Integral of 1/x?

Question

If you take the integral of 1/x from 1 to infinity, it is an undefined integral. But when you find the rotational volume of 1/x around the x-axis you come up with a finite number (3.14). Why is it that the value under the graph can be undefined when the volume if defined perfectly well?

ok, maybe not perfectly well...
but if you solve both of these cases as improper integrals, the area is still undefined while the rotational volume is pi

Answer 1

First of all, the integral of 1/x dx from 1 to infinity is infinity. It's not a real number, but it's defined. So, the area betwen the curve f(x) = 1/x and the x axis is infinite for the interval (1, oo) .

But to compute the rotational volume, the, according to calculus, you compute the integral V= Int (1 to oo) pi [f(x)]2 dx = Int (1 to oo) pi (1/x^2) dx = = pi * Int (1 to oo) (1/x^2) dx . What happens is that, contrarily to what happens to the integral from 1 to oo of 1/x, the integral of 1/x^2 is finite. In fact, we have Int (1 to oo) (1/x^2) dx = [-1/x] (1 to oo) = 1, so that, like you staed, your volume is FINITe and is V = pi =~ 3.14.

There´s no contradiction here. Try to visualize this geometrically. The araa of the cross section of the rotational solid is infinite. But it's getting so "thinner" as we walk along the x -axis that it's volume tends to a finite limit.

Answer 2

the 1/x function tends towards 0 at infinity.
However, it never reaches "0"...
therefore the volume cannot be defined "perfectly well" as you stated.
this is why the integral of 1/x from 1 to infinity is undefined.