Answer
$$19$$
Work Step by Step
$$\eqalign{
& \int_1^2 {{{\left( {4 - 3x} \right)}^8}} dx \cr
& u = 4 - 3x,{\text{ then }}du = - 3dx,{\text{ }}dx = - \frac{1}{3}du \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = 2,{\text{ }}u = - 2 \cr
& {\text{if }}x = 1,{\text{ }}u = 1 \cr
& {\text{so}} \cr
& \int_1^2 {{{\left( {4 - 3x} \right)}^8}} dx = \int_1^{ - 2} {{u^8}} \left( { - \frac{1}{3}du} \right) \cr
& = - \frac{1}{3}\int_1^{ - 2} {{u^8}} du \cr
& = - \frac{1}{3}\left( {\frac{{{u^9}}}{9}} \right)_1^{ - 2} \cr
& = - \frac{1}{{27}}\left( {{u^9}} \right)_1^{ - 2} \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = - \frac{1}{{27}}\left( {{{\left( { - 2} \right)}^9} - {{\left( 1 \right)}^9}} \right) \cr
& = - \frac{1}{{27}}\left( { - 512 - 1} \right) \cr
& = \frac{{513}}{{27}} \cr
& = 19 \cr} $$
