Proving Properties of Double Integrals

[_N3WTON_]

Homework Statement

Prove the following property:
If $m <= f(x,y) <= M \hspace{2 mm} \forall (x,y) \in D,$ then:
$mA(D) <= \int\int f(x,y)\,dA <= MA(D)$

Homework Equations

I use a few other known properties in the proof (see below)

The Attempt at a Solution

First, I should state that this problem is for Multivariable Calc. My ability to do proof's is severely lacking, and it is a skill I would like to learn (although I have yet to take a course on it, perhaps next semester). So I am hoping you all will be as critical as possible so I can learn from my mistakes. Also, I am not sure if I am using correct terminology, so please inform me if I am not. This is my attempt:
Given that m and M are real numbers such that:
$m <= f(x,y) <=M \hspace{2 mm}\forall (x,y) \in D$
It is known that:
$\int\int mdA <= \int\int f(x,y) \,dA <= \int\int MdA$
Lemma:
${if} \hspace{2 mm} f(x,y)>=g(x,y) \hspace{2 mm}\forall (x,y) \in D$
then:
$\int\int f(x,y)\,dA >= \int\int g(x,y)\,dA$
Lemma:
$\int\int cf(x,y)\,dA = c\int\int f(x,y)\,dA$
then:
$m\int\int\,dA <= \int\int f(x,y)\,dA <= M\int\int\,dA$
Lemma:
$\int\int\,dA = A(D)$
Therefore:
$mA(D) <= \int\int f(x,y)\,dA <= MA(D)$
Proved.
Again, please be very critical, I want to learn how I can become better at this. :)

 

[Mark44]

Homework Statement

Prove the following property:
If $m <= f(x,y) <= M \hspace{2 mm} \forall (x,y) \in D,$ then:
$mA(D) <= \int\int f(x,y)\,dA <= MA(D)$

What are m(A) and M(A)?

Homework Equations

I use a few other known properties in the proof (see below)

The Attempt at a Solution

First, I should state that this problem is for Multivariable Calc. My ability to do proof's is severely lacking, and it is a skill I would like to learn (although I have yet to take a course on it, perhaps next semester). So I am hoping you all will be as critical as possible so I can learn from my mistakes. Also, I am not sure if I am using correct terminology, so please inform me if I am not. This is my attempt:
Given that m and M are real numbers such that:
$m <= f(x,y) <=M \hspace{2 mm}\forall (x,y) \in D$
It is known that:
$\int\int mDA <= \int\int f(x,y) \,dA <= \int\int MdA$
Lemma:
${if} \hspace{2 mm} f(x,y)>=g(x,y) \hspace{2 mm}\forall (x,y) \in D$
then:
$\int\int f(x,y)\,dA >= \int\int g(x,y)\,dA$
Lemma:
$\int\int cf(x,y)\,dA = c\int\int f(x,y)\,dA$
then:
$m\int\int\,dA <= \int\int f(x,y)\,dA <= M\int\int\,dA$
Lemma:
$\int\int\,dA = A(D)$
Therefore:
$mA(D) <= \int\int f(x,y)\,dA <= MA(D)$
Proved.
Again, please be very critical, I want to learn how I can become better at this. :)

 

[_N3WTON_]

What are m(A) and M(A)?

m and M are real numbers...A(D) is the area of some closed, bounded region in the xy-plane

 

[Mark44]

Homework Statement

Prove the following property:
If $m <= f(x,y) <= M \hspace{2 mm} \forall (x,y) \in D,$ then:
$mA(D) <= \int\int f(x,y)\,dA <= MA(D)$

Homework Equations

I use a few other known properties in the proof (see below)

The Attempt at a Solution

First, I should state that this problem is for Multivariable Calc. My ability to do proof's is severely lacking, and it is a skill I would like to learn (although I have yet to take a course on it, perhaps next semester). So I am hoping you all will be as critical as possible so I can learn from my mistakes. Also, I am not sure if I am using correct terminology, so please inform me if I am not. This is my attempt:
Given that m and M are real numbers such that:
$m <= f(x,y) <=M \hspace{2 mm}\forall (x,y) \in D$
It is known that:
$\int\int mdA <= \int\int f(x,y) \,dA <= \int\int MdA$

Is the above given information? If not, how is it "known that ..."? You have cited the lemmas below, so I can see where they came from. I don't doubt that it's true, though.

Lemma:
${if} \hspace{2 mm} f(x,y)>=g(x,y) \hspace{2 mm}\forall (x,y) \in D$
then:
$\int\int f(x,y)\,dA >= \int\int g(x,y)\,dA$
Lemma:
$\int\int cf(x,y)\,dA = c\int\int f(x,y)\,dA$
then:
$m\int\int\,dA <= \int\int f(x,y)\,dA <= M\int\int\,dA$
Lemma:
$\int\int\,dA = A(D)$
Therefore:
$mA(D) <= \int\int f(x,y)\,dA <= MA(D)$
Proved.
Again, please be very critical, I want to learn how I can become better at this. :)

I would organize things differently. Put the lemmas at the top so that they don't interrupt the logic of your argument.
It's given that m <= f(x, y) <= M, for all (x, y) in D,
so $\int\int mdA \leq \int\int f(x, y) dA \leq \int\int MdA$,
hence $m \int\int dA \leq \int\int f(x, y) dA \leq M\int\int dA$,
and the conclusion follows from this.

 

[_N3WTON_]

Is the above given information? If not, how is it "known that ..."? You have cited the lemmas below, so I can see where they came from. I don't doubt that it's true, though.

The above is not given information, thanks for pointing that out. So maybe it would work better if I stated that I am "taking the double integral over the entire region D" instead of saying "it is known"?

 

[_N3WTON_]

I would organize things differently. Put the lemmas at the top so that they don't interrupt the logic of your argument.
It's given that m <= f(x, y) <= M, for all (x, y) in D,
so $\int\int mdA \leq \int\int f(x, y) dA \leq \int\int MdA$,
hence $m \int\int dA \leq \int\int f(x, y) dA \leq M\int\int dA$,
and the conclusion follows from this.

Thanks for the advice! :)

 

[Mark44]

The above is not given information, thanks for pointing that out. So maybe it would work better if I stated that I am "taking the double integral over the entire region D" instead of saying "it is known"?

I think that's better.

 

[_N3WTON_]

I think that's better.

Awesome, thanks for the advice!