Area bounded by a parabola and a curve.

By inspection we see that $f(1) = g(1) = \dfrac{4}{5}$ and $f(0) = g(0) = \dfrac{-4}{5}$ and for $x > 0$

$x^2 + x - 1 = 0 \implies x = \dfrac{\sqrt{5} - 1}{2}$ and $\sqrt{x} - \sqrt{1 - x^2} = 0 \implies x^2 + x - 1 = 0 \implies x = \dfrac{\sqrt{5} - 1}{2} \implies$

$f(\dfrac{\sqrt{5} - 1}{2}) = g(\dfrac{\sqrt{5} - 1}{2}) = 0$ .

Since there are only three points of intersection we have them all.

Let $x_{0} = \dfrac{\sqrt{5} - 1}{2}$

$A_{1} = \int_{x_{0}}^{1} (g(x) - f(x)) dx = \dfrac{4}{5}\int_{x_{0}}^{1} (\sqrt{1 - x^2} - \sqrt{x} + x^2 + x - 1) dx$

Let $x = \sin(\theta) \implies dx = \cos(\theta) \implies$

$A_{1} = \dfrac{4}{5}(\dfrac{\pi}{4} - \dfrac{1}{2}\arcsin(x_{0}) - \dfrac{1}{2}x_{0}\sqrt{1 - x_{0}^2} - \dfrac{5}{6} + \dfrac{2}{3}x_{0}^{\dfrac{3}{2}} - \dfrac{1}{3}x_{0}^{3} - \dfrac{1}{2}x_{0}^2 + x_{0})$ .

In a similar fashion $A_{2} = \int_{0}^{x_{0}} (f(x) - g(x)) dx = \int_{0}^{x_{0}} (-\sqrt{1 - x^2} + \sqrt{x} - x^2 - x + 1) dx =$ $\dfrac{4}{5}(\dfrac{-1}{2}\arcsin(x_{0}) - \dfrac{1}{2}x_{0}\sqrt{1 - x_{0}^2} + \dfrac{2}{3}x_{0}^{\dfrac{3}{2}} - \dfrac{1}{3}x_{0}^3 - \dfrac{1}{2}x_{0}^2 + x_{0})$

The desired area $A = A_{1} + A_{2} = \dfrac{4}{5}(\dfrac{\pi}{4} - \arcsin(x_{0}) - x_{0}\sqrt{1 - x_{0}^2} - \dfrac{5}{6} + \dfrac{4}{3}x_{0}^{\dfrac{3}{2}} - \dfrac{2}{3}x_{0}^{3} - x_{0}^2 + 2x_{0}) \approx \boxed{0.11560}$ .