When I was answering number of integrable functions is greater than number of differentiable functions I got to wonder if the inequality was strict.
So with $\mathcal I$ being the set of integrable functions $\mathbb R\to\mathbb R$ and $\mathcal D$ being the set of differentiable functions $\mathbb R\to\mathbb R$: Is $\operatorname{card}(\mathcal I)=\operatorname{card}(\mathcal D)$? Does it matter whether we're talking Riemann- or Lebesgue integrability (or one of the variants I'm not very familiar with, and don't remember the name of)? Would the answer be different if either $\mathbb R$ was replaced with another set?
$\geq$ is obvious, and integrable functions that aren't differentiable are well-known too, but I imagine that both sets have cardinality $2^{2^{\aleph_0}}$?
Assuming that you mean Lebesgue integrable, note that if a function is non-zero only on a subset of the Cantor set, then it is integrable.
This alone gives you $2^{2^{\aleph_0}}$ integrable functions whose integral is $0$. (Note that this works for Riemann integrable functions as well, since on $[0,1]$ the function is integrable, since the points of discontinuity are measure $0$, and outside $[0,1]$ it is a constant function.)
On the other hand, a differentiable function is continuous, and there are only $2^{\aleph_0}$ of those.