How to prove $\int f (x)dx\times \int g(y)dy =\iint f(x)\cdot g(y)\, dx\, dy$ How to we prove $$\int f(x) \,dx\times \int g(y) \,dx=\iint f(x)\cdot g(y)\, dx\, dy$$
Also will this hold for definite too ?
 A: Since $\int f(x)\,\mathrm{d}x$ is constant in $y$ and since integration is linear we have
$$\left(\int f(x)\,\mathrm{d}x\right)\int g(y)\,\mathrm{d}y = \int\left(\int f(x)\,\mathrm{d}x\right)g(y)\,\mathrm{d}y.$$
Next, since $g(y)$ is constant in $x$, we similarly have
$$\int\left(\int f(x)\,\mathrm{d}x\right)g(y)\,\mathrm{d}y = \int\int f(x)g(y)\,\mathrm{d}x\,\mathrm{d}y.$$
This holds for definite integrals too, provided the range of integration is constant with respect to $y$ as well.
A: (I'm expanding on my comment.)
Given an interval $I\subset{\mathbb R}$ and a continuous function $f:\ I\to{\mathbb R}$ the expression
$$\int f(x)\ dx$$
denotes the set of all primitives of $f$ on $I$, i.e. the set of all functions $F:\ I\to{\mathbb R}$ with the property that $F'(x)=f(x)$ for all $x\in I$. Given one such primitive $x\mapsto F_0(x)$ the full set of primitives is given by
$$\int f(x)\ dx=\{x\mapsto F_0(x)+C\ |\ C\in{\mathbb R}\}$$
(I'm sure you knew that). 
Given another interval $J$ and a second continuous function $g:\ J\to{\mathbb R}$ one obtains a second such set of functions. A priori the product of the two function sets is undefined. With some stretch of the imagination one could say that the product $\int f(x)\ dx\cdot\int g(y)\ dy$ is the set of all functions of the form
$$(x,y)\mapsto (F(x)+C)(G(y)+C')\ .$$
So far, so good; but I just cannot interpret the right side of your formula.
