\f2 SET UP the definite integral that gives the AREA of the region. f(x) = 3x3 - x2 - 10x and g(x) = -x2 + 2x. - (0, 0) (2, 0) -1 -6 (-2, -8) -8 -10 8(x) =-x2+2x f(x) = 3x3 -x2 - 10x O A= [[(x-1) -(x-1) Jax A = mf [(3x)-x2 - 10x) -(-x3 +2x) Jdx A= f'(x) - 8'(x) A = ] [(3x3 -x2 -10x) -(-x2 + 2x) ]dx O J.[(-x2 +2x)-(3x3-x2 -10x) ]dx O A = [ ", ( ) + (8 )3 Find the volume of the solid that results when the region enclosed by the curves is revolved about the [x]-axis. y =-x-+1, y=0, x=-1, x=0 * ( - x ? + 1)2 dx O -1 8 JU ~ 1.676 15 (-x2 + 1)2 dx O 13 IT ~ 2.723 15 (-x2 + 1)'dx O -1 31 -JI ~ 3.246 30 (-x2 + 1)2 dx O -1 7 It ~ 1.466 15