Calculus question about the existence of antiderivative Let $f(x)$ be the function equal $0$ on $\mathbb R_{&lt;0}$ and equal $1$ otherwise. Is clear to me that $f$ has derivative undefined at $x=0$. Is intuitive because the derivative would be $\infty$ as $f$ increases by $1$ in an infinitely small space. 

Now I was wondering whether antiderivative exists at $0$. My thoughts: antiderivative corresponds to area under the function. Before zero the are is always zero and after zero the area is $1$ so that like in derivative case the area increases by $1$ in an infinitely small space. But I don't know if intuition can be used in this case. 

When does antiderivatives of functions exists in general and when not?
For such function does not exist any antiderivative; that is because even if $F$ is derivable with derivative discontinuous $F'$ always respect the intermediate value property, so assuming $0$ and $1$ and never $\frac{1}{2}$, $f$ is not a derivative of any function.

What exists is its definite integral, in any derivable point of this integral the derivative is the value of $f$ but such integral is not derivable in $0$ only continuous.
The function F(x) = 0 for x &lt; 0, F(x) = x for x &ge; 0 is an antiderivative for the function f that you describe.  Or if you prefer F(x) = c for x &lt; 0 and F(x) = x + c for x &ge; 0, where c is any real number. Notice that F(x) is continuous on the entire real line, but not differentiable at x = 0 (which is as it should be).

Any piecewise continuous function on the interval (a,b) will have an antiderivative there, and most of the functions you are likely to run into will be at least piecewise continuous.  To avoid having an antiderivative, the function has to avoid being piecewise continuous, which is very discontinuous indeed, but can happen.  For example, the function g(x) = 0 where x is rational and 1 where x is irrational does not have an antiderivative on the interval (a,b). 

In general integration is much more forgiving than differentiation.  That is because addition is additive, in fact is really an average, which tends to smooth things out. (That's why people use 4 week running averages to analyze trends-- they are smoother than the underlying daily data.) 

Whereas, the existence of a f'(c)  (c any number) implies that f is locally linear in the neighborhood of c.  That is, f and the tangent line to f at c are very similar in a small enough interval of c.  Obviously a function with a discontinuity cannot be locally linear at that point, nor can a function like |x| (at zero).