# Area under the curves $1/x$ and $1/x^2$

Area under $1/x$ curve is considered to be infinite. Area under $1/x^2$ curve is considered to be finite. Why is it so?

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To be clear, you're talking about bounds of $x=1$ and $x =\infty$, right? – Jason DeVito Sep 10 '12 at 14:55
From x=1 to x=∞, yes. – David Hoffman Sep 10 '12 at 14:57

I guess you are referring to the fact that

$$\int_1^\infty \frac{1}{x^2}dx = 1$$

whilst

$$\int_1^\infty \frac{1}{x} dx = \infty$$

This is the way that "area under a curve" is normally defined. These integrals follow from calculus, for example

$$\int_1^\infty \frac{1}{x^2}dx = \left[ -\frac{1}{x} \right]_1^\infty = 0 - (-1) = 1$$

and

$$\int_1^\infty \frac{1}{x}dx = \left[ \log x\right]_1^\infty = \infty$$

since log x is unbounded as $x\to\infty$.

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I guess you meant lnx not logx. – David Hoffman Sep 10 '12 at 15:06
@DavidHoffman When mathematicians write $\log x$ they generally mean the natural logarithm. If they want the logarithm in some other base, say base 10, they will write $\log_{10} x$. – Chris Taylor Sep 10 '12 at 15:19

By the definition of area itself. What is the area of an unbounded region of the plane? It is usually defined by an improper integral, as in your case: the areas are $$\lim_{b \to +\infty} \int_1^b \frac{dx}{x} = +\infty$$ and $$\lim_{b \to +\infty} \int_1^b \frac{dx}{x^2} < +\infty.$$

It is a matter of definitions, rather than of intuition. We can say that first comes the integral, then comes the area.

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That seems to be like a reasonable justification, thank you. – David Hoffman Sep 10 '12 at 15:01
@DavidHoffman: Even without mentioning an integral, it is not hard to argue that the area is infinite. If you are familiar with the standard argument that $1+\frac{1}{2}+\frac{1}{3}+\cdots$ diverges, it is basically the same. – André Nicolas Sep 10 '12 at 16:03