For A1: intersection of y=x2+2 and x+y=8 at x=2,y=6
A1=∫02(8−x−(x2+2))dx=∫02(6−x−x2)dx=[6x−2x2−3x3]02=12−2−38=322
For A2: bounded by y=x2+2, y2=x (i.e., x=y2), x=2, and y-axis.
In first quadrant, y2=x goes from (0,0) to (2,2).
The parabola y=x2+2 passes through (0,2) and (2,6).
A2=∫02(2−y2)dy=[2y−3y3]02=22−322=342
A1−A2=322−342=32(22+1)