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I am given the following:

D=[0,1]×[0,1] and f(x,y)=[itex]\frac{1}{1+x+y}[/itex] on D.

a) why is f integrable on D?

I know that the function is integrable because it is bounded and has a finite amount of points where it is discontinuous.

That f is bounded follows directly from the given information.

I also know where f is discontinuous, obviously in the point (0,0)

And I can determine the upper and lower bounds for y and x.

by looking at the boundaries: [itex]\frac{1}{1+x+y}[/itex]=0

and [itex]\frac{1}{1+x+y}[/itex]=1

so for y:

y=[itex]\varphi[/itex][itex]_{1}[/itex](x)=0

y=[itex]\varphi[/itex][itex]_{2}[/itex](x)=-x

and for x:

x=[itex]\varphi[/itex][itex]_{1}[/itex](y)=0

x=[itex]\varphi[/itex][itex]_{1}[/itex](y)=-y

So I thought that this mend that f(x,y) is discontinuous at the points: (0,0),y=-x,x=-y

but I am not sure because by writing x and y in terms of the other variable I can draw some graphs within D and there the function for y=-x is discontinuous at the boundary x=1 and x=0 and for x=-y is this function discontinuous at y=1 and y=0.

So I'm having a bit of a problem with understanding when a function is discontinuous when there are 2 variables. And I am also wondering, if I can find the points where f is discontinuous, then is it enough to just state that f is discontinuous at those points or do I have to give some kind of proof?

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# Homework Help: Proof function is integrable

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