6.1. Integrating a Function that Blows UpWhen we integrate a function that blows up to infinity at some point in the interval we're integrating, the result may be either finite or infinite. Example 75◊ Integrate the function \(y = 1/\sqrt{x}\) from x=0 to x=1. ◊ The function blows up to infinity at one end of the region of integration, but let's just try evaluating it, and see what happens. \begin{align} \int_0^1 x^{1/2}dx &= \left.2x^{1/2}\right_0^1 \\ &= 2 \end{align} The result turns out to be finite. Intuitively, the reason for this is that the spike at x=0 is very skinny, and gets skinny fast as we go higher and higher up. Figure A. The integral \(\int_0^1 dx/\sqrt{x}\) is finite. Example 76◊ Integrate the function y = 1/x^{2} from x=0 to x=1. ◊ \begin{align} \int_0^1 x^{2}dx &= \left.x^{1}\right_0^1 \\ &= 1+\frac{1}{0} \end{align} Division by zero is undefined, so the result is undefined. Another way of putting it, using the hyperreal number system, is that if we were to integrate from ε to 1, where ε was an infinitesimal number, then the result would be 1+1/ε, which is infinite. The smaller we make ε, the bigger the infinite result we get out. Intuitively, the reason that this integral comes out infinite is that the spike at x=0 is fat, and doesn't get skinny fast enough. Figure B. The integral \(\int_0^1 dx/x^2\) is infinite. These two examples were examples of improper integrals.
