The required region is bounded by 2 green lines (x=1 and y=-x+2) and a red curve, y=x2-1.
The picture shows that the curve and the lines intersect, the vertical x=1 when y=x2-1=1-1=0, the point (1,0).
The other intersection point is when -x+2=x2-1, x2+x-3=0, x=(-1+√13)/2 (the negative solution of this quadratic is ignored because it's outside the region).
So the area extends from x=1 to x=(-1+√13)/2. Imagine thin rectangular strips constituting the region so that the sum of the areas of the strips is the area of the region.
The leftmost strip has a length of 1. If we take some arbitrary value of x between 1 and (-1+√13)/2, the length of the strip will be 2-x-(x2-1)=3-x-x2. The width of the strip is dx for each strip.
The area of the strip is (3-x-x2)dx which is the integrand for a summation of infinitesimally thin strips. Integrating: 3x-½x2-⅓x3 for 1≤x≤(-1+√13)/2.
The upper value of this expression is 2.32268 approx. For the lower value it's 13/6=2.16667. The difference is the area of the region: 0.156 square units approx.