Question

what is the area enclosed by the curved y=4x-x^2,and y=x^2-2x along the x-axis

thanks so much help

Answer 1

Two parabolas, one opening up the other down

Intersection:
x^2 - 2x = 4x - x^2
2x^2 - 6x = 0
2x(x - 3) = 0
x = 0 or x = 3

effective height of region = 4x-x^2 - (x^2 - 2x)
= 6x - 2x^2

area = ? (6x - 2x^2) dx from 0 to 3
= [3x^2 - (2/3)x^3] from 0 to 3
= 27 - (2/3)(27) - 0
= 9

nice confirmation by Wolfram:
http://www.wolframalpha.com/input/?i=area+between+y%3D4x-x%5E2,and+y%3Dx%5E2-2x