Answer
$$80$$
Work Step by Step
$$\eqalign{
& \int_1^2 {{{\left( {4x - 2} \right)}^3}} dx \cr
& u = 4x - 2,{\text{ then }}du = 4dx,{\text{ }}dx = \frac{1}{4}du \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = 2,{\text{ }}u = 6 \cr
& {\text{if }}x = 1,{\text{ }}u = 2 \cr
& {\text{so}} \cr
& \int_1^2 {{{\left( {4x - 2} \right)}^3}} dx = \int_2^6 {{u^3}} \left( {\frac{1}{4}du} \right) \cr
& = \frac{1}{4}\int_2^6 {{u^3}} du \cr
& = \frac{1}{4}\left( {\frac{{{u^4}}}{4}} \right)_2^6 \cr
& = \frac{1}{{16}}\left( {{u^4}} \right)_2^6 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \frac{1}{{16}}\left( {{{\left( 6 \right)}^4} - {{\left( 2 \right)}^4}} \right) \cr
& = \frac{1}{{16}}\left( {1296 - 16} \right) \cr
& = 80 \cr} $$
