Solve the integral and mark the correct option for given function $I={\int }_{\infty }^R\frac{GMm}{x^2}dx.$ [Here M,m and G are constants]

$I={\int }_{\infty }^R\frac{GMm}{x^2}dx=GMm{\int }_{\infty }^R\frac{dx}{x^2}=GMm{[-\frac{1}{x}]}_{\infty }^R=GMm[-\frac{1}{R}+\frac{1}{\infty }]=\frac{-GMm}{R}$

The above expression represents gravitational potential energy which is obtained by integrating $GMm/x^2$ between the limits infinity $\left(\infty \right)$ and radius of the earth (R).