Double Integral Over a Rectangle

Basic Definition of a Double Integral

Let f be a real-valued function

Assume RR2 is a closed rectangle where R=[a,b]×[c,d]
(open would be something like (a,b]×[c,d])

Take small segments along the two axes so that

xj+1xj=ban,yk+1yk=dcn

A continuous function on a closed rectangle is always bounded where M>0 such that M<f(x)<M

Let Rjk be the rectangular partition [xj,xj+1]×[yk,yk+1]

The double integral can be defined at the sum

Sn=j,k=0n1f(cjk) Δx Δy=j,k=0n1f(cjk) ΔA

Where cjk is an arbitrary point in Rjk

where we take limn , if the limit converges to the same value no matter the choice of cjk (for example the left-most point versus the right-most point) then f is integrable over R

and we write this limit as

Rf(x,y)dA=Rf(x,y)dxdy=Rfdxdy

Properties

  1. Any continuous function defined on a closed rectangle is integrable
  2. A real-valued function is also integrable if it is discontinuous only a finite region but still bounded

Linearity

R[f(x,y)+g(x,y)]dA=Rf(x,y)dA+R,g(x,y)dA.

Homogeneity

Rcf(x,y)dA=cRf(x,y)dA

Monotonicity

f(x,y)g(x,y)Rf(x,y)dARg(x,y)dA

Additivity

if Rii=1,,m are partitioned rectangles such that f is bounded and integrable over each Ri and their union is also a rectangle R than:

Rf(x,y)dA=i=1mRif(x,y)dA

Fubini's Theorem

Let f be bounded and continuous in each partition of R=[a,b]×[c,d]:

Rf(x,y)dA=abcdf(x,y)dydx=cdabf(x,y)dxdy

Basic idea of the proof

Rf(x,y)dA=limnj=0n1k=0n1f(cjk)(yk+1yk)(xj+1xj)by MVT cjk=(x,y)Riy1yk+1ykykyk+1f(x,y)dy=f(x,y)soykyk+1f(cjk)dy=f(x,y)(yk+1yk)Rf(x,y)dA=limnj=0n1f(x,y)(yk+1yk)(xj+1xj)=abcdf(x,y)dydx