A weaker but easier result is that the differentiation operator is uncomputable: Define f_t(x) = t arctan(x/t). As t -> 0, f_t -> 0. But notice that f'_t(0) = 1.
By contrast, the integration operator is computable. Why? Because a piece of code that defines a function f over the real numbers can be generalised to interval numbers (see "interval arithmetic") -- it then follows that the upper and lower Darboux sums are both computable. QED
I think you don't even need arctan for this. Just put one of Myhill's "bumps" at x=0 and shrink it down. That function approaches f(x)=0 for all x but its derivative at x=0 remains 1.
By contrast, the integration operator is computable. Why? Because a piece of code that defines a function f over the real numbers can be generalised to interval numbers (see "interval arithmetic") -- it then follows that the upper and lower Darboux sums are both computable. QED
[edit] Changed oo to 0.