Is the derivative of a continuously differentiable function always integrable? I know that $f(x)$ is bounded by the extreme value theorem, because $f(x)$ is defined on a closed bounded interval. Now, how can I prove that its derivative is integrable knowing that it is continuous? Or, is there a counterexample?
 A: Any continuous function on a closed and bounded interval $[a,b]$ is integrable. Since, $f$ is given to be continuously differentiable, its derivative is continuous on the domain, and hence integrable.
If you only want to assume that $f'$ is continuous on $(a,b)$ then it is not so simple. Consider the function $f : [0,1] \to \mathbb{R}$ defined by
$$
f(x) = \begin{cases}
0, & x=0;\\
x^2 \sin (1/x^2), & x \neq 0.
\end{cases}
$$
It is differentiable everywhere, and its derivative is continuous on $(0,1)$, but the derivative is unbounded as $x \to 0^+$. You can take a look at the answers to this question for more examples of bounded functions with unbounded derivatives: Can the graph of a bounded function ever have an unbounded derivative?
A: If, according to your question, $f'$ is continuous on a bounded interval (i.e. a compact set), it makes no doubt it is Riemann-integrable (and hence also Lebesgue integrable):

A bounded function on a compact interval $[a, b]$ is Riemann integrable
  if and only if it is continuous almost everywhere (the set of its
  points of discontinuity has measure zero, in the sense of Lebesgue
  measure).

See: Riemann integral
