Calculus: Application of Integration
posted by Anonymous .
Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label ts height and width. Then find the area of the region.
It is easy to see that the two curves intersect at (0,0) and (1,1)
I will integrate with respect to x
area = [integral] (x^(1/2) - x^2) dx from 0 to 1
= [(2/3)x^(3/2) - (1/3)x^3] from 0 to 1
= 2/3 - 1/3 - 0