#### Answer

$${\text{A}} = {e^4} + {e^2}$$

#### Work Step by Step

$$\eqalign{
& {\text{let }}y = \left( {x - 2} \right){e^x},{\text{ the }}x{\text{ - axis from }}x = 2{\text{ to }}x = 4 \cr
& {\text{The area between two curves is defined by }}\int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{where }}f{\text{ and }}g{\text{ are continuous functions and }}f\left( x \right) \geqslant g\left( x \right){\text{ on }}\left[ {a,b} \right]{\text{for this exercise }} \cr
& g\left( x \right) = 0{\text{ }}\left( {x{\text{ - axis}}} \right){\text{ and the interval }}x = 2{\text{ to }}x = 4{\text{ implies that }}a = 2{\text{ and }}b = 4 \cr
& {\text{then}} \cr
& {\text{A}} = \int_2^4 {\left[ {\left( {x - 2} \right){e^x} - 0} \right]} dx \cr
& {\text{A}} = \int_2^4 {\left( {x - 2} \right){e^x}} dx \cr
& {\text{setting }}\,\,\,\,\,\,u = x - 2{\text{ then }}du = dx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^x}dx{\text{ then }}v = {e^x} \cr
& {\text{Substituting these values into the formula for integration by parts}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {\left( {x - 2} \right){e^x}} dx = \left( {x - 2} \right){e^x} - \int {{e^x}} dx \cr
& {\text{integrate}} \cr
& \int {\left( {x - 2} \right){e^x}} dx = \left( {x - 2} \right){e^x} - {e^x} + C \cr
& \cr
& {\text{Now find the definite integral}} \cr
& {\text{A}} = \int_2^4 {\left( {x - 2} \right){e^x}} dx = \left. {\left( {\left( {x - 2} \right){e^x} - {e^x}} \right)} \right|_2^4 \cr
& {\text{evaluating the limits}} \cr
& {\text{A}} = \left( {\left( {4 - 2} \right){e^4} - {e^4}} \right) - \left( {\left( {2 - 2} \right){e^2} - {e^2}} \right) \cr
& {\text{simplifying}} \cr
& {\text{A}} = 2{e^4} - {e^4} + {e^2} \cr
& {\text{A}} = {e^4} + {e^2} \cr} $$