Answer
$\dfrac{32}{3}$ $units^2$
Work Step by Step
The area under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by:
$Area, A= \int_{0}^{4} [2y-((y-1)^2-1)] \ dy \\ = \int_{0}^{4} (2y-y^2+2y) \ dy\\= [2y^2 -\dfrac{1}{3} y^3]_0^4 \\= 2(4^2) -\dfrac{1}{3} (4^3) \\=\dfrac{32}{3}$
