Derivative of an accumulation function via FTC Part 1

By the Fundamental Theorem of Calculus Part 1, since $g$ is continuous and the upper limit of integration is $x$ itself, $F'(x) = g(x)$ for all $x$ in $(0,2)$. No chain rule is needed here because the upper limit is exactly $x$, not a composite function of $x$.

We need $F'(1) = g(1)$. From the graph, $g(1)$ is the height of the semicircle at its peak, which occurs at the center $t=1$ of the semicircle of radius 1. Substituting into $g(t)=\sqrt{1-(t-1)^2}$ gives $g(1) = \sqrt{1-0} = 1$. Therefore $F'(1) = 1$.