Question

Calculus II- Double Integral

[see attachment]

Answer 1

simple, first do this integration

\[\int\limits_{}^{} y dy\]

Answer 2

It's supposed to integrated first with respect to x and then with respect to y. But there is no x value.

Answer 3

?

you do the inner integral first
which is respect to y

Answer 4

unless the problem is asking you to change the order in which you integrate

then you need to draw the picture out and redefine the bounds

Answer 5

are there any other instructions other than evaluate the integral?

Answer 6

no... just find the double integral... so the inner integral is y^2/ 2 right?

Answer 7

yes
and its from 2x to x^2

Answer 8

so you substitute those in

Answer 9

but those are for x values I thought. so it wouldn't go into y?

Answer 10

and 5 to 1 would go into y^2/2

Answer 11

the integral to put it in another way
\[\int\limits_{y= -2x}^{y=x^2}y dy\]

Answer 12

\[(x^2)^2/2 - (-2x)^2/2\]

Answer 13

you cant change the order

ex.

\[\int\limits_{a}^{b} \int\limits_{c}^{d} xy dx dy = \int\limits_{c}^{d} \int\limits_{a}^{b} xy dy dx\]

Answer 14

but you cant just switch the cd and ab around without switching the dy and dx

Answer 15

ok anyways assuming you're right

\[\int\limits_{0}^{2} \frac{ (x^2)^2}{2} - \frac{ (-2x)^2}{2} dx\]

Answer 16

then you take the outer integral

Answer 17

... now I'm lost.

Answer 18

@Loser66
his problem was he was trying to take the integral of

\[\int\limits_{-2x}^{x^2}\int\limits_{0}^{2}y dy dx\]

Answer 19

So with my equation I gave you I subtracted them \[\frac{ x^4 }{ 2 } - 2x^2\]

Answer 20

k, then you do the other integral

Answer 21

so the integral

\[\int\limits_{0}^{2} \frac{x^4}{2} -2 x^2 dx\]

Answer 22

outer not other...

Answer 23

I get... 0?

Answer 24

really?