 Solution

The fundamental theorem of calculus says first that if $h$ is continuous on $[a,b]$ and $H(x)={\int }_a^{x}h(t)dt$, then $H^{\prime }(x)=h(x)$. Indeed,whose absolute value tends to zero by continuity. Consequently, if $F$ is differentiable with continuous derivative, applying the first part to $h=F^{\prime }$ shows that $F(x)-{\int }_a^x{F}^{\prime }(t)dt$ has zero derivative and is constant. Hence

A derivative need not be Riemann integrable. Define $f(0)=0$ and $f(x)=x^2\sin (x^{-2})$ for $x\ne 0$. Then $f$ is differentiable at zero with $f^{\prime }(0)=0$, while for $x\ne 0$,which is unbounded near zero. Since every Riemann-integrable function on a compact interval is bounded, $g=f^{\prime }{|}_{[0,1]}$ need not be Riemann integrable.