Answer
$$\sin (4)$$
Work Step by Step
We calculate the integral as follows:
$$I=\int_{0}^{1} \int_{2y}^{2} 4 \cos x^2 \ dx \ dy \\=\int_{0}^{2} \int_{0}^{x/2} 4 \times \cos x^2 \ dy \ dx \\=\int_{0}^{2} 2x \times \cos x^2 \ dx \\=[\sin x^2]_0^2 \\=\sin (2)^2-\sin 0 \\=\sin (4)$$
