Answer
$\dfrac{32}{3}$
Work Step by Step
Let us consider that two continuous functions $f(x)$ and $g(x)$ with $f(x)\geq g(x)$ on the interval $[a,b]$ . The area (A) of the region bounded by the graph of $f(x)$ and $g(x)$ on the interval $[a,b]$ can be calculated as: $Area(A)=\int_a^b [f(x)-g(x)] \ dx$
Thus, the area of the region is:
$A\int_a^b [f(x)-g(x)] \ dx= \int_{0}^{4} [3y-y(y-1)] \ dy \\=-2 \int_0^1 y(y-1) \ dy \\= \int_0^4 (3y-y^2+y) \ dy \\=\int_0^4 [4y-y^2] \ dy \\= [\dfrac{4y^2}{2}-\dfrac{y^3}{2}]_0^4\\=32 -\dfrac{64}{3} \\=\dfrac{32}{3}$