posted by AIRA .
Region A that on xy-plane is bounded by two (2) curves and a line. The curves are y=x^3-2x+3 and y=-x^2+3 while the line is x=0. It is located in the first quadrant of xy-plane. Determine the area of region A.
Make a sketch and then solve for the intersection of the two curves
x^2 - 2x + 3 = -x^2 + 3
x = 0 or x = 1
area = [integral] (-x^2 + 3 -(x^2 - 2x+3)) dx from 0 to 1
= integral (-2x^2 + 2x)dx from 0 to 1
= (-2/3)x^3 + x^2 | from 0 to 1
= -2/3 + 1 - 0