Well, for one thing, you can't always require that x be postive!

For example, find the area bounded by the graphs of y= 1/x, y= 0, x= -2, and x= -1.

That is [tex]\int_{-2}^{-1} (0-\frac{1}{x})dx= -\int_{-2}^{-1}\frac{dx}{x}[/itex].

Now, we have

That is the same as we would get for the area bounded by y= 1/x, y= 0, x=1, and x= 2, which, because of the symmetry is what we should get.