Example of bounded f:[0,1]->R which has primitive but not Riemann-integrability in closed interval?

Question

This must be a function which is antiderivative but it has no Riemann-integral. These conditions must be fulfilled in closed interval from 0 to 1.

Answer 1

Let $$F (x)=x^2\sin (\frac {1}{x^2}) $$ for $x\ne 0$ and $F (0)=0$.

$F $ is differentiable at $[0,1]. $

Let $f=F'$.

$f $ has an antiderivative but $f $ is unbounded near $0$ and non Riemann- integrable.

For $x\ne 0,$ $$f (x)=2x\sin (\frac {1}{x^2})-2\frac {1}{x}\cos (\frac {1}{x^2}) $$

If $u_n=\frac {1}{\sqrt {2n\pi}} $ then $$f (u_n)=2\sqrt {2n\pi}\to +\infty . $$