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I am reviewing/browsing through Stewart'sEarly Transcendentalsand had a question about the section on Double Integrals over General Regions. It says that in order to integrate a functionfover regionsDof a general shape, we supposeDis a bounded region that can be enclosed in a rectangular regionR. Then a new functionFwith domainRis defined :

F(x,y) =f(x,y) if (x,y) is inD, and F(x,y) = 0 if (x,y) is inRbut not inD.

Then, a theorem is stated:

[tex]\int[/tex][tex]\int_{D}[/tex]f(x,y)dA=[tex]\int[/tex][tex]\int_{R}[/tex]F(x,y)dA.

And then (finally, getting to my question) it says that iffis continuous on D and the boundary curve of D is "well behaved" (in a sense outside the scope of this book), then it can be shown that [tex]\int[/tex][tex]\int_{R}[/tex]F(x,y)dA exists and therefore [tex]\int[/tex][tex]\int_{D}[/tex]f(x,y)dA exists. What does it mean by well-behaved, and can someone offer an elementary explanation as to why the boundary curve of D must be well behaved in order for [tex]\int[/tex][tex]\int_{R}[/tex]F(x,y)dA to exist, and why that implies that [tex]\int[/tex][tex]\int_{D}[/tex]f(x,y)dA exists?

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# Well-Behaved curve

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