 Solution

Fix $x\in (0,1)$. For any sufficiently small nonzero $h$, the additivity of the Riemann integral givesTaking absolute values, for either sign of $h$,Because $f$ is continuous at $x$, the right side tends to zero. Thus the fundamental theorem of calculus here follows directly from the difference quotient:Only continuity at the point under consideration was used.

Without continuity, the assertion is false. Take the Riemann integrable step function $f(t)=0$ for $t<1/2$ and $f(t)=1$ for $t\ge 1/2$. Its primitive is $F(x)=0$ for $x\le 1/2$ and $F(x)=x-1/2$ for $x\ge 1/2$. The left and right difference quotients at $1/2$ are respectively zero and one, so the primitive need not be differentiable at every interior point. A function with this single jump is Riemann integrable; changing its value at the jump does not change the integral or the counterexample.