Answer
$4$
Work Step by Step
Re-arrange the given integral as follows:
$V=\int_{0}^{2} [\dfrac{xy}{2}]_{0}^{4}dx$
This implies that
$\int_{0}^{2} (\dfrac{4y}{2}) dy=\int_{0}^{2} 2y dy$
Thus, $V=[y^2]_0^2$
Hence, $V=2^2-0=4$
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