Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_0^4 {x\left( {x - 2} \right)\left( {x - 4} \right)} dx \cr
& {\text{multiply the integrand}} \cr
& = \int_0^4 {x\left( {{x^2} - 6x + 8} \right)} dx \cr
& = \int_0^4 {\left( {{x^3} - 6{x^2} + 8x} \right)} dx \cr
& {\text{using power rule for integration}} \cr
& = \left. {\left( {\frac{{{x^4}}}{4} - 6\left( {\frac{{{x^3}}}{3}} \right) + 8\left( {\frac{{{x^2}}}{2}} \right)} \right)} \right|_0^4 \cr
& = \left. {\left( {\frac{{{x^4}}}{4} - 2{x^3} + 4{x^2}} \right)} \right|_0^4 \cr
& {\text{Using The Fundamental Theorem}} \cr
& = \left( {\frac{{{{\left( 4 \right)}^4}}}{4} - 2{{\left( 4 \right)}^3} + 4{{\left( 4 \right)}^2}} \right) - \left( {\frac{{{{\left( 0 \right)}^4}}}{4} - 2{{\left( 0 \right)}^3} + 4{{\left( 0 \right)}^2}} \right) \cr
& {\text{simplify}} \cr
& = \left( {64 - 128 + 64} \right) - \left( {0 - 0 + 0} \right) \cr
& = 0 \cr} $$
