Step $3$

Now integrate and simplify $v={\displaystyle \underset{1}{\overset{2}{}}}\frac{16}{x^4}dx.$

$V=16{\displaystyle \underset{1}{\overset{2}{}}}{x}^{-}dx$

$=16[\frac{x^{-}}{-3}{]}_1^2$

$=16(\frac{-3}{-3})$

$=-\frac{16}{3}(\frac{1}{8}-1)$

$=-\frac{16}{3}*(-\frac{-3}{8})$

$=\frac{14}{3}$