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I have two questions that i'm curious about.

  1. If $f$ is differentiable real function on its domain, then $f'$ is Riemann integrable.

  2. If $g$ is a real function with intermediate value property, then $g$ is Riemann integrable.

Thank you in advance.

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Hints:

1) The function $f$ defined by $$f(x)=\cases{ x^2\sin(1/x^2),&$x\ne0$ \cr 0,&$x=0$}$$ is differentiable on $[-1,1]$; but its derivative is unbounded on $[-1,1]$.

2) Derivatives enjoy the Intermediate Value Property (by Darboux's Theorem).

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  • $\begingroup$ Note that one can verify that $f'$, with $f$ is as above, has the Intermediate Value Property directly, rather than appealing to Darboux. $\endgroup$ – David Mitra Dec 12 '12 at 14:30
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  1. A Riemann integrable function $f$ on an interval $[a,b]$ must be bounded on that interval. So if you take $f(x) = x^{\frac{3}{2}} \sin(\frac{1}{x})$ 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.
  2. Again, you take a function such as $f(x) = \frac{1}{x} \sin(\frac{1}{x})$ on $[0,1]$ with $f(0) = 0$. 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.
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