Answer
$$A = \frac{{81}}{5}$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{{x^2}}},{\text{ }}y = 4,{\text{ }}x = 5 \cr
& {\text{Let }}y = y \cr
& \frac{1}{{{x^2}}} = 4 \cr
& x = \frac{1}{2} \cr
& {\text{From the graph we can note that }}4 \geqslant \frac{1}{{{x^2}}}{\text{ on }}\left[ {\frac{1}{2},5} \right] \cr
& {\text{Then, the area is given by:}} \cr
& A = \int_{1/2}^5 {\left( {4 - \frac{1}{{{x^2}}}} \right)} dx \cr
& {\text{Integrate and evaluate}} \cr
& A = \left[ {4x + \frac{1}{x}} \right]_{1/2}^5 \cr
& A = \left[ {4\left( 5 \right) + \frac{1}{5}} \right] - \left[ {4\left( {1/2} \right) + \frac{1}{{1/2}}} \right] \cr
& A = \frac{{101}}{5} - 4 \cr
& A = \frac{{81}}{5} \cr} $$
