Answer
$\dfrac{128}{3}$
Work Step by Step
The area under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by:
$Area, A= \int_{0}^{16} [8-\sqrt x-\sqrt x] \ dx \\ =\int_0^{16} [8-2 \sqrt x] \ dx \\= [8x-\dfrac{4 x^{3/2}}{3}]_0^{16} \\= 8(16)-\dfrac{4 (16^{3/2})}{3} \\=\dfrac{128}{3}$
