Question

If:

Answer 1

\[\int\limits_{-3}^{0} f(x)dx = 2 \] and \[\int\limits_{2}^{3} g(x)dx = -4\] what is the value of \[\int\limits_{?}^{?}\int\limits_{?}^{?} f(x)g(y) dA\] where D is the square:

Answer 2

\[-3\le x \le 0\]

Answer 3

\[2 \le y \le 3\]

Answer 4

it depends on the order of integration. If you integrate x first, then the limits are from -3 to 0 and they apply to the inner most integral

\[\large \int\limits_{?}^{?}\int\limits_{-3}^{0} f(x)g(y) dA\]

Answer 5

the limits of integration for y are applied to the outer integral

\[\large \int\limits_{2}^{3}\int\limits_{-3}^{0} f(x)g(y) dA\]

Answer 6

and you change the dA to dxdy

the dx comes first to say you integrate wrt x first, then y comes next

\[\large \int\limits_{2}^{3}\int\limits_{-3}^{0} f(x)g(y)dxdy\]

Answer 7

hopefully you see how to finish up

Answer 8

I'm not seeing what function(s) to integrate . . .

Answer 9

It might help to pull out g(y). This entire function is a constant because g(y) has no x's present in it. It's a function of y.

That allows you to do this

\[\large \int\limits_{2}^{3}\int\limits_{-3}^{0} f(x)g(y)dxdy\]
\[\large \int\limits_{2}^{3}g(y)\left(\int\limits_{-3}^{0} f(x)dx\right)dy\]

Answer 10

you can't pull g(y) out entirely because g(y) is no longer a constant for the outer integral (wrt y)

Answer 11

Okay so when you take the antiderivative of g(y) would it just be y? or would it be (y^2)/2 . . . I'm unsure what function I'm actually integrating.

Answer 12

don't worry about the outer integral right now

Answer 13

focus on the inner integral

\[\large \int\limits_{-3}^{0} f(x)dx\]

Answer 14

normally you'd integrate using rules you learn in calc 2, but you're not given the definition of f(x). However, they do provide the value of the definite integral above.

Answer 15

okay, so the value is 2

Answer 16

correct, allowing you to say

\[\large \int\limits_{2}^{3}g(y)\left(\int\limits_{-3}^{0} f(x)dx\right)dy = \int\limits_{2}^{3}g(y)\left(2\right)dy\]

Answer 17

then the 2 is a constant which you can pull out to get

\[\large 2\int\limits_{2}^{3}g(y)dy\]

Answer 18

alright

Answer 19

Ooh, okay. I think I see where this problem is going. Basically you can multiply the two values of the integrals together to get the answer?

Answer 20

yeah pretty much, so

\[\large \int\limits_{2}^{3}\int\limits_{-3}^{0} f(x)g(y)dxdy = -8\]
which means
\[\large \int\limits\int f(x)g(y)dA = -8\]
where A is the square created by \(-3 \le x \le 0\) and \(2 \le y \le 3\)

Answer 21

And you can reverse the order of integration if you wanted. You could start off integrating with respect to y first, then move onto x. You'd get the same answer either way.

Answer 22

okay, thanks for the help and sorry about the confusion

Answer 23

but you would have dydx if you did that

Answer 24

that's ok