Answer
$27$
Work Step by Step
The area under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by:
$Area, A= \int_{-3}^{3} [|x|-x^2+6) \ dx \\ = 2 \int_0^{3} [x-x^2+6] \ dx \\= 2 [\dfrac{ x^{2}}{2}-\dfrac{x^3}{3}+6x]_0^{3} \\= 2 [\dfrac{ (3^{2})}{2}-\dfrac{(3^3)}{3}+(6)(3) ] \\= 27$
