Show that for $0\leq b\leq1$,
$$\int_{0}^{b} \sqrt{1-x^2} dx = \frac{1}{2}b\sqrt{1-b^2}+\frac{1}{2}\sin^{-1}b$$
by using geometry.
This is what I have thus far: $\int_{0}^{b} \sqrt{1-x^2} dx$ $$=y=\sqrt{1-x^2}$$ $$=y^2=1-x^2$$ $$=x^2+y^2=1$$
From the picture above, to find the area we add the area of the sector and the area of the triangle.
Area of sector: $\frac{θ}{2}r^2$
Area of triangle: $\frac{1}{2}bh$
Side note: $(h=a)$
I have no clue what to do next. Please help.
The area of the triangle with sides $b$ and $a=\sqrt{1-b^2}$ forming the right angle is $$ \frac 12b\sqrt{1-b^2}\ . $$ The angle, marked in black, can be extracted from the triangle with sides $a,b,r=1$, it is $\arcsin (b/1)$. So the sector has area $$\frac 12\arcsin b\cdot 1^2\ . $$ Adding the two areas we get the answer.