Answer
$${\text{A}} = \frac{{5504}}{7}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {3x + 2} \right)^6}{\text{ on the interval }}\left[ { - 2,0} \right] \cr
& {\text{The area between two curves is defined by }}\int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{where }}f{\text{ and }}g{\text{ are continuous functions and }}f\left( x \right) \geqslant g\left( x \right){\text{ on }}\left[ {a,b} \right]{\text{for this exercise }} \cr
& g\left( x \right) = 0{\text{ }}\left( {x{\text{ - axis}}} \right){\text{ and the interval }}\left[ { - 2,0} \right]{\text{ implies that }}a = - 2{\text{ and }}b = 0 \cr
& {\text{then}} \cr
& {\text{A}} = \int_{ - 2}^0 {\left[ {{{\left( {3x + 2} \right)}^6} - 0} \right]} dx \cr
& {\text{A}} = \int_{ - 2}^0 {{{\left( {3x + 2} \right)}^6}} dx \cr
& {\text{use substitution}}{\text{. Let }}u = 3x + 2,{\text{ so that }}du = 3dx,\,\,\,\,\frac{1}{3}du = dx \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}u = 3\left( 0 \right) + 2 = 2 \cr
& \,\,\,\,\,\,{\text{If }}x = - 2,{\text{ then }}u = 3\left( { - 2} \right) + 2 = - 4 \cr
& {\text{Then}} \cr
& {\text{A}} = \int_{ - 2}^0 {{{\left( {3x + 2} \right)}^6}} dx = \int_{ - 4}^2 {{u^6}\left( {\frac{1}{3}du} \right)} \cr
& {\text{A}} = \frac{1}{3}\int_{ - 4}^2 {{u^6}du} \cr
& {\text{use }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& {\text{A}} = \frac{1}{3}\left( {\frac{{{u^7}}}{7}} \right)_{ - 4}^2 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& {\text{A}} = \frac{1}{3}\left( {\frac{{{{\left( 2 \right)}^7}}}{7} - \frac{{{{\left( { - 4} \right)}^7}}}{7}} \right) \cr
& {\text{simplifying}} \cr
& {\text{A}} = \frac{1}{3}\left( {\frac{{128}}{7} + \frac{{16384}}{7}} \right) \cr
& {\text{A}} = \frac{{5504}}{7} \cr} $$
