# how to break it into dx and dy

• Aug 14th 2009, 12:18 AM
transgalactic
how to break it into dx and dy
x is from 0 to 1
y is from 3 to 3

$
\int_{0}^{3}\int_{0}^{1}(2x+10y+4)dxdy=\int_{0}^{3 }\int_{0}^{1}(2xdxdy+10ydxdy+4dxdy)=
$

$
=\int_{0}^{3}2xdx\int_{0}^{1}(+10y+4)dy
$

correct?
• Aug 14th 2009, 12:54 AM
Amer
Quote:

Originally Posted by transgalactic
x is from 0 to 1
y is from 3 to 3

$
\int_{0}^{3}\int_{0}^{1}(2x+10y+4)dxdy=\int_{0}^{3 }\int_{0}^{1}(2xdxdy+10ydxdy+4dxdy)=
$

$
=\int_{0}^{3}2xdx\int_{0}^{1}(+10y+4)dy
$

correct?

you can't break integral that's not correct

$\int_{0}^{3}\int_{0}^{1} (2x+10y+4)dxdy$ integrate with respect to x first

$\int_{0}^{3} \frac{2x^2}{2} + 10yx +4x \mid_{0}^{1}$

$\int_{0}^{3} 1+10y+4 dy$ integrate with respect to y
• Aug 14th 2009, 12:56 AM
transgalactic
thanks :) i understan now