Answer
$$A = 5 - \frac{3}{{\ln 2}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {2^x},{\text{ }}g\left( x \right) = \frac{3}{2}x + 1 \cr
& \frac{3}{2}x + 1 \geqslant {2^x}{\text{ on the interval }}0 \leqslant x \leqslant 2 \cr
& {\text{From the graph shown below, we can define the enclosed area}} \cr
& {\text{as:}} \cr
& A = \int_0^2 {\left( {\frac{3}{2}x + 1 - {2^x}} \right)} dx \cr
& {\text{Integrate and evaluate}} \cr
& A = \left[ {\frac{3}{4}{x^2} + x - \frac{{{2^x}}}{{\ln 2}}} \right]_0^2 \cr
& A = \left[ {\frac{3}{4}{{\left( 2 \right)}^2} + \left( 2 \right) - \frac{{{2^2}}}{{\ln 2}}} \right] - \left[ {\frac{3}{4}{{\left( 0 \right)}^2} + \left( 0 \right) - \frac{{{2^0}}}{{\ln 2}}} \right] \cr
& A = 5 - \frac{4}{{\ln 2}} + \frac{1}{{\ln 2}} \cr
& A = 5 - \frac{3}{{\ln 2}} \cr
& A \approx 0.6719 \cr} $$
