Draw the line from the center of the circle to your point $(b,f(b))$. That line splits your big blue region into two parts: a triangle below the line, and a sector above.
The area of the triangle is clearly $\frac{1}{2}b\sqrt{1-b^2}$.
For the area of the sector, the angle of that sector is $\arcsin b$. This is because the complementary angle (below the line) has cosine equal to $b$. Or else you can see directly, by drawing a perpendicular from $(b,f(b))$ to the $y$-axis, that the angle of the sector has "opposite" side, and therefore sine, equal to $b$. So the area of the sector is $\frac{1}{2}\arcsin b$.
By integration, the area of the blue region is $\int_0^b\sqrt{1-x^2}dx$. We conclude that $\int_0^b\sqrt{1-x^2}\,dx=\frac{1}{2}b\sqrt{1-b^2}+\frac{1}{2}\arcsin b.\qquad\qquad(\ast)$
Now in $(\ast)$, change the $b$ to $x$, and (to make me feel good) the dummy variable of integration to $t$. We have $\int_0^x\sqrt{1-t^2}\,dt=\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}\arcsin x.$ This says that the right-hand side is an antiderivative of $\sqrt{1-x^2}$. All antiderivatives are obtained by adding a constant of integration.
Note that the geometric derivation is not quite complete, since the picture does not deal directly with negative $b$. But that is not hard to do. The simplest way is to make the change of variable $z=-x$.