double integral difficulty

[thebenji]

∫∫R (6x+2y+12)dA where R is the region: [0,1] x [0,3]

Here's how I do it:

=∫∫R (6x+2y+12)dydx

=∫(6x+y²+12y] evaluated from 0 to 3 * dx

=∫ (6x+9+36)-(6x+0+0)dx

=∫45dx

=45x evaluated from 0 to 1

=45-0

=45

What am I doing wrong?

 

[galactus]

Your notation is wacky. Is this it:

\(\displaystyle \L\\\int_{0}^{1}\int_{0}^{3}(6x+2y+12)dydx\)


\(\displaystyle \L\\\int_{0}^{3}(6x+2y+12)dy=|_{0}^{3}(y^{2}+6xy+12y)=18x+45\)


\(\displaystyle \L\\\int_{0}^{1}(18x+45)dx=54\)

 

[thebenji]

Oh! I didn't truly treat the 6x as a constant when I integrated...I pulled it out as 6x instead of making it 6xy like any other constant. Thanks!