Question

The area enclosed by the curve y^2 = x(1 − x) is given by

Answer 1

$y^2 = x(1-x) = x- x^2$

$x^2 + y^2 = x$

That's a circle, as we can see by the usual completing of the square,

$x^2 - x + y^2 = 0$

$x^2 - x + \frac 1 4 + y^2 = \frac 1 4$

$(x - \frac 1 2)^2 + y^2 = (\frac 1 2 )^2$

That's the circle of radius $\frac 1 2$ centered at $(\frac 1 2, 0)$

It's a pretty good substitute for the unit circle with some interesting trigonometry of its own.

Anyway its radius is a half so its area is $\pi r^2 = \pi (\frac 1 2 )^2 = \frac{\pi}{4}$

Answer: $\quad \pi / 4$