Answer
$\displaystyle \frac{40}{27\ln 3}$
Work Step by Step
The method of example 4, as suggested by the hint, involves substitution.
Remember to express the LIMITS of integration accordingly.
$\left[\begin{array}{ll}
u=2x-1 & du=2dx\\
dx=du/2 & \\
& \\
x=-1\Rightarrow & u=-3\\
x=1\Rightarrow & u=1
\end{array}\right]$
$\displaystyle \int_{0}^{2}3^{2x+1}dx=\int_{-3}^{1}3^{u}(\frac{du}{2})=\frac{1}{2}\int_{-3}^{1}3^{u}du$
... Apply the FTC,
... def.integral = [antiderivative$]_{a}^{b}$
$=\displaystyle \frac{1}{2}\left[\frac{3^{u}}{\ln 3}\right]_{-3}^{1}$
$=\displaystyle \frac{1}{2}(\frac{3}{\ln 3}-\frac{3^{-3}}{\ln 3})$
$=\displaystyle \frac{1}{2\ln 3}(3-\frac{1}{27})$
$=\displaystyle \frac{1}{2\ln 3}\cdot\frac{80}{27}=\frac{40}{27\ln 3}$
