The correct answer is $\boxed{\frac{1}{2}{\left( {{{\text{e}}^2} – 1} \right)^2}}$.
To evaluate the integral, we can substitute in the limits of integration. We get:
$$\int_0^2 {\int_0^{\text{x}} {{{\text{e}}^{{\text{x}} + {\text{y}}}}} } {\text{dy dx}} = \int_0^2 {\text{e}^{\text{x}} \int_0^{\text{x}} {{{\text{e}}^{\text{y}}}}} } {\text{dy dx}} = \int_0^2 {\text{e}^{\text{x}} \left[ {{\text{e}}^{\text{y}}} \right]_0^{\text{x}} } {\text{dx}} = \int_0^2 {\text{e}^{2\text{x}}} {\text{dx}}$$
We can then evaluate the integral using the following formula:
$$\int {\text{e}^{kx}} {\text{dx}} = \frac{\text{e}^{kx}}{k} + C$$
where $C$ is an arbitrary constant.
We get:
$$\int_0^2 {\text{e}^{2\text{x}}} {\text{dx}} = \frac{\text{e}^{2\text{x}}}{2} \bigg|_0^2 = \frac{\text{e}^{4} – \text{e}^0}{2} = \frac{1}{2}{\left( {{{\text{e}}^2} – 1} \right)^2}$$