Computing the arc length of the graph $y=\sqrt{x-x^2}+\arcsin(\sqrt{x})$
Is this done right?
I found the interval of this function which is $[0,1]$ I know the arc length formula is $$ L=\int_a^b\sqrt{1+(y')^2}dx$$
The derivative of the function is $$y'=\sqrt{\frac{1-x}{x}}$$ and the the square of $y'$ is $$(y')^2=\frac{1-x}{x}$$
When I plug in these results on the integral I get $$L=\int_0^1\sqrt{\frac{1}{x}d}x$$ with the substitution of $\sqrt{x}=t$ we get the result $L=2$
This is the graph of the function
Thanks, @DougM for pointing out my mistake