Answer
$${\text{A}} = \frac{1}{2}\left( {{e^4} - 1} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = x{e^{{x^2}}}{\text{ on the interval }}\left[ {0,2} \right] \cr
& {\text{The area between two curves is defined by }}\int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{where }}f{\text{ and }}g{\text{ are continuous functions and }}f\left( x \right) \geqslant g\left( x \right){\text{ on }}\left[ {a,b} \right]{\text{for this exercise }} \cr
& g\left( x \right) = 0{\text{ }}\left( {x{\text{ - axis}}} \right){\text{ and the interval }}\left[ {0,2} \right]{\text{ implies that }}a = 0{\text{ and }}b = 2 \cr
& {\text{then}} \cr
& {\text{A}} = \int_0^2 {\left[ {x{e^{{x^2}}} - 0} \right]} dx \cr
& {\text{A}} = \int_0^2 {x{e^{{x^2}}}} dx \cr
& {\text{use substitution}}{\text{. Let }}u = {x^2},{\text{ so that }}du = 2xdx,\,\,\,\,\frac{1}{2}du = xdx \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 2,{\text{ then }}u = {\left( 2 \right)^2} = 4 \cr
& \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}u = {\left( 0 \right)^2} = 0 \cr
& {\text{Then}} \cr
& {\text{A}} = \int_0^2 {x{e^{{x^2}}}} dx = \int_0^4 {{e^u}\left( {\frac{1}{2}du} \right)} \cr
& {\text{A}} = \frac{1}{2}\int_0^4 {{e^u}du} \cr
& {\text{use }}\int_a^b {{e^u}} du = \left( {{e^u}} \right)_a^b \cr
& {\text{A}} = \frac{1}{2}\left( {{e^u}} \right)_0^4 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& {\text{A}} = \frac{1}{2}\left( {{e^4} - {e^0}} \right) \cr
& {\text{simplifying}} \cr
& {\text{A}} = \frac{1}{2}\left( {{e^4} - 1} \right) \cr} $$
