Answer
$222$
Work Step by Step
Apply formula (2),
$\displaystyle \int_{1}^{4}\int_{0}^{2}(6x^{2}y-2x)dy dx=\displaystyle \int_{1}^{4}[\int_{0}^{2}(6x^{2}y-2x)dy] dx$
...treat $x$ as a constant in the inner integral.
$=\displaystyle \int_{1}^{4}[(3x^{2})y^{2}-(2x)y]_{y=0}^{y=2}\ \ dx$
$=\displaystyle \int_{1}^{4}[(3x^{2}\cdot 2^{2}-2x\cdot 2)-(0-0)]dx$
$=\displaystyle \int_{1}^{4}(12x^{2}-4x)dx$
$=[12\displaystyle \cdot\frac{x^{3}}{3}-2\cdot\frac{x^{2}}{2}]_{1}^{4}$
$=[4x^{3}-2x^{2}]_{1}^{4}$
$=[4(4)^{3}-2(4)^{2}]-[4(1)^{4}-2(1)^{2}]$
$=(256-32)-(4-2)$
$=222$
