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?
Area under the curves $1/x$ and $1/x^2$
-
0From$x=1$to x=∞, yes. – 2019-03-09
2 Answers
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 $x\to\infty$.
-
0@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 $\log_{10} x$. – 2012-09-10
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.
-
0@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. – 2012-09-10