Answer
$$
\int_{0}^{\ln 3}\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{\mathrm{x}}\right) \mathrm{d} \mathrm{x} =2$$
Work Step by Step
Given $$
\int_{0}^{\ln 3}\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{\mathrm{x}}\right) \mathrm{d} \mathrm{x} $$
So, we have
\begin{aligned}
I&=\int_{0}^{\ln 3}\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{\mathrm{x}}\right) \mathrm{dx}\\
&=\left[\frac{\mathrm{e}^{2 \mathrm{x}}}{2}-\mathrm{e}^{\mathrm{x}}\right]_{0}^{\ln 3}\\
&=\left(\frac{\mathrm{e}^{2 \mathrm{\ln3} }}{2}-\mathrm{e}^{\ln 3}\right)-\left(\frac{\mathrm{e}^{0}}{2}-\mathrm{e}^{0}\right)\\
&=\left(\frac{9}{2}-3\right)-\left(\frac{1}{2}-1\right)\\
&=\frac{8}{2}-2\\
&=2
\end{aligned}
