## Thomas' Calculus 13th Edition

$$e-1$$
We calculate the integral as follows: $$\int_{0}^{2} \int_{y/2}^{1} e^{x^2} \ dx \ dy \\=\int_{0}^{1} \int_{0}^{2x} e^{ x^2} \ dy \ dx \\= 2 \int_{0}^{1} x \times e^{x^2} \ dx \\=[e^{x^2}]_0^1 \\=e^{1}-e^0 \\=e-1$$