The area is given by the integral $\int_{0}^{2} (2x + 3) dx$. Integrating term by term: $\int 2x dx = x^2$ and $\int 3 dx = 3x$. Thus, we have $[x^2 + 3x]_{0}^{2}$. Substituting the upper limit $x=2$: $2^2 + 3(2) = 4 + 6 = 10$. Substituting the lower limit $x=0$: $0^2 + 3(0) = 0$. The area is $10 - 0 = 10$ sq units. Geometrically, this is a trapezoid with height 2 and parallel sides of lengths 3 and 7 ($Area = \frac{1}{2}(3+7) \times 2 = 10$). Options 2, 3, and 4 are results of common arithmetic mistakes in integration or substitution.