Integrals are limits of sums and as such they inherit most of their
properties. For example if you multiply each term of a
sum of terms by a number (the number 2 say) then the result of the
sum is the same as if we multiply the original sum by 2. This is true
for sums of a finite number of terms but it is still true for limits
of these sums (when the limit exists) since the limits also have the
property that
the limit of a product is the product of the limits.
Using familiar properties of sums and limits it is not difficult
to show that integrals satisfy:

The integral of a linear combination of functions is the linear combination of the integrals of these functions. 
> #
/ /     a f(x, y) + b g(x, y) dx dy =   / / D / / / /     a   f(x, y) dx dy + b   g(x, y) dx dy     / / / / D D

> #
2 1 / /   2   2 x y  3 y dx dy = 16   / / 0 1
and this is clearly equal to: 
> #
2 1 2 1 / / / /     2 2   x y dx dy  3   y dx dy = 2(0)  3(16/3)     / / / / 0 1 0 1

When, f(x,y) > g(x,y) for all the (x,y) in D then 
> #
/ / / /       g dA <   f dA     / / / /

On the unit square R=[0,1]x[0,1] we have,

> Int(Int(2*x*y,x=0..1),y=0..1) < Int(Int((x+y+1)^2,x=0..1),y=0..1);
1 1 1 1 / / / /     2   2 x y dx dy <   (x + y + 1) dx dy     / / / / 0 0 0 0> #
1/2 < 25/6

Area of bounded regions D, denoted here by A(D) can be computed by just integrating the constant function 1 over D. 
> #
/ /   A(D) =   1 dA   / / D

The area of a circle of radius a is given by, 
> Int(Int(1,y=sqrt(a^2x^2)..sqrt(a^2x^2)),x=a..a) =
> int(int(1,y=sqrt(a^2x^2)..sqrt(a^2x^2)),x=a..a);
2 2 1/2 a (a  x ) / /   2   1 dy dx = a Pi   / /  a 2 2 1/2  (a  x )

From the previous three properties it follows that, if 
> #
/ /   m A(D) <   f dA < M A(D)   / / D

Over the rectangle R=[0,1]x[1,2] 
> #
2 2 0 < x + y < 4
hence, 
> #
2 1 / /   2 2 0 <   x + y dx dy = 4 < 4 (10)(2+1) = 12   / / 1 0

If we split the domain of integration D into two pieces, D1 and D2 then, 
> #
/ / / / / /         f dA =   f dA +   g dA       / / / / / / D = D1 U D2 D1 D2