Area on my Valentine!

For the $x$ intercept of $y_{1} = \dfrac{4}{5} (\sqrt{x} - \sqrt{1 - x^2})$ we obtain:

$x^2 + x - 1 = 0 \implies x = \dfrac{-1 \pm \sqrt{5}}{2}$ , since $x = -\dfrac{1 + \sqrt{5}}{2} = -\phi$ results in a complex valued square root $\implies x = \dfrac{\sqrt{5} - 1}{2}$ is the x intercept of $y_{1} = \dfrac{4}{5} (\sqrt{x} - \sqrt{1 - x^2})$ .

The equation of the line passing thru $A: (0,\dfrac{4}{5})$ and $B: (\dfrac{\sqrt{5} - 1}{2},0)$ is:

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

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

For $\int \sqrt{1 - x^2} dx$

Let $x = \sin(\theta) \implies dx = \cos(\theta) \implies \int \sqrt{1 - x^2} dx = \int \cos^2(\theta) d\theta = \dfrac{1}{2}(\theta + \sin(\theta)\cos(\theta))$ .

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

$\implies \int_{0}^{\beta} \sqrt{1 - x^2} dx = \dfrac{1}{2}\arcsin(\beta) + \dfrac{1}{2} \beta^{\frac{3}{2}} \implies$ $A_{R_{1}} = \dfrac{4}{5} (\dfrac{1}{2}\arcsin(\beta) + \dfrac{1}{2} \beta^{\frac{3}{2}} + (-\dfrac{2}{3} x^{\frac{3}{2}} + x - \dfrac{x^2}{\sqrt{5} - 1})|_{0}^{\beta} =$ $\dfrac{2}{5}(\arcsin(\beta) - \dfrac{1}{3}\beta^{\frac{3}{2}} + \beta)$

$\beta = \dfrac{\sqrt{5} - 1}{2} = \dfrac{1 + \sqrt{5}}{2} - 1 = \phi - 1 \implies A_{R_{1}} = \dfrac{2}{5}(\arcsin(\phi - 1) - \dfrac{1}{3}(\phi - 1)^{\frac{3}{2}} + \phi - 1) \implies$ $A_{R} = 2A_{R_{1}} = \dfrac{4}{5}(\arcsin(\phi - 1) - \dfrac{1}{3}(\phi - 1)^{3/2} + \phi - 1) =$ $\dfrac{a^2}{b}(\arcsin(\phi - 1) - \dfrac{1}{c}(\phi - 1)^{c/a} + (\phi - 1)) \implies a + b + c = \boxed{10}.$