Answer
$$2$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\frac{4}{{{v^2}}}} dv \cr
& = \int_1^2 {4{v^{ - 2}}} dv \cr
& {\text{integrate by using the power rule}} \cr
& = \left( {\frac{{4{v^{ - 1}}}}{{ - 1}}} \right)_1^2 \cr
& = - 4\left( {\frac{1}{v}} \right)_1^2 \cr
& {\text{Evaluating, we get:}} \cr
& = - 4\left( {\frac{1}{2} - \frac{1}{1}} \right) \cr
& = - 4\left( { - \frac{1}{2}} \right) \cr
& = 2 \cr} $$
