Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_0^1 {{{\left( {2x - 1} \right)}^3}} dx \cr
& u = 2x - 1,{\text{ then }}du = 2dx,{\text{ }}dx = \frac{1}{2}du \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = 1,{\text{ }}u = 1 \cr
& {\text{if }}x = 0,{\text{ }}u = - 1 \cr
& {\text{so}} \cr
& \int_0^1 {{{\left( {2x - 1} \right)}^3}} dx = \int_{ - 1}^1 {{u^3}} \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int_{ - 1}^1 {{u^3}} du \cr
& = \frac{1}{2}\left( {\frac{{{u^4}}}{4}} \right)_{ - 1}^1 \cr
& = \frac{1}{8}\left( {{u^4}} \right)_{ - 1}^1 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \frac{1}{8}\left( {{{\left( 1 \right)}^4} - {{\left( { - 1} \right)}^4}} \right) \cr
& = \frac{1}{8}\left( 0 \right) \cr
& = 0 \cr} $$
