Answer
$${A_{R1}} = \frac{{341}}{{10}},\,\,\,\,{A_{R2}} = \frac{{257}}{{30}}\,$$
Work Step by Step
$$\eqalign{
& {\text{The area of the region }}{R_1}{\text{ is given by}} \cr
& {A_{R1}} = \int_0^{8/5} {\left( {16 - {x^2}} \right)} dx + \int_{8/5}^3 {\left( {16 - {x^2} - 5x + 8} \right)} dx \cr
& {A_{R1}} = \int_0^{8/5} {\left( {16 - {x^2}} \right)} dx + \int_{8/5}^3 {\left( {24 - {x^2} - 5x} \right)} dx \cr
& {\text{Integrating}} \cr
& {A_{R1}} = \left[ {16x - \frac{{{x^3}}}{3}} \right]_0^{8/5} + \left[ {24x - \frac{{{x^3}}}{3} - \frac{{5{x^2}}}{2}} \right]_{8/5}^3 \cr
& {A_{R1}} = \left[ {16\left( {\frac{8}{5}} \right) - \frac{1}{3}{{\left( {\frac{8}{5}} \right)}^3}} \right] - 0 + \left[ {24\left( 3 \right) - \frac{{{{\left( 3 \right)}^3}}}{3} - \frac{{5{{\left( 3 \right)}^2}}}{2}} \right] \cr
& \,\,\,\,\,\,\,\,\, - \left[ {24\left( 3 \right) - \frac{1}{3}{{\left( {\frac{8}{5}} \right)}^3} - \frac{5}{2}{{\left( {\frac{8}{5}} \right)}^2}} \right] \cr
& {\text{Simplify}} \cr
& {A_{R1}} = \frac{{9088}}{{375}} + \frac{{81}}{2} - \frac{{11488}}{{375}} \cr
& {A_{R1}} = \frac{{81}}{2} - \frac{{2400}}{{375}} \cr
& {A_{R1}} = \frac{{341}}{{10}} \cr
& \cr
& {\text{The total area enclosed is given by}} \cr
& {A_T} = \int_0^4 {\left( {16 - {x^2}} \right)} dx \cr
& {A_T} = \left[ {16x - \frac{{{x^3}}}{3}} \right]_0^4 \cr
& {A_T} = 16\left( 4 \right) - \frac{{{{\left( 4 \right)}^3}}}{3} \cr
& {A_T} = \frac{{128}}{3} \cr
& \cr
& {\text{The area of the region }}{R_2}{\text{ is given by}} \cr
& {A_{R2}} = {A_T} - {A_{R1}} \cr
& {A_{R2}} = \frac{{128}}{3} - \frac{{341}}{{10}} \cr
& {A_{R2}} = \frac{{257}}{{30}} \cr} $$
