Answer
$$A = \frac{1}{6}$$
Work Step by Step
$$\eqalign{
& y = \left( {1 - x} \right)\left( {x - 2} \right) \cr
& y = x - 2 - {x^2} + 2x \cr
& y = - {x^2} + 3x - 2 \cr
& {\text{The root of the functions are }}x = 1{\text{ and }}x = 2 \cr
& {\text{The function is of the form of the polynomial }}a{x^2} + bx + c,\,\,a < 0 \cr
& {\text{then the area above the }}x{\text{ axis is given by:}} \cr
& A = \int_1^2 {\left( { - {x^2} + 3x - 2} \right)} dx \cr
& A = \left[ { - \frac{1}{3}{x^3} + \frac{3}{2}{x^2} - 2x} \right]_1^2 \cr
& {\text{Evaluating}} \cr
& A = \left[ { - \frac{2}{3}} \right] - \left[ { - \frac{5}{6}} \right] \cr
& A = \frac{1}{6} \cr} $$
