Next, we need to find where the two curves intersect, which will be the endpoints of the
region. We do this by setting the two curves equal to each other. We get y^3 − y^2 = 2y. The
solutions are (4, 2), (0, 0), and (−2, −1). Notice that in the region from y = −1 to y = 0, the right
curve is f(y) = y^3 − y^2 and the left curve is g(y) = 2y, but from y = 0 to y = 2 the situation is
reversed, so the right curve is f(y) = 2y and the left curve is g(y) = y^3 − y^2 . Thus, we split the
region into two pieces and find the area by evaluating two integrals and adding the answers:
(y^3 − y^2 − 2y) dy and (2y − (y^3 − y^2 )) dy = (2y − y^3 + y^2 ) dy We get (y^3 − y^2 − 2y)
dy = = 0 − = and (2y − y^3 + y^2 ) dy =
= . Therefore, the area of the region is + = .
3.
