Separating double integral Let's say I have the following integral:
$$\int_0^1\int_1^3 x+y\: dy\,dx $$
Can I separate this (like with multiplication, i.e x*y) to:
$$\int_0^1 x\:dx +\int_1^3 y\: dy $$
Is this correct?
 A: No, it is not.
The property which you may be interested in is given by
\begin{align*}
\int_{a}^{b}\int_{c}^{d}f(x)g(y)\mathrm{d}y\mathrm{d}x = \int_{a}^{b}f(x)\mathrm{d}x\int_{c}^{d}g(y)\mathrm{d}y
\end{align*}
At the given example, one has that
\begin{align*}
\int_{0}^{1}\int_{1}^{3}(x + y)\mathrm{d}y\mathrm{d}x & = \int_{0}^{1}\left(xy + \frac{y^{2}}{2}\right)\Biggr|_{y=1}^{y=3}\mathrm{d}x\\\\
& = \int_{0}^{1}(2x + 4)\mathrm{d}x\\\\
& = \left(x^{2} + 4x\right)\Biggr|_{0}^{1} = 5
\end{align*}
On the other hand, the other integrals result into the following value:
\begin{align*}
\int_{0}^{1}x\mathrm{d}x + \int_{1}^{3}y\mathrm{d}y = \frac{1}{2} + 4 = \frac{9}{2} 
\end{align*}
Hopefully this helps !
A: No.
$$\int_0^1\int_1^3 (x+y) dydx = \int_0^1\int_1^3 x dydx + \int_0^1\int_1^3 ydydx = \int_1^3\int_0^1 x dxdy + \int_0^1\int_1^3 ydydx$$
is how you use linearity for this integral.
A: In general \begin{eqnarray*}\int_a^b \int_c^d f(x)+g(y) \quad dy\, dx&=&
\int_a^b \int_c^d f(x) \quad dy\, dx+\int_a^b \int_c^dg(y) \quad dy\, dx\\\\
&=&(d-c)\int_a^b  f(x) \quad dx +(b-a)\int_c^d g(y) \quad dy.
\end{eqnarray*}
Thus in your case:
$$\int_0^1\int_1^3 x+y\: dy\,dx =2\int_0^1 x\:dx +\int_1^3 y\: dy,$$
which is almost exactly what you conjectured.
