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?

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- Aug 13th 2009, 11:18 PMtransgalactichow to break it into dx and dy
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? - Aug 13th 2009, 11:54 PMAmer
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 - Aug 13th 2009, 11:56 PMtransgalactic
thanks :) i understan now