- #1
davidbenari
- 466
- 18
This summer I was studying multivariable calculus and it did bother me how multiple integrals were for some reason reducible to iterated integrals. I thought of a proof, which in my opinion is fairly simple. I think it contains the same information that fubini's theorem states (at least to my knowledge). I wanted someone to tell me if all the steps are valid.
Let ##R=[a,b]\times[c,d]##
For simplicity let ##lim_{i,j} \to \infty \sum_{i}^{} \sum_{j}^{} f(x,y) \Delta x \Delta y = \sum_{i}^{} \sum_{j}^{} f(x,y) \Delta x \Delta y##
Fubini's theorem says:
##\int\int_R f(x,y)dA = \int_{c}^{d} \int_{a}^{b} f(x,y) dx dy##
This can be shown by the associative property of addition in the following manner:
##\sum_{i}^{} \sum_{j}^{} f(x,y) \Delta x \Delta y = \sum_{j}^{} \Big(\sum_{i}^{} f(x,y) \Delta x \Big) \Delta y ##By Riemann definition of integration
##\Big(\sum_{i}^{} f(x,y) \Delta x \Big)=\int_{a}^{b} f(x,y)dx=A(y) ##
Then
##\sum_{j}^{} \Big(\sum_{i}^{} f(x,y) \Delta x \Big) \Delta y =\sum_{j}^{} A(y) \Delta y = \int_{c}^{d} A(y) dy##
Therefore
##\sum_{j}^{} \sum_{i}^{} f(x,y) \Delta x \Delta y = \int_{c}^{d} \Big(\int_{a}^{b} f(x,y)dx \Big) dy##
Which by definition of iterated integrals ##= \int_{c}^{d} \int_{a}^{b} f(x,y)dx dy##
This procedure can also be used to show that the order of integration can be flipped around.
I wanted to know if this is valid, and if it in any way simplifies stuff?
I haven't seen and don't have the skills to understand Fubini's actual proof. But I just read in Stewarts calculus book that the proof is "too complicated". Is my reasoning correct? Can I follow this line of thought in the meanwhile?
Thanks.
Let ##R=[a,b]\times[c,d]##
For simplicity let ##lim_{i,j} \to \infty \sum_{i}^{} \sum_{j}^{} f(x,y) \Delta x \Delta y = \sum_{i}^{} \sum_{j}^{} f(x,y) \Delta x \Delta y##
Fubini's theorem says:
##\int\int_R f(x,y)dA = \int_{c}^{d} \int_{a}^{b} f(x,y) dx dy##
This can be shown by the associative property of addition in the following manner:
##\sum_{i}^{} \sum_{j}^{} f(x,y) \Delta x \Delta y = \sum_{j}^{} \Big(\sum_{i}^{} f(x,y) \Delta x \Big) \Delta y ##By Riemann definition of integration
##\Big(\sum_{i}^{} f(x,y) \Delta x \Big)=\int_{a}^{b} f(x,y)dx=A(y) ##
Then
##\sum_{j}^{} \Big(\sum_{i}^{} f(x,y) \Delta x \Big) \Delta y =\sum_{j}^{} A(y) \Delta y = \int_{c}^{d} A(y) dy##
Therefore
##\sum_{j}^{} \sum_{i}^{} f(x,y) \Delta x \Delta y = \int_{c}^{d} \Big(\int_{a}^{b} f(x,y)dx \Big) dy##
Which by definition of iterated integrals ##= \int_{c}^{d} \int_{a}^{b} f(x,y)dx dy##
This procedure can also be used to show that the order of integration can be flipped around.
I wanted to know if this is valid, and if it in any way simplifies stuff?
I haven't seen and don't have the skills to understand Fubini's actual proof. But I just read in Stewarts calculus book that the proof is "too complicated". Is my reasoning correct? Can I follow this line of thought in the meanwhile?
Thanks.
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