Answer
$2$
Work Step by Step
An equation of the line through $(2,0)$ and $(0,2)$ is $y=-x+2$; through $(2,0)$ and $(-1,1)$ is $y=-\frac{1}{3}x$$+\frac{2}{3}$;through $(0,2)$ and $(-1,1)$ is $y=x+2$
Now,
$A=\int_{-1}^{0}[(x+2)-(-\frac{1}{3}x+\frac{2}{3})]dx$$+\int_{0}^{2}[(-x+2)-(-\frac{1}{3}x+\frac{2}{3})]dx$
$=\int_{-1}^{0}(\frac{4}{3}x+\frac{4}{3})dx + \int_{0}^{2}(-\frac{2}{3}x+\frac{4}{3})dx$
$=[\frac{2}{3}x^2+\frac{4}{3}x]_{-1}^{0}$$ $$+[-\frac{1}{3}x^2+\frac{4}{3}x]_{0}^{2}$
$=0-(\frac{2}{3}-\frac{4}{3})+(-\frac{4}{3}+\frac{8}{3})-0$
$=2$
