Answer
$$\frac{1}{2}\left( {e - \frac{1}{e}} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^1 {{e^{2x - 1}}} dx,\,\,\,u = 2x - 1 \cr
& u = 2x - 1,\,\,\,du = 2dx,\,\,dx = \frac{1}{2}du \cr
& {\text{for }}x = 1,\,\,u = 2\left( 1 \right) - 1 = 1 \cr
& {\text{for }}x = 0,\,\,u = 2\left( 0 \right) - 1 = - 1 \cr
& {\text{Thus}}{\text{,}} \cr
& \int_0^1 {{e^{2x - 1}}} dx = \int_{ - 1}^1 {{e^u}} \left( {\frac{1}{2}} \right)du\, \cr
& {\text{drop out the constant}} \cr
& = \frac{1}{2}\int_{ - 1}^1 {{e^u}} du\, \cr
& {\text{integrating}} \cr
& = \frac{1}{2}\left[ {{e^u}} \right]_{ - 1}^1 \cr
& {\text{evaluating the limits}} \cr
& = \frac{1}{2}\left[ {{e^1} - {e^{ - 1}}} \right] \cr
& = \frac{1}{2}\left( {e - \frac{1}{e}} \right) \cr} $$
