Answer
$$A = \frac{{19}}{3}$$
Work Step by Step
$$\eqalign{
& {\text{The intersection point between }}y = 8x{\text{ and }}y = 9 - {x^2},{\text{ is}} \cr
& 8x = 9 - {x^2} \cr
& {x^2} + 8x - 9 = 0 \cr
& \left( {x + 9} \right)\left( {x - 1} \right) = 0 \cr
& {x_1} = - 9,{\text{ }}{x_2} = 1 \cr
& {\text{The intersection point between }}y = \frac{5}{2}x{\text{ and }}y = 9 - {x^2},{\text{ is}} \cr
& \frac{5}{2}x = 9 - {x^2} \cr
& {x^2} + \frac{5}{2}x - 9 = 0 \cr
& 2{x^2} + 5x - 18 = 0 \cr
& \left( {2x + 9} \right)\left( {x - 2} \right) = 0 \cr
& {x_1} = - \frac{9}{2}{\text{ and }}x = 2 \cr
& {\text{The area of the region is given by:}} \cr
& A = \int_0^1 {\left( {8x - \frac{5}{2}x} \right)} dx + \int_1^2 {\left( {9 - {x^2} - \frac{5}{2}x} \right)} dx \cr
& A = \int_0^1 {\frac{{11}}{2}x} dx + \int_1^2 {\left( {9 - {x^2} - \frac{5}{2}x} \right)} dx \cr
& A = \frac{{11}}{4}\left[ {{x^2}} \right]_0^1 + \left[ {9x - \frac{1}{3}{x^3} - \frac{5}{4}{x^2}} \right]_1^2 \cr
& {\text{Evaluating}} \cr
& A = \frac{{11}}{4} + \frac{{31}}{3} - \frac{{89}}{{12}} \cr
& {\text{Simplifying}} \cr
& A = \frac{{19}}{3} \cr} $$
