Answer
$ \displaystyle \frac{1}{2}(e-\frac{1}{e^{3}})$
Work Step by Step
The method of example 4, as suggested by the hint, involves substitution.
Remember to express the LIMITS of integration accordingly.
$\left[\begin{array}{ll}
u=2x-1 & du=2dx\\
dx=du/2 & \\
& \\
x=-1\Rightarrow & u=-3\\
x=1\Rightarrow & u=1
\end{array}\right]$
$\displaystyle \int_{0}^{1}e^{2x-1}dx=\int_{-3}^{1}e^{u}(\frac{du}{2})=\frac{1}{2}\int_{-3}^{1}e^{u}du$
... Apply the FTC,
... def.integral = [antiderivative$]_{a}^{b}$
$=\left[\dfrac{e^{u}}{2}\right]_{-3}^{1}$
$=\displaystyle \frac{e}{2}-\frac{e^{-3}}{2}$
$=\displaystyle \frac{1}{2}(e-e^{-3})$
$=\displaystyle \frac{1}{2}(e-\frac{1}{e^{3}})$
