TITLE: What is an example that a function is differentiable but derivative is not Riemann integrable
QUESTION [37 upvotes]: I have two questions that i'm curious about.

If $f$ is differentiable real function on its domain, then $f'$ is Riemann integrable.
If $g$ is a real function with intermediate value property, then $g$ is Riemann integrable.

Thank you in advance.

REPLY [19 votes]: A Riemann integrable function $f$ on an interval $[a,b]$ must be bounded on that interval. So if you take $$f(x)=\cases{x^{\frac{3}{2}} \sin(\frac{1}{x}),&$x\ne0$ \cr 0,&$x=0$}$$ on $[0,1]$, you can check that this is a continuous and differentiable function but with an unbounded derivative and so it is not integrable. You can even construct an example of a differentiable function whose derivative is bounded but is still not Riemann integrable - see Volterra's function.
Again, you take a function such as $$f(x)=\cases{\frac{1}{x} \sin(\frac{1}{x}),&$x\ne0$ \cr 0,&$x=0$}$$ on $[0,1]$. This is a discontinuous, unbounded function that satisfies the intermediate value property, but not Riemann integrable. A bounded example is given by the derivative of Volterra's function.