Answer
$16$
Work Step by Step
Re-arrange the given integral as follows:
$\int_{0}^{3}[4y-\dfrac{y^3}{3}]_{0}^{2}]dx=\int_{0}^{3}[4(2)-\dfrac{8}{3}) dx$
This implies that
$\int_{0}^{3}[\dfrac{16}{3} dx= (\dfrac{16}{3}x)_{0}^{3}$
Thus, we have, $\dfrac{16}{3}(3-0)=16$
