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 ?
Asked
Active
Viewed 3,192 times
3
-
The solutions to this are origin of my doubt. – ABC Apr 08 '13 at 17:20
-
4Your grand formula only makes sense for "definite" (short for "definite integrals"?): $$\int_A f(x)\ {\rm d}x \cdot \int_B g(y)\ {\rm d}y=\int_{A\times B} f(x)\cdot g(y)\ {\rm d}(x,y)\ .$$ – Christian Blatter Apr 08 '13 at 18:13
-
Please Give me more answers $\downarrow $ for $\uparrow$ – ABC Apr 09 '13 at 12:31
2 Answers
3
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.
Abel
- 7,312
-
What would this simplify down to if dealing with the following: ∫f(x)dx×∫g(x)dx ? I have a feeling it will be the same answer since $y$ is a dummy variable, right? – Matthew James Feb 01 '23 at 10:34
1
(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.
Christian Blatter
- 226,825