Answer
$$2$$
Work Step by Step
$$\iint_{R} dA= \int_0^{\sqrt {\ln 3}} \int_0^{2x} e^{x^2} dy dx \\= \int_0^{\sqrt {\ln 3}} [ y e^{x^2}]_0^{2x} dy \\= \int_0^{\sqrt {\ln 3}}2x e^{x^2} dy $$
Suppose $ x^2=u$ and $2x dx =du$
Now, $$\int_0^{\ln 3} e^{u} du =[e^u]_0^{\ln 3}\\=3-1\\=2$$
