Answer
$$A = \frac{{53}}{4}$$
Work Step by Step
$$\eqalign{
& x = 1,\,\,\,\,\,x = 2,\,\,\,\,\,y = 3{x^3} + 2,\,\,\,\,\,\,y = 0 \cr
& {\text{Find the area using the formula for the area between two curves }}\left( {{\text{page 399}}} \right) \cr
& {\text{If }}f{\text{ and }}g{\text{ are continuous functions and }}f\left( x \right) \geqslant g\left( x \right){\text{ on }}\left[ {a,b} \right] \cr
& A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{then }} \cr
& f\left( x \right) = 3{x^3} + 2,\,\,\,\,g\left( x \right) = 0,\,\,\,\,a = 1,\,\,\,\,b = 2 \cr
& A = \int_1^2 {\left( {3{x^3} + 2 - 0} \right)} dx \cr
& A = \int_1^2 {\left( {3{x^3} + 2} \right)} dx \cr
& {\text{integrating by the power rule}} \cr
& A = \left[ {\frac{{3{x^4}}}{4} + 2x} \right]_1^2 \cr
& A = \left( {\frac{{3{{\left( 2 \right)}^4}}}{4} + 2\left( 2 \right)} \right) - \left( {\frac{{3{{\left( 1 \right)}^4}}}{4} + 2\left( 1 \right)} \right) \cr
& A = 12 + 4 - \frac{3}{4} - 2 \cr
& A = \frac{{53}}{4} \cr} $$
