Answer
$$36$$
Work Step by Step
$$\eqalign{
& \int_{ - 3}^3 {\left( {{x^3} + 4{x^2} - 3x - 6} \right)} dx \cr
& {\text{Write the integral as a sum of integrals }} \cr
& = \int_{ - 3}^3 {{x^3}} dx + \int_{ - 3}^3 {4{x^2}} dx - \int_{ - 3}^3 {3x} dx - \int_{ - 3}^3 6 dx \cr
& {\text{Where}} \cr
& {x^3}:{\text{ Odd function}} \cr
& 4{x^2}:{\text{Even function}} \cr
& 3x:{\text{Odd function}} \cr
& 6:{\text{Constant function}} \cr
& {\text{Using }}\left( {{\text{THEOREM 4}}.{\text{16}}} \right) \cr
& = 0 + 2\int_0^3 {4{x^2}dx} - 0 - 2\int_0^3 6 dx \cr
& {\text{Integrating}} \cr
& = 2\left[ {\frac{4}{3}{x^3}} \right]_0^3 - 2\left[ {6x} \right]_0^3 \cr
& = 2\left[ {\frac{4}{3}{{\left( 3 \right)}^3}} \right] - 2\left[ {6\left( 3 \right)} \right] \cr
& = 36 \cr} $$
