Answer
$$\frac{{28}}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\left( {5{x^2} - 4x + 2} \right)} dx \cr
& {\text{integrate by using }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C{\text{ and }}\int {dx} = x + C \cr
& = \left( {5\left( {\frac{{{x^{2 + 1}}}}{{2 + 1}}} \right) - 4\left( {\frac{{{x^{1 + 1}}}}{{1 + 1}}} \right) + 2\left( x \right)} \right)_0^2 \cr
& = \left( {\frac{5}{3}{x^3} - 2{x^2} + 2x} \right)_0^2 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \left( {\frac{5}{3}{{\left( 2 \right)}^3} - 2{{\left( 2 \right)}^2} + 2\left( 2 \right)} \right) - \left( {\frac{5}{3}{{\left( 0 \right)}^3} - 2{{\left( 0 \right)}^2} + 2\left( 0 \right)} \right) \cr
& {\text{simplifying}} \cr
& = \left( {\frac{{40}}{3} - 8 + 4} \right) - \left( 0 \right) \cr
& = \frac{{28}}{3} \cr} $$
