how to break it into dx and dy

**transgalactic** · August 13th 2009, 11:18 PM

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?

**Amer** · August 13th 2009, 11:54 PM

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

**transgalactic** · August 13th 2009, 11:56 PM

thanks

i understan now

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