Double integral on one function of x and another of y

[carlosbgois]

Hi there. I think I have proved on little theorem on double integrals, showed below. Are my arguments 'correct' (I mean, rigorous enough)?

Let f be a function of x, f(x), g be a function depending only on y, g(y), and last, let A be the set determined by a≤x≤b and c≤y≤d. By Fubini's theorem, \int\int_{A}f(x)g(y)dxdy=\int^{d}_{c}[\int^{b}_{a}f(x)g(y)dx]dy. Knowing g is a function only dependent on y, it is a constant when integrating over x, hence we have \int^{d}_{c}[g(y)\int^{b}_{a}f(x)dx]dy. Still, as f depends only on x, so does any integral over f, then finally \int^{d}_{c}g(y)dy \int^{b}_{a}f(x)dx.

Many thanks.

 

[DonAntonio]

Hi there. I think I have proved on little theorem on double integrals, showed below. Are my arguments 'correct' (I mean, rigorous enough)?

Let f be a function of x, f(x), g be a function depending only on y, g(y), and last, let A be the set determined by a≤x≤b and c≤y≤d. By Fubini's theorem, \int\int_{A}f(x)g(y)dxdy=\int^{d}_{c}[\int^{b}_{a}f(x)g(y)dx]dy. Knowing g is a function only dependent on y, it is a constant when integrating over x, hence we have \int^{d}_{c}[g(y)\int^{b}_{a}f(x)dx]dy. Still, as f depends only on x, so does any integral over f, then finally \int^{d}_{c}g(y)dy \int^{b}_{a}f(x)dx.

Many thanks.

Kudos if you reached this result by yourself, but in fact it follows in a pretty simple way from well known properties of integrals.

DonAntonio

 

[carlosbgois]

Yes it does follows easily, nice to know it is correct. It can simplify many problems as I was testing just now. Thanks!