Answer
$1$
Work Step by Step
Re-arrange the given integral as follows:
$V=\int_{0}^{1} [2y-xy-\dfrac{y^2}{2}]_{0}^{1}dx$
This implies that
$\int_{0}^{1} (2-x-\dfrac{1}{2}] dx=\int_{0}^{1} [\dfrac{3}{2}-x] dx$
Thus, $V=[\dfrac{3}{2}x-\dfrac{x^2}{2}]_0^1$
Hence, $V=\dfrac{3}{2}-\dfrac{1}{2}=1$
