Answer
$$\int_{1}^2\int_{0}^4(x^2-2y^2)\hspace{0.5mm}dxdy=\frac{8}{3}$$
Work Step by Step
We will begin with integral with respect to x:
$\int_{1}^2\int_{0}^4(x^2-2y^2)\hspace{0.5mm}dxdy=\int_1^2\left(\frac{x^3}{3}-2y^2x\right)\bigg\vert_0^4dy=\int_1^2\left(\left(\frac{64}{3}-8y^2\right)-(0-0)\right)dy=\int_1^2\left(\frac{64}{3}-8y^2\right)dy=\left(\frac{64}{3}y-\frac{8}{3}y^3\right)\bigg\vert_1^2=\left(\frac{128}{3}-\frac{64}{3}\right)-\left(\frac{64}{3}-\frac{8}{3}\right)$
$=\frac{128}{3}-\frac{128}{3}+\frac{8}{3}=\frac{8}{3}$
