How can an improper integral approach a value?

Question

I have $\int_1^\infty\frac {1}{x^2} dx$, which converges to 1. If the region under the curve from 1 to infinity on the function $f(x) = \frac {1}{x^2}$ has infinite area when graphed (even though the function approaches zero as x goes to infinity), how can the integral approach a singular value?

Answer 1

The definition of the improper integral is as follows:

$$ \int_{0}^{\infty} f(x)dx =\lim_{n\rightarrow \infty} \int_{0}^{n} f(x)dx$$

So just as the limit of some function can have a finite value, the improper integral can also have a finite value.