\int\limits_{0}^{1} x \sin(x^2) dx\]
Should I do int by parts now?

Answer 6

just let u = x^2 take du , change the limit respect to u and....done. hehehe

Answer 7

\[\int\limits_{}^{}xcos(y)dydx\]

Answer 8

double intergral then you put your values

you first intergrate holding x as a constant

then you do with dx holding y as constant

Answer 9

\[\int\limits_{?}^{?}\int\limits_{?}^{?}xcos(y)dydx\]

Answer 10

Doing the u substitution sounds a little easier to me.
\[u=x^2\\du=2x dx\] and x=sqrt(u) right?

So then
\[=\frac{1}{2}\int\limits_{0}^{1}\sin(u) du\]
How do I change the limits again? Because it can't be left as it was for the x, right?

Answer 11

Oh wait
\[\frac{1}{2}\int\limits_{?}^{?}\sqrt{u} \sin (u) du\]

Answer 12

no friend, from limit. when x = 0 u = 0. when x =1 fortunately u =1 , so no need to change the limit, but if you have other, it has other limit.

Answer 13

1/2 du = x dx replace only, you don't have sqr (u) recheck

Answer 14

what is all this you guys are doing

Answer 15

So the limits stay 0 and 1 for the u as well?

Ahh I see what you meant with the dx/du --> x dx is already in the int, so I can subs it in as is
So then we have this?
\[0.5 \times \int\limits_{?}^{?}\sin(u) du\]

Answer 16

I don't know, this friend mess something up

Answer 17

yeap

Answer 18

i can t really start doing this now but first start \[\int\limits_{?}^{?}xcos(y) dy\]

Answer 19

@telijahmed you take it over? I can run?

Answer 20

Hhahaha! Thanks you guys:)
Will you please @telijahmed

Answer 21

make sure that this friend get the right answer, please

Answer 22

@taljaards I have a bunch of homework to do, because it is in hand so stop by and help

Answer 23

hey, where telijahmed? @telijahmed

Answer 24

Okay no problem @Hoa , thanks for helping! I'm studying for a test as well

Answer 25

ok @taljaards i need one value of x because your question only gives me x=1

Answer 26

I have 2 tests next week

Answer 27

x=0

Answer 28

from y = x^2 guys

Answer 29

@telijahmed I also though something looked wrong... but took a look again and saw that I had typed correctly. So that's why I though of graphic the functions, as in my second pic

Answer 30

ok now \[\int\limits_{0}^{1}\int\limits_{0}^{x^2}xcos(y)dydx\]

Answer 31

does this make sense since i am starting with dy my dy is infront

Answer 32

my dy values are infront

Answer 33

That makes sense, that is also how I started. So now?

Answer 34

\[\int\limits_{0}^{x^2}xcos(y)dy\]

Answer 35

we start with this this means x is like a constant here and so that means

intergrate cos(y) alone and that gives us xsiny

Answer 36

\[=x (\sin(y))_{0}^{x^2}\]

Answer 37

\[=\int\limits_{0}^{1}x \sin(x^2) dx\]
right?

Answer 38

you put in your values now and that should give us
\[x \sin(x^2)-xsin(0)\]

Answer 39

please @taljaards follow my steps and check for any algeabric errors its been long since i solved maths

Answer 40

Hahahah @telijahmed , what you did looks okay to me

So I did that and since sin(0) = 0 that gives only x sinx^2
and that has to be put back into the original integral?

So then we have the int i last posted, right?

Answer 41

you got it up there
\[\int\limits_{0}^{1}xsin(x^2)dx\]

Answer 42

now you go ahead with your substitution method right

Answer 43

give telijahmed a medal for being patient, hehehe...

Answer 44

let x^2=u

du/dx=2x

Answer 45

\[\int\limits_{0}^{1}xsin(u)(du/2x)\]