Answer
$\dfrac{53}{20}$
Work Step by Step
Consider $V=\int_{0}^{1} \int_{y+1}^{4-2y} x^2y dx \space dy$
or, $=\int_{0}^{1} (\dfrac{x^3y}{3}|_{y+1}^{4-2y} dy$
or, $=\int_0^1 (y/3) [(4-2y)^3-(y+1)^3] dy$
or, $=\int_0^1 -3y^4+15y^3-33y^2+21 y dy$
or, $=\dfrac{-3y^5}{5}+\dfrac{15y^4}{4}-11y^3+\dfrac{21y^2}{2}|_0^1$
or, $Volume =\dfrac{53}{20}$
