Since the antiderivative of $\frac{1}{x}$ is $\ln(|x|)$, the surface under the graph of $\frac{1}{x}$ with $x>1$ is $\infty$.
However, the antiderivative of $\frac{1}{x^2}$ is $-\frac{1}{x}$, so the surface under the graph of $\frac{1}{x^2}$ with $x>1$ is $1$.
Although I understand how one can calculate this, I can't really imagine it. Both graphs have an asymptote, so how can the surface under $\frac{1}{x}$ be $\infty$, whilst the surface under $\frac{1}{x^2}$ is $1$? It doesn't look like the surface under $\frac{1}{x}$ is infinite.
Perhaps my question is a little argumentative but I'm looking for a way to imagine that this is true. So, is there any way to intuitively "see" that the surface under $\frac{1}{x}$ is $\infty$?