y^2=-x^2+2x
y^2 -1 =-(x^2-2x+1)
Y^2+(x-1)^2=1
Isn't that a circle of radius 1, centered at 1,0?
The area enclosed by the curve y^2 = x(2 − x) is given by what definite integral?
Should I begin by square rooting both sides? Not really sure.
2 answers
yes, so using symmetry, the area is
a = 4∫[0,1] √(1-(x-1)^2) dx = π
a = 4∫[0,1] √(1-(x-1)^2) dx = π
6 answers