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Is it true that

$\int_a^b \int_c^d f(x)g(y)dydx = \int_a^b f(x)dx \cdot \int_c^dg(y)dy $

My intuition says it is true, but I also have the feeling that I am missing something, but I cannot prove it. In my application $a,b,c,d$ are length variables. $f(x)$ and $g(y)$ are sinusoidal functions.

I would like to prove the general case, can somebody help me out?

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    $\begingroup$ both functions are independent so it can be split up like this $\endgroup$
    – Henry Lee
    Jan 9 '19 at 11:34
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    $\begingroup$ in the same way $\left(\int_a^bf(x)dx\right)^2=\int_a^b\int_a^bf(x)f(y)dxdy$ $\endgroup$
    – Henry Lee
    Jan 9 '19 at 11:35
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Note that for any integrable $f \colon \def\R{\mathbb R}\R \to \R$ and any $\zeta \in \R$ we have $$ \int_\R \zeta f(x)\, dx = \zeta \int_\R f(x) \, dx $$ Now note that with $\zeta := \int_\R f(y)\, dy$ which depends on $f$, but since $f$ is a constant this gives $$ \int_\R \left(\int_\R f(y)\, dy\right)\, f(x)\, dx = \int_\R f(y)\, dy \cdot \int_\R f(x)\, dx $$ Now, for every fixed $x \in \R$, just apply the above result, now for $\zeta = f(x)$ (not dependent on $y$): $$ \left(\int_\R f(y)\, dy\right)\, f(x) = \int_\R f(x)f(y)\, dy $$ Thus $$ \int_\R \int_\R f(y)f(x)\, dy\, dx = \int_\R f(y)\, dy \cdot \int_\R f(x)\, dx $$

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    $\begingroup$ Thank you very much for the quick response, this (and the other answers) helped me out! $\endgroup$
    – seaver
    Jan 9 '19 at 11:48
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Here is a try: $$I=\int_a^b\int_c^df(x)g(y)dxdy$$ where $\int f(x)dx=F(x)$ and $\int g(y)dy=G(y)$ we can start by saying: $$I=\int_a^b\int_c^df(x)g(y)dxdy=\int_c^d\left[F(x)\right]_a^b g(y)dy=\left[F(x)\right]_a^b \left[G(x)\right]_c^d$$ as $\left[F(x)\right]_a^b$ is a constant. They are separable

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The integral is linear so for any constant $K$ you have: $$\int_a^b K \, f(x)\,\mbox{d}x=K\int_a^b f(x)\,\mbox{d}x$$

and with respect to $y$, $f(x)=K_1$ is a constant and likewise the expression $\int_c^d g(y)\,\mbox{d}y = K_2$ is a constant: $$\int_a^b \int_c^d \overbrace{f(x)}^{K_1} g(y)\,\mbox{d}y\,\mbox{d}x =\int_a^b \left( \overbrace{f(x)}^{K_1}\underbrace{\int_c^d g(y)\,\mbox{d}y}_{K_2} \right)\,\mbox{d}x=\underbrace{\int_c^d g(y)\,\mbox{d}y}_{K_2}\int_a^b f(x)\,\mbox{d}x$$

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