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