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6.1. Integrating a Function that Blows Up

When 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/x2 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.