x is from 0 to 1
y is from 3 to 3
$\displaystyle
\int_{0}^{3}\int_{0}^{1}(2x+10y+4)dxdy=\int_{0}^{3 }\int_{0}^{1}(2xdxdy+10ydxdy+4dxdy)=
$
$\displaystyle
=\int_{0}^{3}2xdx\int_{0}^{1}(+10y+4)dy
$
correct?
you can't break integral that's not correct
$\displaystyle \int_{0}^{3}\int_{0}^{1} (2x+10y+4)dxdy $ integrate with respect to x first
$\displaystyle \int_{0}^{3} \frac{2x^2}{2} + 10yx +4x \mid_{0}^{1}$
$\displaystyle \int_{0}^{3} 1+10y+4 dy $ integrate with respect to y