Answer
$A \approx 0.25142$
Work Step by Step
$$\eqalign{
& y = {\tan ^2}x,{\text{ }}y = \sqrt x \cr
& {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr
& {\text{Find the intersection points using a graphing calculator}} \cr
& x \approx 0.75 \cr
& {\text{We have the interval }}\left[ {0,0.75} \right] \cr
& {\text{We can find the area integrating with respect to }}x \cr
& A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx{\text{ }}\left( {\bf{1}} \right){\text{ }}\left( {{\text{see page 439}}} \right) \cr
& {\text{From the graph }} \cr
& \sqrt x \geqslant {\tan ^2}x{\text{ for the interval }}\left[ {0,0.75} \right] \cr
& {\text{Therefore}} \cr
& A = \int_0^{0.75} {\left( {\sqrt x - {{\tan }^2}x} \right)} dx \cr
& {\text{Integrating by a graphing calculator we obtain}} \cr
& A \approx 0.25142 \cr} $$
