Answer
$${\text{Diverges}}$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\frac{1}{{{x^2}}}} dx \cr
& {\text{The integrand }}\frac{1}{{{x^2}}}{\text{ is not continuous at }}x = 0,{\text{ then by the}} \cr
& {\text{definition of improper integrals}} \cr
& \int_0^2 {\frac{1}{{{x^2}}}} dx = \mathop {\lim }\limits_{r \to {0^ + }} \int_r^2 {\frac{1}{{{x^2}}}} dx \cr
& {\text{Integrating }} \cr
& = \mathop {\lim }\limits_{r \to {0^ + }} \left[ { - \frac{1}{x}} \right]_r^2 \cr
& = - \mathop {\lim }\limits_{r \to {0^ + }} \left[ {\frac{1}{2} - \frac{1}{r}} \right] \cr
& {\text{Evaluate the limit when }}r \to {0^ + } \cr
& = - \left[ {\frac{1}{2} - \frac{1}{{{0^ + }}}} \right] \cr
& = - \left[ {\frac{1}{2} - \infty } \right] \cr
& = \infty \cr
& {\text{Diverges}} \cr} $$
