Difference of Areas of an ellipse and a curve

To find $(x_{0},y_{0})$ :

$\dfrac{4}{5}(\sqrt{x} - \sqrt{1 - x^2}) = \dfrac{4}{5}(2x - 1) \implies x - 2\sqrt{x}\sqrt{1 - x^2} + 1 - x^2 = 4x^2 - 4x + 1 \implies 5x - 5x^2 = 2\sqrt{x}\sqrt{1 - x^2}$

$\implies 25x^2(1 - 2x + x^2) = 4x - 4x^3 \implies 25x^4 - 50x^3 + 25x^2 = 4x - 4x^3 \implies x(25x^3 - 46x^2 + 25x - 4) = 0 \implies$

$x(x - 1)(25x^2 - 21x + 4) = 0 \implies x = 0, x = 1, x = \dfrac{21 \pm \sqrt{41}}{50}$ .

From the graph above two points of intersection are $(1,\dfrac{4}{5})$ and $(0,\dfrac{-4}{5})$ . Since there are only three points of intersection $\implies$ only one of the values of $x = \dfrac{21 \pm \sqrt{41}}{50}$ are valid.

Choosing $\boxed{x_{0} = \dfrac{21 - \sqrt{41}}{50}} \implies y(x_{0}) = f(x_{0}) = \dfrac{-4}{125}(4 + \sqrt{41})$ .

$A^{*}_{1} = 2\int_{x_{0}}^{1} (y - f(x)) dx = \dfrac{8}{5}\int_{x_{0}}^{1} (\sqrt{1 - x^2} - \sqrt{x} + (2x - 1))dx$ .

Let $x = \sin(\theta) \implies dx = \cos(\theta)$ $\implies A^{*}_{1} = \dfrac{8}{5}(\dfrac{\pi}{4} - \dfrac{1}{2}\arcsin(x_{0}) - \dfrac{1}{2}x_{0}\sqrt{1 - x_{0}^2} - \dfrac{2}{3} + \dfrac{2}{3}{x_{0}}^{\dfrac{3}{2}} - {x_{0}}^2 + x_{0})$

In a similar fashion:

$A^{*}_{2} = 2\int_{0}^{x_{0}} (f(x) - y) dx = \dfrac{8}{5}\int_{0}^{x_{0}} (-\sqrt{1 - x^2} + \sqrt{x} - 2x + 1) dx =$

$\dfrac{8}{5}(\dfrac{-1}{2}\arcsin(x_{0}) - \dfrac{1}{2}x_{0}\sqrt{1 - {x_{0}}^2} + \dfrac{2}{3}{x_{0}}^{\dfrac{3}{2}} - {x_{0}}^2 + x_{0}).$

The desired area $A_{1} = A^{*}_{1} + A^{*}_{2} = \dfrac{8}{5}(\dfrac{\pi}{4} - \arcsin(x_{0}) - x_{0}\sqrt{1 - x_{0}^2} - \dfrac{2}{3} + \dfrac{4}{3}{x_{0}}^{\dfrac{3}{2}} - 2{x_{0}}^2 + 2x_{0}) \approx \boxed{.2671939}$

The ellipse $36(x - \dfrac{1}{2})^2 + 25y^2 = 25 \implies y = \dfrac{1}{5}\sqrt{25 - 36(x - \dfrac{1}{2})^2}$ , where $\dfrac{-1}{3} \leq x \leq \dfrac{4}{3} \implies$ $A_{2} = \dfrac{2}{5}\int_{\dfrac{-1}{3}}^{\dfrac{4}{3}} \sqrt{25 - 36(x - \dfrac{1}{2})^2}$ .

Letting $6(x - \dfrac{1}{2}) = 5\sin(\theta) \implies A_{2} = \dfrac{5}{6}\int_{\dfrac{-\pi}{2}}^{\dfrac{\pi}{2}} (1 + \cos(2\theta)) = \dfrac{5}{6} (\theta + \dfrac{1}{2}\sin(2\theta))|_{\dfrac{-\pi}{2}}^{\dfrac{\pi}{2}} = \boxed{\dfrac{5\pi}{6}} \implies$

$A_{2} - A_{1} \approx \boxed{2.3508}$