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Assuming unitary square, the demicircles have equations $$y=\frac{1}{2} \left(2 \sqrt{x-x^2}+1\right);\;y=\sqrt{1-x^2}$$ Intersection point is at $x_1=\frac{1}{8} \left(5-\sqrt{7}\right)$
Intersections with the diagonal are at $$x_2=\frac{1}{\sqrt{2}};\;x_3=\frac{1}{4} \left(\sqrt{2}+2\right)$$
The requested area is twice the area colored in the picture below that is $$A/2=\int_{{x_1}}^{{x_2}}\left(\frac{1}{2} \left(2 \sqrt{x-x^2}+1\right)-\sqrt{1-x^2}\right)\,dx+\int_{{x_2}}^{{x_3}}\left(\frac{1}{2} \left(2 \sqrt{x-x^2}+1\right)-x\right)\,dx$$ $$A=\frac{1}{16} \left(2 \sqrt{7}-5 \pi +16 \arcsin \left(\frac{1}{8} \left(5-\sqrt{7}\right)\right)+8 \text{ arcsec}\left(\frac{2 \sqrt{\sqrt{7}+5}}{3}\right)\right)\approx 0.14638$$
$$..$$
Assume unit square. The area is $A_{square}-A_{purple}-A_{green}-2A_{grey}$, i.e.
$$Area =\frac34 - \frac{3\pi}{16} -2A_{grey} \tag1 $$
where $A_{grey}$ can be obtained by evaluating the areas of the circular sectors ABP, OPE, and the triangles AFP and FPO; then subtracting them from the half square, as done below.
Let the vertex A being the origin, the polar equation of the two circles are $r=1; \space \space r(\sin\theta + \cos\theta)=r^2+\frac{1}{4}$. Eliminate $r$ to obtain $\sin\beta = \frac{5+\sqrt{7}}{8}$ and then $\sin\alpha =\frac{\sqrt{7}-1}{4}$. The area of two triangles AFP and FPO is
$$A_q = \frac{1}{4}\sin\beta + \frac{1}{4}\left(\frac{1}{2}-\cos\beta \right)=\frac{2+\sqrt{7}}{16} $$
and the area of the two circular sectors is
$$A_c=\frac{1}{2}\left( \frac{\pi}{2}-\beta\right) + \frac{\alpha}{8} =\frac{\pi}{8}-\frac{1}{8}\sin^{-1} \left(\frac{23\sqrt{7}+67}{128}\right). $$
As a result,
$$A_{grey}=\frac{1}{2}-A_c-A_q=\frac{6-\sqrt{7}-2\pi}{16}+\frac{1}{8}\sin^{-1} \left(\frac{23\sqrt{7}+67}{128}\right) $$ Substitute into (1) to obtain
$$Area = \frac{\pi+2\sqrt7}{16} - \frac{1}{4}\sin^{-1} \left(\frac{23\sqrt{7}+67}{128}\right)$$