Answer
$$\int_1^\infty x^{-2} dx=1$$
Work Step by Step
To evaluate the improper integral recall that
$$\int_a^\infty f(x) dx =\lim\limits_{b \to \infty} \int_a^bf(x)dx$$
So for our problem we have:
$$\int_1^\infty x^{-2} dx =\lim\limits_{b \to \infty} \int_1^bx^{-2}dx$$
Compute the integral:
$$\lim\limits_{b \to \infty} \int_1^bx^{-2}dx=\lim\limits_{b \to \infty}\frac{x^{-1}}{-1}\bigg |_1^b=\lim\limits_{b \to \infty}\left(-\frac{1}{b}+1 \right)=0+1=1$$
