Double integral:

$$\int \int_D e^{-2x-3y} dxdy$$
$$D={(x,y): x\geq 0, y\geq 0, 2x+3y\leq 6}$$

Question

Double integral: 

$$\int \int_D e^{-2x-3y} dxdy$$
$$D={(x,y): x\geq 0, y\geq 0, 2x+3y\leq 6}$$

anonymous · Answer

I recommend you to first draw a sketch of the region or in this case maybe Dimension denoted as D.

anonymous · Answer

They already give you the order of dimensions you're integrating with, $dxdy$.

So you want to see how $x$ depends on $y$.

anonymous · Answer

|dw:1363274930063:dw|

anonymous · Answer

Almost. Divide by 3 on both sides and 
$$y\leq2-\frac{2}{3}x$$

anonymous · Answer

Exactly that is your region, so what you usually do in this step is fix a region and then see how x changes with respect to y. For the double integral to work out, the innermost integral has to be a function of the outer must integral after integration.

Meaning like that:

|dw:1363275368375:dw|

hartnn · Answer

$\large e^{-2x-3y} =e^{-2x}e^{-3y}$ 
integrating first w.r.t.x
|dw:1363275993204:dw||dw:1363276120323:dw|
sorry for bad drawing :P