Consider the region in the plane consisting of points (x, y) satisfying x > 0, y > 0, and lying between the curves y=x^2 +1and y=2x^2 −2.
(b) Calculate the area of this region.
We need their intersection:
2x^2 - 2 = x^2 + 1
x^2 = 3
x = ± √3
we want only the part in the first quadrant
effective height = x^2 + 1 - (2x^2 - 2)
= 3 - x^2
area = ∫(3 - x^2) dx from 0 to √3
= [3x - (1/3)x^3] from 0 to √3
= 3√3 - (1/3)(3√3) - 0
= 3√3 - √3
= 2√3