Calculate the double integral (Calculus 3)

Calculate the double integral (http://imageshack.us/photo/myimages/542/1c99b14df9f17164197fea0.png/) where is the region R is :
http://imageshack.us/photo/my-images/41/83188845.png/

Because the region is a rectangle, you can simply take the integral to be from 0 to 4 on y and 0 to 1 on x (you don't need to find any relation as you would if it was a function line which bounded). You can also do it in either order. I'll integrate over x first, as the 0 to 1 range will tend to simply be easier to look at.

The last thing to remember is that when you're integrating with respect to x, y is treated as a constant (and similarly x when integrating with respect to y).

So the integral from 0 to 1 of (8x+2y+16) can be seen as evaluating the anti derivative: 4x^2+2yx+16x = F(x).

This gives you F(1)-F(0) = (4(1)^2 + 2y(1) + 16(1)) - (4(0)^2 + 2y(0) + 16(0)) = 4+2y+16 = 20+2y.

Now integrate this over y and evaluate from 0 to 4: G(x) = anti derivative of 20+2y = 20y+y^2
G(4) = 80+16 - 96
G(0) = 0

So the final answer is 96.